null and alternative hypothesis left tailed

Statistics Made Easy

How to Identify a Left Tailed Test vs. a Right Tailed Test

In statistics, we use hypothesis tests to determine whether some claim about a population parameter is true or not.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis , which take the following forms:

H 0 (Null Hypothesis): Population parameter = ≤, ≥ some value

H A (Alternative Hypothesis): Population parameter <, >, ≠ some value

There are three different types of hypothesis tests:

• Two-tailed test: The alternative hypothesis contains the “≠” sign
• Left-tailed test: The alternative hypothesis contains the “<” sign
• Right-tailed test: The alternative hypothesis contains the “>” sign

Notice that we only have to look at the sign in the alternative hypothesis to determine the type of hypothesis test.

Left-tailed test: The alternative hypothesis contains the “<” sign   Right-tailed test: The alternative hypothesis contains the “>” sign

The following examples show how to identify left-tailed and right-tailed tests in practice.

Example: Left-Tailed Test

Suppose it’s assumed that the average weight of a certain widget produced at a factory is 20 grams. However, one inspector believes the true average weight is less than 20 grams.

To test this, he weighs a simple random sample of 20 widgets and obtains the following information:

• n = 20 widgets
• x = 19.8 grams
• s = 3.1 grams

He then performs a hypothesis test using the following null and alternative hypotheses:

H 0 (Null Hypothesis): μ ≥ 20 grams

H A (Alternative Hypothesis): μ < 20 grams

The test statistic is calculated as:

• t  = ( x – µ) / (s/√ n )
• t = (19.8-20) / (3.1/√ 20 )

According to the t-Distribution table , the t critical value at α = .05 and n-1 = 19 degrees of freedom is – 1.729 .

Since the test statistic is not less than this value, the inspector fails to reject the null hypothesis. He does not have sufficient evidence to say that the true mean weight of widgets produced at this factory is less than 20 grams.

Example: Right-Tailed Test

Suppose it’s assumed that the average height of a certain species of plant is 10 inches tall. However, one botanist claims the true average height is greater than 10 inches.

To test this claim, she goes out and measures the height of a simple random sample of 15 plants and obtains the following information:

• n = 15 plants
• x = 11.4 inches
• s = 2.5 inches

She then performs a hypothesis test using the following null and alternative hypotheses:

H 0 (Null Hypothesis): μ ≤ 10 inches

H A (Alternative Hypothesis): μ > 10 inches

• t = (11.4-10) / (2.5/√ 15 )

According to the t-Distribution table , the t critical value at α = .05 and n-1 = 14 degrees of freedom is 1.761 .

Since the test statistic is greater than this value, the botanist can reject the null hypothesis. She has sufficient evidence to say that the true mean height for this species of plant is greater than 10 inches.

Additional Resources

How to Read the t-Distribution Table One Sample t-test Calculator Two Sample t-test Calculator

Featured Posts

Hey there. My name is Zach Bobbitt. I have a Masters of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

2 Replies to “How to Identify a Left Tailed Test vs. a Right Tailed Test”

This was so helpful you are a life saver. Thank you so much

Left-tailed test example, -2.885 is less than -1.729, why you mention -2.885 is “not lesss” …

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Join the Statology Community

Sign up to receive Statology's exclusive study resource: 100 practice problems with step-by-step solutions. Plus, get our latest insights, tutorials, and data analysis tips straight to your inbox!

By subscribing you accept Statology's Privacy Policy.

Hypothesis Testing for Means & Proportions

•   1  
• |   2  
• |   3  
• |   4  
• |   5  
• |   6  
• |   7  
• |   8  
• |   9  
• |   10  

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

Type i and type ii errors.

All Modules

Z score Table

t score Table

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.                  

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.

return to top | previous page | next page

Content ©2017. All Rights Reserved. Date last modified: November 6, 2017. Wayne W. LaMorte, MD, PhD, MPH

How to Identify a Left Tailed Test vs. a Right Tailed Test

In statistics, we use hypothesis tests to determine whether some claim about a population parameter is true or not.

