hypothesis mean difference

9.1 Null and Alternative Hypotheses

The actual test begins by considering two hypotheses . They are called the null hypothesis and the alternative hypothesis . These hypotheses contain opposing viewpoints.

H 0 , the — null hypothesis: a statement of no difference between sample means or proportions or no difference between a sample mean or proportion and a population mean or proportion. In other words, the difference equals 0.

H a —, the alternative hypothesis: a claim about the population that is contradictory to H 0 and what we conclude when we reject H 0 .

Since the null and alternative hypotheses are contradictory, you must examine evidence to decide if you have enough evidence to reject the null hypothesis or not. The evidence is in the form of sample data.

After you have determined which hypothesis the sample supports, you make a decision. There are two options for a decision. They are reject H 0 if the sample information favors the alternative hypothesis or do not reject H 0 or decline to reject H 0 if the sample information is insufficient to reject the null hypothesis.

Mathematical Symbols Used in H 0 and H a :

H 0 always has a symbol with an equal in it. H a never has a symbol with an equal in it. The choice of symbol depends on the wording of the hypothesis test. However, be aware that many researchers use = in the null hypothesis, even with > or < as the symbol in the alternative hypothesis. This practice is acceptable because we only make the decision to reject or not reject the null hypothesis.

Example 9.1

H 0 : No more than 30 percent of the registered voters in Santa Clara County voted in the primary election. p ≤ 30 H a : More than 30 percent of the registered voters in Santa Clara County voted in the primary election. p > 30

A medical trial is conducted to test whether or not a new medicine reduces cholesterol by 25 percent. State the null and alternative hypotheses.

Example 9.2

We want to test whether the mean GPA of students in American colleges is different from 2.0 (out of 4.0). The null and alternative hypotheses are the following: H 0 : μ = 2.0 H a : μ ≠ 2.0

We want to test whether the mean height of eighth graders is 66 inches. State the null and alternative hypotheses. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 66
• H a : μ __ 66

Example 9.3

We want to test if college students take fewer than five years to graduate from college, on the average. The null and alternative hypotheses are the following: H 0 : μ ≥ 5 H a : μ < 5

We want to test if it takes fewer than 45 minutes to teach a lesson plan. State the null and alternative hypotheses. Fill in the correct symbol ( =, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : μ __ 45
• H a : μ __ 45

Example 9.4

An article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third of the students pass. The same article stated that 6.6 percent of U.S. students take advanced placement exams and 4.4 percent pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6 percent. State the null and alternative hypotheses. H 0 : p ≤ 0.066 H a : p > 0.066

On a state driver’s test, about 40 percent pass the test on the first try. We want to test if more than 40 percent pass on the first try. Fill in the correct symbol (=, ≠, ≥, <, ≤, >) for the null and alternative hypotheses.

• H 0 : p __ 0.40
• H a : p __ 0.40

Collaborative Exercise

Bring to class a newspaper, some news magazines, and some internet articles. In groups, find articles from which your group can write null and alternative hypotheses. Discuss your hypotheses with the rest of the class.

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Hypothesis Testing for Means & Proportions

Lisa Sullivan, PhD

Professor of Biostatistics

Boston University School of Public Health

Introduction

This is the first of three modules that will addresses the second area of statistical inference, which is hypothesis testing, in which a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The process of hypothesis testing involves setting up two competing hypotheses, the null hypothesis and the alternate hypothesis. One selects a random sample (or multiple samples when there are more comparison groups), computes summary statistics and then assesses the likelihood that the sample data support the research or alternative hypothesis. Similar to estimation, the process of hypothesis testing is based on probability theory and the Central Limit Theorem.  

This module will focus on hypothesis testing for means and proportions. The next two modules in this series will address analysis of variance and chi-squared tests. 

Learning Objectives

After completing this module, the student will be able to:

• Define null and research hypothesis, test statistic, level of significance and decision rule
• Distinguish between Type I and Type II errors and discuss the implications of each
• Explain the difference between one and two sided tests of hypothesis
• Estimate and interpret p-values
• Explain the relationship between confidence interval estimates and p-values in drawing inferences
• Differentiate hypothesis testing procedures based on type of outcome variable and number of sample

Introduction to Hypothesis Testing

Techniques for hypothesis testing  .

The techniques for hypothesis testing depend on

• the type of outcome variable being analyzed (continuous, dichotomous, discrete)
• the number of comparison groups in the investigation
• whether the comparison groups are independent (i.e., physically separate such as men versus women) or dependent (i.e., matched or paired such as pre- and post-assessments on the same participants).

In estimation we focused explicitly on techniques for one and two samples and discussed estimation for a specific parameter (e.g., the mean or proportion of a population), for differences (e.g., difference in means, the risk difference) and ratios (e.g., the relative risk and odds ratio). Here we will focus on procedures for one and two samples when the outcome is either continuous (and we focus on means) or dichotomous (and we focus on proportions).

General Approach: A Simple Example

The Centers for Disease Control (CDC) reported on trends in weight, height and body mass index from the 1960's through 2002. 1 The general trend was that Americans were much heavier and slightly taller in 2002 as compared to 1960; both men and women gained approximately 24 pounds, on average, between 1960 and 2002.   In 2002, the mean weight for men was reported at 191 pounds. Suppose that an investigator hypothesizes that weights are even higher in 2006 (i.e., that the trend continued over the subsequent 4 years). The research hypothesis is that the mean weight in men in 2006 is more than 191 pounds. The null hypothesis is that there is no change in weight, and therefore the mean weight is still 191 pounds in 2006.  

In order to test the hypotheses, we select a random sample of American males in 2006 and measure their weights. Suppose we have resources available to recruit n=100 men into our sample. We weigh each participant and compute summary statistics on the sample data. Suppose in the sample we determine the following:

Do the sample data support the null or research hypothesis? The sample mean of 197.1 is numerically higher than 191. However, is this difference more than would be expected by chance? In hypothesis testing, we assume that the null hypothesis holds until proven otherwise. We therefore need to determine the likelihood of observing a sample mean of 197.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true or under the null hypothesis). We can compute this probability using the Central Limit Theorem. Specifically,

(Notice that we use the sample standard deviation in computing the Z score. This is generally an appropriate substitution as long as the sample size is large, n > 30. Thus, there is less than a 1% probability of observing a sample mean as large as 197.1 when the true population mean is 191. Do you think that the null hypothesis is likely true? Based on how unlikely it is to observe a sample mean of 197.1 under the null hypothesis (i.e., <1% probability), we might infer, from our data, that the null hypothesis is probably not true.

Suppose that the sample data had turned out differently. Suppose that we instead observed the following in 2006:

How likely it is to observe a sample mean of 192.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true)? We can again compute this probability using the Central Limit Theorem. Specifically,

There is a 33.4% probability of observing a sample mean as large as 192.1 when the true population mean is 191. Do you think that the null hypothesis is likely true?  

Neither of the sample means that we obtained allows us to know with certainty whether the null hypothesis is true or not. However, our computations suggest that, if the null hypothesis were true, the probability of observing a sample mean >197.1 is less than 1%. In contrast, if the null hypothesis were true, the probability of observing a sample mean >192.1 is about 33%. We can't know whether the null hypothesis is true, but the sample that provided a mean value of 197.1 provides much stronger evidence in favor of rejecting the null hypothesis, than the sample that provided a mean value of 192.1. Note that this does not mean that a sample mean of 192.1 indicates that the null hypothesis is true; it just doesn't provide compelling evidence to reject it.