Whenever we perform a hypothesis test, we always write a null hypothesis and an alternative hypothesis , which take the following forms:

H 0 (Null Hypothesis): Population parameter = ≤, ≥ some value

H A (Alternative Hypothesis): Population parameter , ≠ some value

There are three different types of hypothesis tests:

• Two-tailed test: The alternative hypothesis contains the “≠” sign
• Left-tailed test: The alternative hypothesis contains the “
• Right-tailed test: The alternative hypothesis contains the “>” sign

Notice that we only have to look at the sign in the alternative hypothesis to determine the type of hypothesis test.

Left-tailed test: The alternative hypothesis contains the “   Right-tailed test: The alternative hypothesis contains the “>” sign

The following examples show how to identify left-tailed and right-tailed tests in practice.

Example: Left-Tailed Test

Suppose it’s assumed that the average weight of a certain widget produced at a factory is 20 grams. However, one inspector believes the true average weight is less than 20 grams.

To test this, he weighs a simple random sample of 20 widgets and obtains the following information:

• n = 20 widgets
• x = 19.8 grams
• s = 3.1 grams

He then performs a hypothesis test using the following null and alternative hypotheses:

H 0 (Null Hypothesis): μ ≥ 20 grams

H A (Alternative Hypothesis): μ

The test statistic is calculated as:

• t  = ( x – µ) / (s/√ n )
• t = (19.8-20) / (3.1/√ 20 )

According to the t-Distribution table , the t critical value at α = .05 and n-1 = 19 degrees of freedom is – 1.729 .

Since the test statistic is not less than this value, the inspector fails to reject the null hypothesis. He does not have sufficient evidence to say that the true mean weight of widgets produced at this factory is less than 20 grams.

Example: Right-Tailed Test

Suppose it’s assumed that the average height of a certain species of plant is 10 inches tall. However, one botanist claims the true average height is greater than 10 inches.

To test this claim, she goes out and measures the height of a simple random sample of 15 plants and obtains the following information:

• n = 15 plants
• x = 11.4 inches
• s = 2.5 inches

She then performs a hypothesis test using the following null and alternative hypotheses:

H 0 (Null Hypothesis): μ ≤ 10 inches

H A (Alternative Hypothesis): μ > 10 inches

• t = (11.4-10) / (2.5/√ 15 )

According to the t-Distribution table , the t critical value at α = .05 and n-1 = 14 degrees of freedom is 1.761 .

Since the test statistic is greater than this value, the botanist can reject the null hypothesis. She has sufficient evidence to say that the true mean height for this species of plant is greater than 10 inches.

Additional Resources

How to Read the t-Distribution Table One Sample t-test Calculator Two Sample t-test Calculator

Joint Frequency: Definition & Examples

G-test of goodness of fit: definition + example, related posts, how to normalize data between -1 and 1, how to interpret f-values in a two-way anova, how to create a vector of ones in..., vba: how to check if string contains another..., how to determine if a probability distribution is..., what is a symmetric histogram (definition & examples), how to find the mode of a histogram..., how to find quartiles in even and odd..., how to calculate sxy in statistics (with example), how to calculate sxx in statistics (with example).

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.

As an Amazon Associate we earn from qualifying purchases.

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-statistics . Changes were made to the original material, including updates to art, structure, and other content updates.

Access for free at https://openstax.org/books/statistics/pages/1-introduction

• Authors: Barbara Illowsky, Susan Dean
• Publisher/website: OpenStax
• Book title: Statistics
• Publication date: Mar 27, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/statistics/pages/1-introduction
• Section URL: https://openstax.org/books/statistics/pages/9-1-null-and-alternative-hypotheses

© Jan 23, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.