In essence, hypothesis testing is a procedure to compute a probability that reflects the strength of the evidence (based on a given sample) for rejecting the null hypothesis. In hypothesis testing, we determine a threshold or cut-off point (called the critical value) to decide when to believe the null hypothesis and when to believe the research hypothesis. It is important to note that it is possible to observe any sample mean when the true population mean is true (in this example equal to 191), but some sample means are very unlikely. Based on the two samples above it would seem reasonable to believe the research hypothesis when x̄ = 197.1, but to believe the null hypothesis when x̄ =192.1. What we need is a threshold value such that if x̄ is above that threshold then we believe that H 1 is true and if x̄ is below that threshold then we believe that H 0 is true. The difficulty in determining a threshold for x̄ is that it depends on the scale of measurement. In this example, the threshold, sometimes called the critical value, might be 195 (i.e., if the sample mean is 195 or more then we believe that H 1 is true and if the sample mean is less than 195 then we believe that H 0 is true). Suppose we are interested in assessing an increase in blood pressure over time, the critical value will be different because blood pressures are measured in millimeters of mercury (mmHg) as opposed to in pounds. In the following we will explain how the critical value is determined and how we handle the issue of scale.

First, to address the issue of scale in determining the critical value, we convert our sample data (in particular the sample mean) into a Z score. We know from the module on probability that the center of the Z distribution is zero and extreme values are those that exceed 2 or fall below -2. Z scores above 2 and below -2 represent approximately 5% of all Z values. If the observed sample mean is close to the mean specified in H 0 (here m =191), then Z will be close to zero. If the observed sample mean is much larger than the mean specified in H 0 , then Z will be large.  

In hypothesis testing, we select a critical value from the Z distribution. This is done by first determining what is called the level of significance, denoted α ("alpha"). What we are doing here is drawing a line at extreme values. The level of significance is the probability that we reject the null hypothesis (in favor of the alternative) when it is actually true and is also called the Type I error rate.

α = Level of significance = P(Type I error) = P(Reject H 0 | H 0 is true).

Because α is a probability, it ranges between 0 and 1. The most commonly used value in the medical literature for α is 0.05, or 5%. Thus, if an investigator selects α=0.05, then they are allowing a 5% probability of incorrectly rejecting the null hypothesis in favor of the alternative when the null is in fact true. Depending on the circumstances, one might choose to use a level of significance of 1% or 10%. For example, if an investigator wanted to reject the null only if there were even stronger evidence than that ensured with α=0.05, they could choose a =0.01as their level of significance. The typical values for α are 0.01, 0.05 and 0.10, with α=0.05 the most commonly used value.  

Suppose in our weight study we select α=0.05. We need to determine the value of Z that holds 5% of the values above it (see below).

The critical value of Z for α =0.05 is Z = 1.645 (i.e., 5% of the distribution is above Z=1.645). With this value we can set up what is called our decision rule for the test. The rule is to reject H 0 if the Z score is 1.645 or more.  

With the first sample we have

Because 2.38 > 1.645, we reject the null hypothesis. (The same conclusion can be drawn by comparing the 0.0087 probability of observing a sample mean as extreme as 197.1 to the level of significance of 0.05. If the observed probability is smaller than the level of significance we reject H 0 ). Because the Z score exceeds the critical value, we conclude that the mean weight for men in 2006 is more than 191 pounds, the value reported in 2002. If we observed the second sample (i.e., sample mean =192.1), we would not be able to reject the null hypothesis because the Z score is 0.43 which is not in the rejection region (i.e., the region in the tail end of the curve above 1.645). With the second sample we do not have sufficient evidence (because we set our level of significance at 5%) to conclude that weights have increased. Again, the same conclusion can be reached by comparing probabilities. The probability of observing a sample mean as extreme as 192.1 is 33.4% which is not below our 5% level of significance.

Hypothesis Testing: Upper-, Lower, and Two Tailed Tests

The procedure for hypothesis testing is based on the ideas described above. Specifically, we set up competing hypotheses, select a random sample from the population of interest and compute summary statistics. We then determine whether the sample data supports the null or alternative hypotheses. The procedure can be broken down into the following five steps.  

• Step 1. Set up hypotheses and select the level of significance α.

H 0 : Null hypothesis (no change, no difference);  

H 1 : Research hypothesis (investigator's belief); α =0.05

• Step 2. Select the appropriate test statistic.  

The test statistic is a single number that summarizes the sample information.   An example of a test statistic is the Z statistic computed as follows:

When the sample size is small, we will use t statistics (just as we did when constructing confidence intervals for small samples). As we present each scenario, alternative test statistics are provided along with conditions for their appropriate use.

• Step 3.  Set up decision rule.  

The decision rule is a statement that tells under what circumstances to reject the null hypothesis. The decision rule is based on specific values of the test statistic (e.g., reject H 0 if Z > 1.645). The decision rule for a specific test depends on 3 factors: the research or alternative hypothesis, the test statistic and the level of significance. Each is discussed below.

• The decision rule depends on whether an upper-tailed, lower-tailed, or two-tailed test is proposed. In an upper-tailed test the decision rule has investigators reject H 0 if the test statistic is larger than the critical value. In a lower-tailed test the decision rule has investigators reject H 0 if the test statistic is smaller than the critical value.  In a two-tailed test the decision rule has investigators reject H 0 if the test statistic is extreme, either larger than an upper critical value or smaller than a lower critical value.
• The exact form of the test statistic is also important in determining the decision rule. If the test statistic follows the standard normal distribution (Z), then the decision rule will be based on the standard normal distribution. If the test statistic follows the t distribution, then the decision rule will be based on the t distribution. The appropriate critical value will be selected from the t distribution again depending on the specific alternative hypothesis and the level of significance.  
• The third factor is the level of significance. The level of significance which is selected in Step 1 (e.g., α =0.05) dictates the critical value.   For example, in an upper tailed Z test, if α =0.05 then the critical value is Z=1.645.  

The following figures illustrate the rejection regions defined by the decision rule for upper-, lower- and two-tailed Z tests with α=0.05. Notice that the rejection regions are in the upper, lower and both tails of the curves, respectively. The decision rules are written below each figure.

Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < 1.645.

Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05

The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.

The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources."

Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."

• Step 4. Compute the test statistic.  

Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.

• Step 5. Conclusion.  

The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely).  

If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 .

Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .  

• Step 1. Set up hypotheses and determine level of significance

H 0 : μ = 191 H 1 : μ > 191                 α =0.05

The research hypothesis is that weights have increased, and therefore an upper tailed test is used.

• Step 2. Select the appropriate test statistic.

Because the sample size is large (n > 30) the appropriate test statistic is

• Step 3. Set up decision rule.  

In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05.   Reject H 0 if Z > 1.645.

We now substitute the sample data into the formula for the test statistic identified in Step 2.  

We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010.                  

Type I and Type II Errors

In all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality).

Table - Conclusions in Test of Hypothesis

In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ).

When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true.

 The most common reason for a Type II error is a small sample size.

Tests with One Sample, Continuous Outcome

Hypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow.

Test Statistics for Testing H 0 : μ= μ 0

• if n > 30
• if n < 30

Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value. 

The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races.  The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured.   The sample data are summarized as follows: n=100, x̄

=$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach. 

• Step 1.  Set up hypotheses and determine level of significance

H 0 : μ = 3,302 H 1 : μ < 3,302           α =0.05

The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used.

This is a lower tailed test, using a Z statistic and a 5% level of significance.   Reject H 0 if Z < -1.645.

•   Step 4. Compute the test statistic.  

We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002.  

Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures.      

The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring?

Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach.

H 0 : μ= 203 H 1 : μ≠ 203                       α=0.05

The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.

•   Step 3. Set up decision rule.  

This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960.

We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002.   Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001.  

Statistical Significance versus Clinical (Practical) Significance

This example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference?  

Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol.   Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows:   n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach. 

H 0 : μ= 203 H 1 : μ< 203                   α=0.05

•  Step 2. Select the appropriate test statistic.  

Because the sample size is small (n<30) the appropriate test statistic is

This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows:   Reject H 0 if t < -2.145.

We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious).

This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here.

Video - Comparing a Sample Mean to Known Population Mean (8:20)

Link to transcript of the video

Tests with One Sample, Dichotomous Outcome

Hypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator.    

In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size,

We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below.

Test Statistic for Testing H 0 : p = p 0

if min(np 0 , n(1-p 0 )) > 5

The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e.,

Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion.

Example:  

The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%.  Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans?

H 0 : p = 0.211 H 1 : p < 0.211                     α=0.05

We must first check that the sample size is adequate.   Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used:

This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.

We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001.  

The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data?

Calculate this on your own before checking the answer.

Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55)

Tests with Two Independent Samples, Continuous Outcome

There are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important.

Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows:

for sample 1:

for sample 2:

The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2.  

In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e.,

The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses.

Test Statistics for Testing H 0 : μ 1 = μ 2

• if n 1 > 30 and n 2 > 30
• if n 1 < 30 or n 2 < 30

NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances.    

The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows:

Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .)

Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.  

Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance.  

H 0 : μ 1 = μ 2

H 1 : μ 1 ≠ μ 2                       α=0.05

Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is

We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation.

Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample.   Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes.  

Now the test statistic:

We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010.  

Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference.  

Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials).  

A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.

Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach.

H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2                         α=0.05

Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is:

This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701.

Now the test statistic,

We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005.

The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug.

Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02)

Tests with Matched Samples, Continuous Outcome

In the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0).  

The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores.

Test Statistics for Testing H 0 : μ d =0

A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)

Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have

The calculations are shown below.  

Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach.

H 0 : μ d = 0 H 1 : μ d > 0                 α=0.05

NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0. 

• Step 2 . Select the appropriate test statistic.

This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145.

We now substitute the sample data into the formula for the test statistic identified in Step 2.

We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks.  

Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study.

Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11)

Tests with Two Independent Samples, Dichotomous Outcome

There are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare.

In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2.      

For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows:

H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 .  

The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:

• For the risk difference, H 0 : p 1 - p 2 = 0 versus H 1 : p 1 - p 2 ≠ 0 which are, by definition, equal to H 0 : RD = 0 versus H 1 : RD ≠ 0.
• If an investigator wants to focus on the risk ratio, the equivalent hypotheses are H 0 : RR = 1 versus H 1 : RR ≠ 1.
• If the investigator wants to focus on the odds ratio, the equivalent hypotheses are H 0 : OR = 1 versus H 1 : OR ≠ 1.  

Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠

1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively.  

The formula for the test of hypothesis for the difference in proportions is given below.

Test Statistics for Testing H 0 : p 1 = p

The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions.

The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.

The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach.

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2                 α=0.05

• Step 2.  Select the appropriate test statistic.  

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used:

Reject H 0 if Z < -1.960 or if Z > 1.960.

We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes:

We now substitute to compute the test statistic.

• Step 5. Conclusion.

We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers.  

A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers.    

Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD?

A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.

We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach.  

H 0 : p 1 = p 2     H 1 : p 1 ≠ p 2              α=0.05

Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2.

We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e.,

In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used

We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever.

A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result. 

Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions).  

Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55)

Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.

• Continuous Outcome, One Sample: H0: μ = μ0
• Continuous Outcome, Two Independent Samples: H0: μ1 = μ2
• Continuous Outcome, Two Matched Samples: H0: μd = 0
• Dichotomous Outcome, One Sample: H0: p = p 0
• Dichotomous Outcome, Two Independent Samples: H0: p1 = p2, RD=0, RR=1, OR=1

Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason.

In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis.    

We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect.

Answers to Selected Problems

Dental services problem - bottom of page 5.

• Step 1: Set up hypotheses and determine the level of significance.

α=0.05

• Step 2: Select the appropriate test statistic.

First, determine whether the sample size is adequate.

Therefore the sample size is adequate, and we can use the following formula:

• Step 3: Set up the decision rule.

Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.

• Step 4: Compute the test statistic
• Step 5: Conclusion.

We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion.

Module 10: Inference for Means

Hypothesis test for a difference in two population means (1 of 2), learning outcomes.

• Under appropriate conditions, conduct a hypothesis test about a difference between two population means. State a conclusion in context.

Using the Hypothesis Test for a Difference in Two Population Means

The general steps of this hypothesis test are the same as always. As expected, the details of the conditions for use of the test and the test statistic are unique to this test (but similar in many ways to what we have seen before.)

Step 1: Determine the hypotheses.

The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0 , is again a statement of “no effect” or “no difference.”

• H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2

The alternative hypothesis, H a , can be any one of the following.

• H a : μ 1 – μ 2 < 0, which is the same as H a : μ 1 < μ 2
• H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2
• H a : μ 1 – μ 2 ≠ 0, which is the same as H a : μ 1 ≠ μ 2

Step 2: Collect the data.

As usual, how we collect the data determines whether we can use it in the inference procedure. We have our usual two requirements for data collection.

• Samples must be random to remove or minimize bias.
• Samples must be representative of the populations in question.

We use this hypothesis test when the data meets the following conditions.

• The two random samples are independent .
• The variable is normally distributed in both populations . If this variable is not known, samples of more than 30 will have a difference in sample means that can be modeled adequately by the t-distribution. As we discussed in “Hypothesis Test for a Population Mean,” t-procedures are robust even when the variable is not normally distributed in the population. If checking normality in the populations is impossible, then we look at the distribution in the samples. If a histogram or dotplot of the data does not show extreme skew or outliers, we take it as a sign that the variable is not heavily skewed in the populations, and we use the inference procedure. (Note: This is the same condition we used for the one-sample t-test in “Hypothesis Test for a Population Mean.”)

Step 3: Assess the evidence.

If the conditions are met, then we calculate the t-test statistic. The t-test statistic has a familiar form.

[latex]T\text{}=\text{}\frac{(\mathrm{Observed}\text{}\mathrm{difference}\text{}\mathrm{in}\text{}\mathrm{sample}\text{}\mathrm{means})-(\mathrm{Hypothesized}\text{}\mathrm{difference}\text{}\mathrm{in}\text{}\mathrm{population}\text{}\mathrm{means})}{\mathrm{Standard}\text{}\mathrm{error}}[/latex]

[latex]T\text{}=\text{}\frac{({\stackrel{¯}{x}}_{1}-{\stackrel{¯}{x}}_{2})-({μ}_{1}-{μ}_{2})}{\sqrt{\frac{{{s}_{1}}^{2}}{{n}_{1}}+\frac{{{s}_{2}}^{2}}{{n}_{2}}}}[/latex]

Since the null hypothesis assumes there is no difference in the population means, the expression (μ 1 – μ 2 ) is always zero.

As we learned in “Estimating a Population Mean,” the t-distribution depends on the degrees of freedom (df) . In the one-sample and matched-pair cases df = n – 1. For the two-sample t-test, determining the correct df is based on a complicated formula that we do not cover in this course. We will either give the df or use technology to find the df . With the t-test statistic and the degrees of freedom, we can use the appropriate t-model to find the P-value, just as we did in “Hypothesis Test for a Population Mean.” We can even use the same simulation.

Step 4: State a conclusion.

To state a conclusion, we follow what we have done with other hypothesis tests. We compare our P-value to a stated level of significance.

• If the P-value ≤ α, we reject the null hypothesis in favor of the alternative hypothesis.
• If the P-value > α, we fail to reject the null hypothesis. We do not have enough evidence to support the alternative hypothesis.

As always, we state our conclusion in context, usually by referring to the alternative hypothesis.

“Context and Calories”

Does the company you keep impact what you eat? This example comes from an article titled “Impact of Group Settings and Gender on Meals Purchased by College Students” (Allen-O’Donnell, M., T. C. Nowak, K. A. Snyder, and M. D. Cottingham, Journal of Applied Social Psychology 49(9), 2011, onlinelibrary.wiley.com/doi/10.1111/j.1559-1816.2011.00804.x/full) . In this study, researchers examined this issue in the context of gender-related theories in their field. For our purposes, we look at this research more narrowly.