6a.4.1 - Making a Decision

In the previous example for Penn State students, we found that assuming the true population proportion is 0.5, a sample proportion of 0.556 is 2.504 standard deviations above the mean, \(p \).

Is it far enough away from the 0.5 to suggest that there is evidence against the null? Is there a cutoff for the number of standard deviations that we would find acceptable?

What if instead of a cutoff, we found a probability? Recall the alternative hypothesis for this example was \(H_a\colon p>0.5 \). So if we found, for example, the probability of a sample proportion being 0.556 or greater, then we get \( P(Z>2.504)=0.0061 \).

This means that, if the true proportion is 0.5, the probability we would get a sample proportion of 0.556 or greater is 0.0061. Very small! But is it small enough to say there is evidence against the null?

To determine whether the probability is small or how many standard deviations are “acceptable”, we need a preset level of significance, which is the probability of a Type I error. Recall that a Type I error is the event of rejecting the null hypothesis when that null hypothesis is true. Think of finding guilty a person who is actually innocent.

When we specify our hypotheses, we should have some idea of what size of a Type I error we can tolerate. It is denoted as \(\alpha \). A conventional choice of \(\alpha \) is 0.05. Values ranging from 0.01 to 0.1 are also common and the choice of \(\alpha \) depends on the problem one is working on.

Once we have this preset level, we can determine whether or not there is significant evidence against the null. There are two methods to determine if we have enough evidence: the rejection region method and the p-value method .

Rejection Region Approach

We start the hypothesis test process by determining the null and alternative hypotheses. Then we set our significance level, \(\alpha \), which is the probability of making a Type I error. We can determine the appropriate cutoff called the critical value and find a range of values where we should reject, called the rejection region.

The graphs below show us how to find the critical values and the rejection regions for the three different alternative hypotheses and for a set significance level, \(\alpha \). The rejection region is based on the alternative hypothesis.

The rejection region is the region where, if our test statistic falls, then we have enough evidence to reject the null hypothesis. If we consider the right-tailed test, for example, the rejection region is any value greater than \(c_{1-\alpha} \), where \(c_{1-\alpha}\) is the critical value.

Left-Tailed Test

Reject \(H_0\) if the test statistics is less than or equal to the critical value (\(c_\alpha\))

Right-Tailed Test

Reject \(H_0\) if the test statistic is greater than or equal to the critical value (\(c_{1-\alpha}\))

Two-Tailed Test

Reject \(H_0\) if the absolute value of the test statistic is greater than or equal to the absolute value of the critical value (\(c_{\alpha/2}\)).

P-Value Approach

As with the rejection region approach, the P-value approach will need the null and alternative hypotheses, the significance level, and the test statistic. Instead of finding a region, we are going to find a probability called the p-value.

The p-value is a probability statement based on the alternative hypothesis. The p-value is found differently for each of the alternative hypotheses.

• Left-tailed : If \(H_a \) is left-tailed, then the p -value is the probability the sample data produces a value equal to or less than the observed test statistic.
• Right-tailed : If \(H_a \) is right-tailed, then the p -value is the probability the sample data produces a value equal to or greater than the observed test statistic.
• Two-tailed : If \(H_a \) is two-tailed, then the p -value is two times the probability the sample data produces a value equal to or greater than the absolute value of the observed test statistic.

So for one-sample proportions we have...

\(P(Z \le z^*)\)

\(P(Z \ge z^*)\)

\(2\) x \(P(Z \ge |z^*|)\)

Once we find the p-value, we compare the p-value to our preset significance level.

• If our p-value is less than or equal to \(\alpha \) , then there is enough evidence to reject the null hypothesis .
• If our p-value is greater than \(\alpha \) , there is not enough evidence to reject the null hypothesis .

Caution! One should be aware that \(\alpha \) is also called level of significance. This makes for a confusion in terminology. \(\alpha \) is the preset level of significance whereas the p -value is the observed level of significance. The p -value, in fact, is a summary statistic which translates the observed test statistic's value to a probability which is easy to interpret.