Step 1: Stating the hypotheses.

In the article, the authors make the following hypothesis. “The attempt to appear feminine will be empirically demonstrated by the purchase of fewer calories by women in mixed-gender groups than by women in same-gender groups.” We translate this into a simpler and narrower research question: Do women purchase fewer calories when they eat with men compared to when they eat with women?

Here the two populations are “women eating with women” (population 1) and “women eating with men” (population 2). The variable is the calories in the meal. We test the following hypotheses at the 5% level of significance.

The null hypothesis is always H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2 .

The alternative hypothesis H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2 .

Here μ 1 represents the mean number of calories ordered by women when they were eating with other women, and μ 2 represents the mean number of calories ordered by women when they were eating with men.

Note: It does not matter which population we label as 1 or 2, but once we decide, we have to stay consistent throughout the hypothesis test. Since we expect the number of calories to be greater for the women eating with other women, the difference is positive if “women eating with women” is population 1. If you prefer to work with positive numbers, choose the group with the larger expected mean as population 1. This is a good general tip.

Step 2: Collect Data.

As usual, there are two major things to keep in mind when considering the collection of data.

• Samples need to be representative of the population in question.
• Samples need to be random in order to remove or minimize bias.

Representative Samples?

The researchers state their hypothesis in terms of “women.” We did the same. But the researchers gathered data by watching people eat at the HUB Rock Café II on the campus of Indiana University of Pennsylvania during the Spring semester of 2006. Almost all of the women in the data set were white undergraduates between the ages of 18 and 24, so there are some definite limitations on the scope of this study. These limitations will affect our conclusion (and the specific definition of the population means in our hypotheses.)

Random Samples?

The observations were collected on February 13, 2006, through February 22, 2006, between 11 a.m. and 7 p.m. We can see that the researchers included both lunch and dinner. They also made observations on all days of the week to ensure that weekly customer patterns did not confound their findings. The authors state that “since the time period for observations and the place where [they] observed students were limited, the sample was a convenience sample.” Despite these limitations, the researchers conducted inference procedures with the data, and the results were published in a reputable journal. We will also conduct inference with this data, but we also include a discussion of the limitations of the study with our conclusion. The authors did this, also.

Do the data met the conditions for use of a t-test?

The researchers reported the following sample statistics.

• In a sample of 45 women dining with other women, the average number of calories ordered was 850, and the standard deviation was 252.
• In a sample of 27 women dining with men, the average number of calories ordered was 719, and the standard deviation was 322.

One of the samples has fewer than 30 women. We need to make sure the distribution of calories in this sample is not heavily skewed and has no outliers, but we do not have access to a spreadsheet of the actual data. Since the researchers conducted a t-test with this data, we will assume that the conditions are met. This includes the assumption that the samples are independent.

As noted previously, the researchers reported the following sample statistics.

To compute the t-test statistic, make sure sample 1 corresponds to population 1. Here our population 1 is “women eating with other women.” So x 1 = 850, s 1 = 252, n 1 =45, and so on.

[latex]T\text{}=\text{}\frac{{\stackrel{¯}{x}}_{1}\text{}\text{−}\text{}{\stackrel{¯}{x}}_{2}}{\sqrt{\frac{{{s}_{1}}^{2}}{{n}_{1}}+\frac{{{s}_{2}}^{2}}{{n}_{2}}}}\text{}=\text{}\frac{850\text{}\text{−}\text{}719}{\sqrt{\frac{{252}^{2}}{45}+\frac{{322}^{2}}{27}}}\text{}\approx \text{}\frac{131}{72.47}\text{}\approx \text{}1.81[/latex]

Using technology, we determined that the degrees of freedom are about 45 for this data. To find the P-value, we use our familiar simulation of the t-distribution. Since the alternative hypothesis is a “greater than” statement, we look for the area to the right of T = 1.81. The P-value is 0.0385.

Generic Conclusion

The hypotheses for this test are H 0 : μ 1 – μ 2 = 0 and H a : μ 1 – μ 2 > 0. Since the P-value is less than the significance level (0.0385 < 0.05), we reject H 0 and accept H a .

Conclusion in context

At Indiana University of Pennsylvania, the mean number of calories ordered by undergraduate women eating with other women is greater than the mean number of calories ordered by undergraduate women eating with men (P-value = 0.0385).

Comment about Conclusions

In the conclusion above, we did not generalize the findings to all women. Since the samples included only undergraduate women at one university, we included this information in our conclusion. But our conclusion is a cautious statement of the findings. The authors see the results more broadly in the context of theories in the field of social psychology. In the context of these theories, they write, “Our findings support the assertion that meal size is a tool for influencing the impressions of others. For traditional-age, predominantly White college women, diminished meal size appears to be an attempt to assert femininity in groups that include men.” This viewpoint is echoed in the following summary of the study for the general public on National Public Radio (npr.org).

• Both men and women appear to choose larger portions when they eat with women, and both men and women choose smaller portions when they eat in the company of men, according to new research published in the Journal of Applied Social Psychology . The study, conducted among a sample of 127 college students, suggests that both men and women are influenced by unconscious scripts about how to behave in each other’s company. And these scripts change the way men and women eat when they eat together and when they eat apart.

Should we be concerned that the findings of this study are generalized in this way? Perhaps. But the authors of the article address this concern by including the following disclaimer with their findings: “While the results of our research are suggestive, they should be replicated with larger, representative samples. Studies should be done not only with primarily White, middle-class college students, but also with students who differ in terms of race/ethnicity, social class, age, sexual orientation, and so forth.” This is an example of good statistical practice. It is often very difficult to select truly random samples from the populations of interest. Researchers therefore discuss the limitations of their sampling design when they discuss their conclusions.

In the following activities, you will have the opportunity to practice parts of the hypothesis test for a difference in two population means. On the next page, the activities focus on the entire process and also incorporate technology.

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• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

Hypothesis Testing (cont...)

Hypothesis testing, the null and alternative hypothesis.

In order to undertake hypothesis testing you need to express your research hypothesis as a null and alternative hypothesis. The null hypothesis and alternative hypothesis are statements regarding the differences or effects that occur in the population. You will use your sample to test which statement (i.e., the null hypothesis or alternative hypothesis) is most likely (although technically, you test the evidence against the null hypothesis). So, with respect to our teaching example, the null and alternative hypothesis will reflect statements about all statistics students on graduate management courses.

The null hypothesis is essentially the "devil's advocate" position. That is, it assumes that whatever you are trying to prove did not happen ( hint: it usually states that something equals zero). For example, the two different teaching methods did not result in different exam performances (i.e., zero difference). Another example might be that there is no relationship between anxiety and athletic performance (i.e., the slope is zero). The alternative hypothesis states the opposite and is usually the hypothesis you are trying to prove (e.g., the two different teaching methods did result in different exam performances). Initially, you can state these hypotheses in more general terms (e.g., using terms like "effect", "relationship", etc.), as shown below for the teaching methods example:

Depending on how you want to "summarize" the exam performances will determine how you might want to write a more specific null and alternative hypothesis. For example, you could compare the mean exam performance of each group (i.e., the "seminar" group and the "lectures-only" group). This is what we will demonstrate here, but other options include comparing the distributions , medians , amongst other things. As such, we can state:

Now that you have identified the null and alternative hypotheses, you need to find evidence and develop a strategy for declaring your "support" for either the null or alternative hypothesis. We can do this using some statistical theory and some arbitrary cut-off points. Both these issues are dealt with next.

Significance levels

The level of statistical significance is often expressed as the so-called p -value . Depending on the statistical test you have chosen, you will calculate a probability (i.e., the p -value) of observing your sample results (or more extreme) given that the null hypothesis is true . Another way of phrasing this is to consider the probability that a difference in a mean score (or other statistic) could have arisen based on the assumption that there really is no difference. Let us consider this statement with respect to our example where we are interested in the difference in mean exam performance between two different teaching methods. If there really is no difference between the two teaching methods in the population (i.e., given that the null hypothesis is true), how likely would it be to see a difference in the mean exam performance between the two teaching methods as large as (or larger than) that which has been observed in your sample?

So, you might get a p -value such as 0.03 (i.e., p = .03). This means that there is a 3% chance of finding a difference as large as (or larger than) the one in your study given that the null hypothesis is true. However, you want to know whether this is "statistically significant". Typically, if there was a 5% or less chance (5 times in 100 or less) that the difference in the mean exam performance between the two teaching methods (or whatever statistic you are using) is as different as observed given the null hypothesis is true, you would reject the null hypothesis and accept the alternative hypothesis. Alternately, if the chance was greater than 5% (5 times in 100 or more), you would fail to reject the null hypothesis and would not accept the alternative hypothesis. As such, in this example where p = .03, we would reject the null hypothesis and accept the alternative hypothesis. We reject it because at a significance level of 0.03 (i.e., less than a 5% chance), the result we obtained could happen too frequently for us to be confident that it was the two teaching methods that had an effect on exam performance.

Whilst there is relatively little justification why a significance level of 0.05 is used rather than 0.01 or 0.10, for example, it is widely used in academic research. However, if you want to be particularly confident in your results, you can set a more stringent level of 0.01 (a 1% chance or less; 1 in 100 chance or less).

One- and two-tailed predictions

When considering whether we reject the null hypothesis and accept the alternative hypothesis, we need to consider the direction of the alternative hypothesis statement. For example, the alternative hypothesis that was stated earlier is:

The alternative hypothesis tells us two things. First, what predictions did we make about the effect of the independent variable(s) on the dependent variable(s)? Second, what was the predicted direction of this effect? Let's use our example to highlight these two points.

Sarah predicted that her teaching method (independent variable: teaching method), whereby she not only required her students to attend lectures, but also seminars, would have a positive effect (that is, increased) students' performance (dependent variable: exam marks). If an alternative hypothesis has a direction (and this is how you want to test it), the hypothesis is one-tailed. That is, it predicts direction of the effect. If the alternative hypothesis has stated that the effect was expected to be negative, this is also a one-tailed hypothesis.

Alternatively, a two-tailed prediction means that we do not make a choice over the direction that the effect of the experiment takes. Rather, it simply implies that the effect could be negative or positive. If Sarah had made a two-tailed prediction, the alternative hypothesis might have been:

In other words, we simply take out the word "positive", which implies the direction of our effect. In our example, making a two-tailed prediction may seem strange. After all, it would be logical to expect that "extra" tuition (going to seminar classes as well as lectures) would either have a positive effect on students' performance or no effect at all, but certainly not a negative effect. However, this is just our opinion (and hope) and certainly does not mean that we will get the effect we expect. Generally speaking, making a one-tail prediction (i.e., and testing for it this way) is frowned upon as it usually reflects the hope of a researcher rather than any certainty that it will happen. Notable exceptions to this rule are when there is only one possible way in which a change could occur. This can happen, for example, when biological activity/presence in measured. That is, a protein might be "dormant" and the stimulus you are using can only possibly "wake it up" (i.e., it cannot possibly reduce the activity of a "dormant" protein). In addition, for some statistical tests, one-tailed tests are not possible.

Rejecting or failing to reject the null hypothesis

Let's return finally to the question of whether we reject or fail to reject the null hypothesis.

If our statistical analysis shows that the significance level is below the cut-off value we have set (e.g., either 0.05 or 0.01), we reject the null hypothesis and accept the alternative hypothesis. Alternatively, if the significance level is above the cut-off value, we fail to reject the null hypothesis and cannot accept the alternative hypothesis. You should note that you cannot accept the null hypothesis, but only find evidence against it.

Module 10: Inference for Means

Hypothesis Test for a Difference in Two Population Means (1 of 2)

Learning outcomes.

• Under appropriate conditions, conduct a hypothesis test about a difference between two population means. State a conclusion in context.

Using the Hypothesis Test for a Difference in Two Population Means

The general steps of this hypothesis test are the same as always. As expected, the details of the conditions for use of the test and the test statistic are unique to this test (but similar in many ways to what we have seen before.)

Step 1: Determine the hypotheses.

The hypotheses for a difference in two population means are similar to those for a difference in two population proportions. The null hypothesis, H 0 , is again a statement of “no effect” or “no difference.”

• H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2

The alternative hypothesis, H a , can be any one of the following.

• H a : μ 1 – μ 2 < 0, which is the same as H a : μ 1 < μ 2
• H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2
• H a : μ 1 – μ 2 ≠ 0, which is the same as H a : μ 1 ≠ μ 2

Step 2: Collect the data.

As usual, how we collect the data determines whether we can use it in the inference procedure. We have our usual two requirements for data collection.

• Samples must be random to remove or minimize bias.
• Samples must be representative of the populations in question.

We use this hypothesis test when the data meets the following conditions.

• The two random samples are independent .
• The variable is normally distributed in both populations . If this variable is not known, samples of more than 30 will have a difference in sample means that can be modeled adequately by the t-distribution. As we discussed in “Hypothesis Test for a Population Mean,” t-procedures are robust even when the variable is not normally distributed in the population. If checking normality in the populations is impossible, then we look at the distribution in the samples. If a histogram or dotplot of the data does not show extreme skew or outliers, we take it as a sign that the variable is not heavily skewed in the populations, and we use the inference procedure. (Note: This is the same condition we used for the one-sample t-test in “Hypothesis Test for a Population Mean.”)

Step 3: Assess the evidence.

If the conditions are met, then we calculate the t-test statistic. The t-test statistic has a familiar form.

[latex]T=\frac{Observeddifferenceinsamplemeans-Hypothesizeddiferenceinpopulationmeans}{ standarderror}[/latex]

[latex]T=\frac{(\bar{x}_{1}-\bar{x}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}}}+\frac{s_{2}^{2}}{n_{2}}}[/latex]

Since the null hypothesis assumes there is no difference in the population means, the expression (μ 1 – μ 2 ) is always zero.

As we learned in “Estimating a Population Mean,” the t-distribution depends on the degrees of freedom (df) . In the one-sample and matched-pair cases df = n – 1. For the two-sample t-test, determining the correct df is based on a complicated formula that we do not cover in this course. We will either give the df or use technology to find the df . With the t-test statistic and the degrees of freedom, we can use the appropriate t-model to find the P-value, just as we did in “Hypothesis Test for a Population Mean.” We can even use the same simulation.

Step 4: State a conclusion.

To state a conclusion, we follow what we have done with other hypothesis tests. We compare our P-value to a stated level of significance.

• If the P-value ≤ α, we reject the null hypothesis in favor of the alternative hypothesis.
• If the P-value > α, we fail to reject the null hypothesis. We do not have enough evidence to support the alternative hypothesis.

As always, we state our conclusion in context, usually by referring to the alternative hypothesis.

“Context and Calories”

Does the company you keep impact what you eat? This example comes from an article titled “Impact of Group Settings and Gender on Meals Purchased by College Students” (Allen-O’Donnell, M., T. C. Nowak, K. A. Snyder, and M. D. Cottingham, Journal of Applied Social Psychology 49(9), 2011, onlinelibrary.wiley.com/doi/10.1111/j.1559-1816.2011.00804.x/full) . In this study, researchers examined this issue in the context of gender-related theories in their field. For our purposes, we look at this research more narrowly.

Step 1: Stating the hypotheses.

In the article, the authors make the following hypothesis. “The attempt to appear feminine will be empirically demonstrated by the purchase of fewer calories by women in mixed-gender groups than by women in same-gender groups.” We translate this into a simpler and narrower research question: Do women purchase fewer calories when they eat with men compared to when they eat with women?