Important note: We can summarize the data by reporting the p -value and let the users decide to reject \(H_0 \) or not to reject \(H_0 \) for their subjectively chosen \(\alpha\) values.

This video will further explain the meaning of the p-value.

Video: Understanding the P-Value

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

8.1.5: Additional Information on Hypothesis Tests

• Last updated
• Save as PDF
• Page ID 10979

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\).
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• \(H_{a}\) never has a symbol that contains an equal sign.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller \(p\)-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

The following examples illustrate a left-, right-, and two-tailed test.

Example \(\PageIndex{1}\)

\(H_{0}: \mu = 5, H_{a}: \mu < 5\)

Test of a single population mean. \(H_{a}\) tells you the test is left-tailed. The picture of the \(p\)-value is as follows:

Exercise \(\PageIndex{1}\)

\(H_{0}: \mu = 10, H_{a}: \mu < 10\)

Assume the \(p\)-value is 0.0935. What type of test is this? Draw the picture of the \(p\)-value.

left-tailed test

Example \(\PageIndex{2}\)

\(H_{0}: \mu \leq 0.2, H_{a}: \mu < 0.2\)

This is a test of a single population proportion. \(H_{a}\) tells you the test is right-tailed . The picture of the p -value is as follows:

Exercise \(\PageIndex{2}\)

\(H_{0}: \mu \leq 1, H_{a}: \mu > 1\)

Assume the \(p\)-value is 0.1243. What type of test is this? Draw the picture of the \(p\)-value.

right-tailed test

Example \(\PageIndex{3}\)

\(H_{0}: \mu = 50, H_{a}: \mu \neq 50\)

This is a test of a single population mean. \(H_{a}\) tells you the test is two-tailed . The picture of the \(p\)-value is as follows.

Exercise \(\PageIndex{3}\)

\(H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5\)

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the \(p\)-value.

two-tailed test

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

Exercise 9.6.8

Assume \(H_{0}: \mu = 9\) and \(H_{a}: \mu < 9\). Is this a left-tailed, right-tailed, or two-tailed test?

This is a left-tailed test.

Exercise 9.6.9

Assume \(H_{0}: \mu \leq 6\) and \(H_{a}: \mu > 6\). Is this a left-tailed, right-tailed, or two-tailed test?

Exercise 9.6.10

Assume \(H_{0}: p = 0.25\) and \(H_{a}: p \neq 0.25\). Is this a left-tailed, right-tailed, or two-tailed test?

This is a two-tailed test.

Exercise 9.6.11

Draw the general graph of a left-tailed test.

Exercise 9.6.12

Draw the graph of a two-tailed test.

Exercise 9.6.13

A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?

Exercise 9.6.14

Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?

a right-tailed test

Exercise 9.6.15

A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?

Exercise 9.6.16

You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?

a left-tailed test

Exercise 9.6.17

If the alternative hypothesis has a not equals ( \(\neq\) ) symbol, you know to use which type of test?

Exercise 9.6.18

Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?

Exercise 9.6.19

Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?

Exercise 9.6.20

Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Contributors and Attributions

Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] .

Module 9: Hypothesis Testing With One Sample

Null and alternative hypotheses, learning outcomes.

• Describe hypothesis testing in general and in practice

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: It is a statement about the population that either is believed to be true or is used to put forth an argument unless it can be shown to be incorrect beyond a reasonable doubt.

H a : The alternative hypothesis : It is 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 adecision. 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 (including one of the co-authors in research work) 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.

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

H a : More than 30% 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%. State the null and alternative hypotheses.