Here the two populations are “women eating with women” (population 1) and “women eating with men” (population 2). The variable is the calories in the meal. We test the following hypotheses at the 5% level of significance.

The null hypothesis is always H 0 : μ 1 – μ 2 = 0, which is the same as H 0 : μ 1 = μ 2 .

The alternative hypothesis H a : μ 1 – μ 2 > 0, which is the same as H a : μ 1 > μ 2 .

Here μ 1 represents the mean number of calories ordered by women when they were eating with other women, and μ 2 represents the mean number of calories ordered by women when they were eating with men.

Note: It does not matter which population we label as 1 or 2, but once we decide, we have to stay consistent throughout the hypothesis test. Since we expect the number of calories to be greater for the women eating with other women, the difference is positive if “women eating with women” is population 1. If you prefer to work with positive numbers, choose the group with the larger expected mean as population 1. This is a good general tip.

Step 2: Collect Data.

As usual, there are two major things to keep in mind when considering the collection of data.

• Samples need to be representative of the population in question.
• Samples need to be random in order to remove or minimize bias.

Representative Samples?

The researchers state their hypothesis in terms of “women.” We did the same. But the researchers gathered data by watching people eat at the HUB Rock Café II on the campus of Indiana University of Pennsylvania during the Spring semester of 2006. Almost all of the women in the data set were white undergraduates between the ages of 18 and 24, so there are some definite limitations on the scope of this study. These limitations will affect our conclusion (and the specific definition of the population means in our hypotheses.)

Random Samples?

The observations were collected on February 13, 2006, through February 22, 2006, between 11 a.m. and 7 p.m. We can see that the researchers included both lunch and dinner. They also made observations on all days of the week to ensure that weekly customer patterns did not confound their findings. The authors state that “since the time period for observations and the place where [they] observed students were limited, the sample was a convenience sample.” Despite these limitations, the researchers conducted inference procedures with the data, and the results were published in a reputable journal. We will also conduct inference with this data, but we also include a discussion of the limitations of the study with our conclusion. The authors did this, also.

Do the data meet the conditions for use of a t-test?

The researchers reported the following sample statistics.

• In a sample of 45 women dining with other women, the average number of calories ordered was 850, and the standard deviation was 252.
• In a sample of 27 women dining with men, the average number of calories ordered was 719, and the standard deviation was 322.

One of the samples has fewer than 30 women. We need to make sure the distribution of calories in this sample is not heavily skewed and has no outliers, but we do not have access to a spreadsheet of the actual data. Since the researchers conducted a t-test with this data, we will assume that the conditions are met. This includes the assumption that the samples are independent.

As noted previously, the researchers reported the following sample statistics.

To compute the t-test statistic, make sure sample 1 corresponds to population 1. Here our population 1 is “women eating with other women.” So x 1 = 850, s 1 = 252, n 1 =45, and so on.

[latex]T=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}}+\frac{s_{2}^{2}}{n_{2}}}= \frac{850-719}{\sqrt{\frac{252^{2}}{45}+\frac{322^{2}}{27}}}\approx \frac{131}{72.47}\approx 1.81[/latex]

Using technology, we determined that the degrees of freedom are about 45 for this data. To find the P-value, we use our familiar simulation of the t-distribution. Since the alternative hypothesis is a “greater than” statement, we look for the area to the right of T = 1.81. The P-value is 0.0385.

Generic Conclusion

The hypotheses for this test are H 0 : μ 1 – μ 2 = 0 and H a : μ 1 – μ 2 > 0. Since the P-value is less than the significance level (0.0385 < 0.05), we reject H 0 and accept H a .

Conclusion in context

At Indiana University of Pennsylvania, the mean number of calories ordered by undergraduate women eating with other women is greater than the mean number of calories ordered by undergraduate women eating with men (P-value = 0.0385).

Comment about Conclusions

In the conclusion above, we did not generalize the findings to all women. Since the samples included only undergraduate women at one university, we included this information in our conclusion. But our conclusion is a cautious statement of the findings. The authors see the results more broadly in the context of theories in the field of social psychology. In the context of these theories, they write, “Our findings support the assertion that meal size is a tool for influencing the impressions of others. For traditional-age, predominantly White college women, diminished meal size appears to be an attempt to assert femininity in groups that include men.” This viewpoint is echoed in the following summary of the study for the general public on National Public Radio (npr.org).

• Both men and women appear to choose larger portions when they eat with women, and both men and women choose smaller portions when they eat in the company of men, according to new research published in the Journal of Applied Social Psychology . The study, conducted among a sample of 127 college students, suggests that both men and women are influenced by unconscious scripts about how to behave in each other’s company. And these scripts change the way men and women eat when they eat together and when they eat apart.

Should we be concerned that the findings of this study are generalized in this way? Perhaps. But the authors of the article address this concern by including the following disclaimer with their findings: “While the results of our research are suggestive, they should be replicated with larger, representative samples. Studies should be done not only with primarily White, middle-class college students, but also with students who differ in terms of race/ethnicity, social class, age, sexual orientation, and so forth.” This is an example of good statistical practice. It is often very difficult to select truly random samples from the populations of interest. Researchers therefore discuss the limitations of their sampling design when they discuss their conclusions.

In the following activities, you will have the opportunity to practice parts of the hypothesis test for a difference in two population means. On the next page, the activities focus on the entire process and also incorporate technology.

National Health and Nutrition Survey

• Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution

Concepts in Statistics Copyright © 2023 by CUNY School of Professional Studies is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.

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10.6: Test of Mean vs. Hypothesized Value – A Complete Example

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• Maurice A. Geraghty
• De Anza College

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Example – Soy sauce production

A food company has a policy that the stated contents of a product match the actual results. A General Question might be “Does the stated net weight of a food product match the actual weight?” The quality control statistician decides to test the 16 ounce bottle of Soy Sauce and must now design the experiment .

The quality‐control statistician has been given the authority to sample 36 bottles of soy sauce and knows from past testing that the population standard deviation is 0.5 ounces. The model will be a test of population mean vs. hypothesized value of 16 oz.  A two‐tailed test is selected since the company is concerned about both overfilling and underfilling the bottles as the stated policy is that the stated weight should match the actual weight of the product.

Research Hypotheses :

\(H_o: \mu =16\)  (The filling machine is operating properly)                                          

\(H_a: \mu \neq 16\) (The filling machine is not operating properly)

Since the population standard deviation is known the test statistic will be \(Z=\dfrac{\overline{X}-\mu}{\sigma / \sqrt{n}}\). This model is appropriate since the sample size assures that the distribution of the sample mean is approximately Normal due to the Central Limit Theorem.

Type I error would be to reject the Null Hypothesis and say that the machine is not running properly when in fact it was operating properly. Since the company does not want to needlessly stop production and recalibrate the machine, the statistician chooses to limit the probability of Type I error by setting the level of significance (\(\alpha\)) to 5%.

The statistician now conducts the experiment and samples 36 bottles over one hour and determines from a box plot of the data that there is one unusual observation of 17.56 ounces. The value is rechecked and kept in the data set.

Next, the sample mean and the test statistic are calculated.

\[\overline{X}=16.12 \text { ounces } \qquad \qquad Z=\dfrac{16.12-16}{0.5 / \sqrt{36}}=1.44 \nonumber \]

The decision rule under the critical value method would be to reject the Null Hypothesis when the value of the test statistic is in the rejection region. In other words, reject \(H_o\) when \(Z >1.96\) or \(Z<‐1.96\).

Based on this result, the decision is fail to reject \(H_o\) , since the test statistic does not fall in the rejection region.

Alternatively (and preferably) the statistician could use the p‐value method of decision rule. The \(p\)‐value for a two‐tailed test must include all values (positive and negative) more extreme than the Test Statistic, so in this example we find the probability that \(Z < ‐1.44\) or \(Z > 1.44\) (the area shaded blue).