H 0 : The drug reduces cholesterol by 25%. p = 0.25

H a : The drug does not reduce cholesterol by 25%. p ≠ 0.25

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:

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

• H 0 : μ = 66
• H a : μ ≠ 66

We want to test if college students take less than five years to graduate from college, on the average. The null and alternative hypotheses are:

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

• H 0 : μ ≥ 45
• H a : μ < 45

In an issue of U.S. News and World Report , an article on school standards stated that about half of all students in France, Germany, and Israel take advanced placement exams and a third pass. The same article stated that 6.6% of U.S. students take advanced placement exams and 4.4% pass. Test if the percentage of U.S. students who take advanced placement exams is more than 6.6%. State the null and alternative hypotheses.

H 0 : p ≤ 0.066

H a : p > 0.066

On a state driver’s test, about 40% pass the test on the first try. We want to test if more than 40% 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

• H 0 : p = 0.40
• H a : p > 0.40

Concept Review

In a  hypothesis test , sample data is evaluated in order to arrive at a decision about some type of claim. If certain conditions about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we: Evaluate the null hypothesis , typically denoted with H 0 . The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤ or ≥) Always write the alternative hypothesis , typically denoted with H a or H 1 , using less than, greater than, or not equals symbols, i.e., (≠, >, or <). If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties.

Formula Review

H 0 and H a are contradictory.

• OpenStax, Statistics, Null and Alternative Hypotheses. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:58/Introductory_Statistics . License : CC BY: Attribution
• Introductory Statistics . Authored by : Barbara Illowski, Susan Dean. Provided by : Open Stax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
• Simple hypothesis testing | Probability and Statistics | Khan Academy. Authored by : Khan Academy. Located at : https://youtu.be/5D1gV37bKXY . License : All Rights Reserved . License Terms : Standard YouTube License

Statistics Tutorial

Descriptive statistics, inferential statistics, stat reference, statistics - hypothesis testing a proportion (left tailed).

A population proportion is the share of a population that belongs to a particular category .

Hypothesis tests are used to check a claim about the size of that population proportion.

Hypothesis Testing a Proportion

The following steps are used for a hypothesis test:

• Check the conditions
• Define the claims
• Decide the significance level
• Calculate the test statistic

For example:

• Population : Nobel Prize winners
• Category : Born in the United States of America

And we want to check the claim:

" Less than 45% of Nobel Prize winners were born in the US"

By taking a sample of 40 randomly selected Nobel Prize winners we could find that:

10 out of 40 Nobel Prize winners in the sample were born in the US

The sample proportion is then: \(\displaystyle \frac{10}{40} = 0.25\), or 25%.

From this sample data we check the claim with the steps below.

1. Checking the Conditions

The conditions for calculating a confidence interval for a proportion are:

• The sample is randomly selected
• There is only two options:
• Being in the category
• Not being in the category
• The sample needs at least:
• 5 members in the category
• 5 members not in the category

In our example, we randomly selected 10 people that were born in the US.

The rest were not born in the US, so there are 30 in the other category.

The conditions are fulfilled in this case.

Note: It is possible to do a hypothesis test without having 5 of each category. But special adjustments need to be made.

2. Defining the Claims

We need to define a null hypothesis (\(H_{0}\)) and an alternative hypothesis (\(H_{1}\)) based on the claim we are checking.

The claim was:

In this case, the parameter is the proportion of Nobel Prize winners born in the US (\(p\)).

The null and alternative hypothesis are then:

Null hypothesis : 45% of Nobel Prize winners were born in the US.

Alternative hypothesis : Less than 45% of Nobel Prize winners were born in the US.

Which can be expressed with symbols as:

\(H_{0}\): \(p = 0.45 \)

\(H_{1}\): \(p < 0.45 \)

This is a ' left tailed' test, because the alternative hypothesis claims that the proportion is less than in the null hypothesis.

If the data supports the alternative hypothesis, we reject the null hypothesis and accept the alternative hypothesis.

3. Deciding the Significance Level

The significance level (\(\alpha\)) is the uncertainty we accept when rejecting the null hypothesis in a hypothesis test.

The significance level is a percentage probability of accidentally making the wrong conclusion.