Using a calculator, computer software or a Standard Normal table, the \(p\)‐value=0.1498 . Since the \(p\)‐value is greater than \(\alpha\) the decision again is fail to reject \(H_o\) .

Finally the statistician must report the conclusions and make a recommendation to the company’s management:

“There is insufficient evidence to conclude that the machine that fills 16 ounce soy sauce bottles is operating improperly. This conclusion is based on 36 measurements taken during a single hour’s production run. I recommend continued monitoring of the machine during different employee shifts to account for the possibility of potential human error”.

The statistician makes the weak statement and is not stating that the machine is running properly, only that there is not enough evidence to state that the machine is running improperly. The statistician also reports concerns about the sampling of only one shift of employees (restricting the inference to the sampled population) and recommends repeating the experiment over several shifts.

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9.2: Comparing Two Independent Population Means (Hypothesis test)

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• The two independent samples are simple random samples from two distinct populations.
• if the sample sizes are small, the distributions are important (should be normal)
• if the sample sizes are large, the distributions are not important (need not be normal)

The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch \(t\)-test. The degrees of freedom formula was developed by Aspin-Welch.

The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, \(\bar{X}_{1} - \bar{X}_{2}\), and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.

Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error , of the difference in sample means , \(\bar{X}_{1} - \bar{X}_{2}\).

The standard error is:

\[\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}\]

The test statistic ( t -score) is calculated as follows:

\[\dfrac{(\bar{x}-\bar{x}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}\]

• \(s_{1}\) and \(s_{2}\), the sample standard deviations, are estimates of \(\sigma_{1}\) and \(\sigma_{1}\), respectively.
• \(\sigma_{1}\) and \(\sigma_{2}\) are the unknown population standard deviations.
• \(\bar{x}_{1}\) and \(\bar{x}_{2}\) are the sample means. \(\mu_{1}\) and \(\mu_{2}\) are the population means.

The number of degrees of freedom (\(df\)) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The \(df\) are not always a whole number. The test statistic calculated previously is approximated by the Student's t -distribution with \(df\) as follows:

Degrees of freedom

\[df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1}-1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2} + \left(\dfrac{1}{n_{2}-1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}\]

We can also use a conservative estimation of degree of freedom by taking DF to be the smallest of \(n_{1}-1\) and \(n_{2}-1\)

When both sample sizes \(n_{1}\) and \(n_{2}\) are five or larger, the Student's t approximation is very good. Notice that the sample variances \((s_{1})^{2}\) and \((s_{2})^{2}\) are not pooled. (If the question comes up, do not pool the variances.)

It is not necessary to compute the degrees of freedom by hand. A calculator or computer easily computes it.

Example \(\PageIndex{1}\): Independent groups

The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table \(\PageIndex{1}\). Each populations has a normal distribution.

Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.

The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, \(\mu_{g}\) is the population mean for girls and \(\mu_{b}\) is the population mean for boys. This is a test of two independent groups, two population means.

Random variable: \(\bar{X}_{g} - \bar{X}_{b} =\) difference in the sample mean amount of time girls and boys play sports each day.

• \(H_{0}: \mu_{g} = \mu_{b}\)  
• \(H_{0}: \mu_{g} - \mu_{b} = 0\)
• \(H_{a}: \mu_{g} \neq \mu_{b}\)  
• \(H_{a}: \mu_{g} - \mu_{b} \neq 0\)

The words "the same" tell you \(H_{0}\) has an "=". Since there are no other words to indicate \(H_{a}\), assume it says "is different." This is a two-tailed test.

Distribution for the test: Use \(t_{df}\) where \(df\) is calculated using the \(df\) formula for independent groups, two population means. Using a calculator, \(df\) is approximately 18.8462. Do not pool the variances.

Calculate the p -value using a Student's t -distribution: \(p\text{-value} = 0.0054\)

\[s_{g} = 0.866\]

\[s_{b} = 1\]

\[\bar{x}_{g} - \bar{x}_{b} = 2 - 3.2 = -1.2\]

Half the \(p\text{-value}\) is below –1.2 and half is above 1.2.

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\). This means you reject \(\mu_{g} = \mu_{b}\). The means are different.

Press STAT . Arrow over to TESTS and press 4:2-SampTTest . Arrow over to Stats and press ENTER . Arrow down and enter 2 for the first sample mean, \(\sqrt{0.866}\) for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER . Arrow down to Pooled: and No . Press ENTER . Arrow down to Calculate and press ENTER . The \(p\text{-value}\) is \(p = 0.0054\), the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.

Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).

Exercise \(\PageIndex{1}\)

Two samples are shown in Table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.

The \(p\text{-value}\) is \(0.4125\), which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.

When the sum of the sample sizes is larger than \(30 (n_{1} + n_{2} > 30)\) you can use the normal distribution to approximate the Student's \(t\).

Example \(\PageIndex{2}\)

A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.

• Is this a test of two means or two proportions?
• Are the populations standard deviations known or unknown?
• Which distribution do you use to perform the test?
• What is the random variable?
• What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
• Is this test right-, left-, or two-tailed?
• What is the \(p\text{-value}\)?
• Do you reject or not reject the null hypothesis?
• Student's t
• \(\bar{X}_{A} - \bar{X}_{B}\)
• \(H_{0}: \mu_{A} \leq \mu_{B}\) and \(H_{a}: \mu_{A} > \mu_{B}\)

• h. Do not reject.
• i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.

Exercise \(\PageIndex{2}\)

A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.

• Are the population standard deviations known?
• Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?
• They are unknown.
• The \(p\text{-value} = 0.0878\). At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.

Example \(\PageIndex{3}\)

A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table \(\PageIndex{3}\) and Table \(\PageIndex{4}\).

Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:

• Are the population standard deviations known or unknown?
• What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
• Is this test right, left, or two tailed?
• At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.

(See the conclusion in Example, and write yours in a similar fashion)

Be careful not to mix up the information for Group 1 and Group 2!

• Student's \(t\)
• \(\bar{X}_{1} - \bar{X}_{2}\)
• \(H_{0}: \mu_{1} = \mu_{2}\) Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
• \(H_{a}: \mu_{1} < \mu_{2}\) Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
• left-tailed

Figure \(\PageIndex{3}\).

• Reject the null hypothesis

At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.

First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to \(\mu_{1}\): and arrow to \(\neq \mu_{1}\) (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.

Cohen's Standards for Small, Medium, and Large Effect Sizes

Cohen's \(d\) is a measure of effect size based on the differences between two means. Cohen’s \(d\), named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.

Cohen's \(d\) is the measure of the difference between two means divided by the pooled standard deviation: \(d = \dfrac{\bar{x}_{2}-\bar{x}_{2}}{s_{\text{pooled}}}\) where \(s_{pooled} = \sqrt{\dfrac{(n_{1}-1)s^{2}_{1} + (n_{2}-1)s^{2}_{2}}{n_{1}+n_{2}-2}}\)

Example \(\PageIndex{4}\)

Calculate Cohen’s d for Example. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.

\(\mu_{1} = 4 s_{1} = 1.5 n_{1} = 11\)

\(\mu_{2} = 3.5 s_{2} = 1 n_{2} = 9\)

\(d = 0.384\)

The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.

Example \(\PageIndex{5}\)

Calculate Cohen’s \(d\) for Example. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.

\(d = 0.834\); Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.

Example 10.2.6

Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in Table and Table, respectively.

Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:

• Calculate Cohen’s d and interpret it.
• Student’s-t
• \(H_{0}: \mu_{1} = \mu_{2}\) Null hypothesis: the means of the weighted alphas are equal.
• \(H_{a}: \mu_{1} \neq \mu_{2}\) Alternative hypothesis : the means of the weighted alphas are not equal.
• \(p\text{-value} = 0.8787\)
• Do not reject the null hypothesis

Figure \(\PageIndex{4}\).

• \(d = 0.040\), Very small, because 0.040 is less than Cohen’s value of 0.2 for small effect size. The size of the difference of the means of the weighted alphas for the two regions of banks is small indicating that there is not a significant difference between their trends in stocks.
• Data from Graduating Engineer + Computer Careers. Available online at www.graduatingengineer.com
• Data from Microsoft Bookshelf .
• Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013).
• “List of current United States Senators by Age.” Wikipedia. Available online at en.Wikipedia.org/wiki/List_of...enators_by_age (accessed June 17, 2013).
• “Sectoring by Industry Groups.” Nasdaq. Available online at www.nasdaq.com/markets/barcha...&base=industry (accessed June 17, 2013).
• “Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013).
• “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013).

Two population means from independent samples where the population standard deviations are not known

• Random Variable: \(\bar{X}_{1} - \bar{X}_{2} =\) the difference of the sampling means
• Distribution: Student's t -distribution with degrees of freedom (variances not pooled)

Formula Review

Standard error: \[SE = \sqrt{\dfrac{(s_{1}^{2})}{n_{1}} + \dfrac{(s_{2}^{2})}{n_{2}}}\]

Test statistic ( t -score): \[t = \dfrac{(\bar{x}_{1}-\bar{x}_{2}) - (\mu_{1}-\mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}\]

Degrees of freedom:

\[df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1} - 1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2}} + \left(\dfrac{1}{n_{2} - 1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}\]

• \(s_{1}\) and \(s_{2}\) are the sample standard deviations, and n 1 and n 2 are the sample sizes.
• \(x_{1}\) and \(x_{2}\) are the sample means.

OR use the   DF to be the smallest of \(n_{1}-1\) and \(n_{2}-1\)

Cohen’s \(d\) is the measure of effect size:

\[d = \dfrac{\bar{x}_{1} - \bar{x}_{2}}{s_{\text{pooled}}}\]

\[s_{\text{pooled}} = \sqrt{\dfrac{(n_{1} - 1)s^{2}_{1} + (n_{2} - 1)s^{2}_{2}}{n_{1} + n_{2} - 2}}\]

• The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if \(X =\) hair color, then the domain is {black, blond, gray, green, orange}.
• We can tell what specific value x of the random variable \(X\) takes only after performing the experiment.

Teach yourself statistics

Hypothesis Test: Difference Between Paired Means

This lesson explains how to conduct a hypothesis test for the difference between paired means . The test procedure, called the matched-pairs t-test , is appropriate when the following conditions are met:

• The sampling method for each sample is simple random sampling .
• The test is conducted on paired data. (As a result, the data sets are not independent .)
• The population distribution is normal.
• The population data are symmetric , unimodal , without outliers , and the sample size is 15 or less.
• The population data are slightly skewed , unimodal, without outliers, and the sample size is 16 to 40.
• The sample size is greater than 40, without outliers.

This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

Every hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa.

The hypotheses concern a new variable d, which is based on the difference between paired values from two data sets.

d = x 1 - x 2

where x 1 is the value of variable x in the first data set, and x 2 is the value of the variable from the second data set that is paired with x 1 .

The table below shows three sets of null and alternative hypotheses. Each makes a statement about how the true difference in population values μ d is related to some hypothesized value D. (In the table, the symbol ≠ means " not equal to ".)

The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use the matched-pairs t-test to determine whether the difference between sample means for paired data is significantly different from the hypothesized difference between population means.

Analyze Sample Data

Using sample data, find the standard deviation, standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic.

s d = sqrt [ (Σ(d i - d ) 2 / (n - 1) ]

SE = s d * sqrt{ ( 1/n ) * [ (N - n) / ( N - 1 ) ] }

SE = s d / sqrt( n )

• Degrees of freedom. The degrees of freedom (DF) is: DF = n - 1 .

t = [ ( x 1 - x 2 ) - D ] / SE = ( d - D) / SE

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the t statistic, having the degrees of freedom computed above. (See the sample problem at the end of this lesson for guidance on how this is done.)

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

Forty-four sixth graders were randomly selected from a school district. Then, they were divided into 22 matched pairs, each pair having equal IQ's. One member of each pair was randomly selected to receive special training. Then, all of the students were given an IQ test. Test results are summarized below.

Σ(d - d ) 2 = 270

Do these results provide evidence that the special training helped or hurt student performance? Use an 0.05 level of significance. Assume that the mean differences are approximately normally distributed.

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

Null hypothesis: μ d = 0

Alternative hypothesis: μ d ≠ 0

• Formulate an analysis plan . For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

s = sqrt [ (Σ(d i - d ) 2 / (n - 1) ]

s = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586

SE = s / sqrt(n) = 3.586 / [ sqrt(22) ]

SE = 3.586/4.69 = 0.765

DF = n - 1 = 22 -1 = 21

t = [ ( x 1 - x 2 ) - D ] / SE

t = ( d - D)/ SE = (1 - 0)/0.765 = 1.307

where d i is the observed difference for pair i , d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test , the P-value is the probability that the t statistic having 21 degrees of freedom is less than -1.307 or greater than 1.307. We use the t Distribution Calculator to find P(t < -1.307) is about 0.103.

• Then, we enter 1.307 as the sample mean in the t Distribution Calculator. We find the that the P(t < 1.307) is about 0.897. Therefore, P(t > 1.307) is 1 minus 0.897 or 0.103. Thus, the P-value = 0.103 + 0.103 = 0.206.
• Interpret results . Since the P-value (0.206) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples consisted of paired data, and the mean differences were normally distributed. In addition, we used the approximation formula to compute the standard error, since the sample size was small relative to the population size.

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5.6.2 - inference for paired means.

When we developed the inference for the independent samples, we depended on the statistical theory to help us. The theory, however, required the samples to be independent. What can we do when the two samples are not independent, i.e., the data is paired?

Consider an example where we are interested in a person’s weight before implementing a diet plan and after. Since the interest is focusing on the difference, it makes sense to “condense” these two measurements into one and consider the difference between the two measurements. For example, if instead of considering the two measures, we take the before diet weight and subtract the after diet weight. The difference makes sense too! It is the weight lost on the diet.

When we take the two measurements to make one measurement (i.e., the difference), we are now back to the one sample case! Now we can apply all we learned for the one sample mean to the difference (Cool!)

Hypothesis Test for the Difference of Paired Means, \(μ_d\) Section  

In this section, we will develop the hypothesis test for the mean difference for paired samples. As we learned in the previous section, if we consider the difference rather than the two samples, then we are back in the one-sample mean scenario.

The possible null and alternative hypotheses are:

Conditions:

We still need to check the conditions and at least one of the following need to be satisfied:

• The differences of the paired follow a normal distribution
• The sample size is large, \(n>30\).

Test Statistics:

If at least one condition is satisfied then...

\(t^*=\dfrac{\bar{d}-0}{\frac{s_d}{\sqrt{n}}}\)

Will follow a t-distribution with \(n-1\) degrees of freedom.

The same process for the hypothesis test for one mean can be applied. The test for the mean difference may be referred to as the paired t-test or the test for paired means.

Minitab ®

Paired t-test section  .

You can use a paired t-test in Minitab to perform the test. Alternatively, you can perform a 1-sample t-test on difference = before and after diet plan.

• Choose Stat > Basic Statistics > Paired t
• Click Options to specify the confidence level for the interval and the alternative hypothesis you want to test. The default null hypothesis is 0.

Diet Plan Section  

The Minitab output for paired T for before-after diet plan is as follows: 

Answer 

95% lower bound for mean difference: 0.0505 

T-Test of mean difference = 0 (vs > 0): T-Value = 4.86 P-Value = 0.000 

Using the p-value to draw a conclusion about our example:

p-value = \(0.000 < 0.05\)

Reject \(H_0\) and conclude that before diet weight is greater than after diet weight.