Typical significance levels are:

• \(\alpha = 0.1\) (10%)
• \(\alpha = 0.05\) (5%)
• \(\alpha = 0.01\) (1%)

A lower significance level means that the evidence in the data needs to be stronger to reject the null hypothesis.

There is no "correct" significance level - it only states the uncertainty of the conclusion.

Note: A 5% significance level means that when we reject a null hypothesis:

We expect to reject a true null hypothesis 5 out of 100 times.

4. Calculating the Test Statistic

The test statistic is used to decide the outcome of the hypothesis test.

The test statistic is a standardized value calculated from the sample.

The formula for the test statistic (TS) of a population proportion is:

\(\displaystyle \frac{\hat{p} - p}{\sqrt{p(1-p)}} \cdot \sqrt{n} \)

\(\hat{p}-p\) is the difference between the sample proportion (\(\hat{p}\)) and the claimed population proportion (\(p\)).

\(n\) is the sample size.

In our example:

The claimed (\(H_{0}\)) population proportion (\(p\)) was \( 0.45 \)

The sample size (\(n\)) was \(40\)

So the test statistic (TS) is then:

\(\displaystyle \frac{0.25-0.45}{\sqrt{0.45(1-0.45)}} \cdot \sqrt{40} = \frac{-0.2}{\sqrt{0.45(0.55)}} \cdot \sqrt{40} = \frac{-0.2}{\sqrt{0.2475}} \cdot \sqrt{40} \approx \frac{-0.2}{0.498} \cdot 6.325 = \underline{-2.543}\)

You can also calculate the test statistic using programming language functions:

With Python use the scipy and math libraries to calculate the test statistic for a proportion.

With R use the built-in math functions to calculate the test statistic for a proportion.

5. Concluding

There are two main approaches for making the conclusion of a hypothesis test:

• The critical value approach compares the test statistic with the critical value of the significance level.
• The P-value approach compares the P-value of the test statistic and with the significance level.

Note: The two approaches are only different in how they present the conclusion.

The Critical Value Approach

For the critical value approach we need to find the critical value (CV) of the significance level (\(\alpha\)).

For a population proportion test, the critical value (CV) is a Z-value from a standard normal distribution .

This critical Z-value (CV) defines the rejection region for the test.

The rejection region is an area of probability in the tails of the standard normal distribution.

Because the claim is that the population proportion is less than 45%, the rejection region is in the left tail:

Choosing a significance level (\(\alpha\)) of 0.01, or 1%, we can find the critical Z-value from a Z-table , or with a programming language function:

With Python use the Scipy Stats library norm.ppf() function find the Z-value for an \(\alpha\) = 0.01 in the left tail.

With R use the built-in qnorm() function to find the Z-value for an \(\alpha\) = 0.01 in the left tail.

Using either method we can find that the critical Z-value is \(\approx \underline{-2.3264}\)

For a left tailed test we need to check if the test statistic (TS) is smaller than the critical value (CV).

If the test statistic is smaller than the critical value, the test statistic is in the rejection region .

When the test statistic is in the rejection region, we reject the null hypothesis (\(H_{0}\)).

Here, the test statistic (TS) was \(\approx \underline{-2.543}\) and the critical value was \(\approx \underline{-2.3264}\)

Here is an illustration of this test in a graph:

Since the test statistic was smaller than the critical value we reject the null hypothesis.

This means that the sample data supports the alternative hypothesis.

And we can summarize the conclusion stating:

The sample data supports the claim that "less than 45% of Nobel Prize winners were born in the US" at a 1% significance level .

The P-Value Approach

For the P-value approach we need to find the P-value of the test statistic (TS).

If the P-value is smaller than the significance level (\(\alpha\)), we reject the null hypothesis (\(H_{0}\)).

The test statistic was found to be \( \approx \underline{-2.543} \)

For a population proportion test, the test statistic is a Z-Value from a standard normal distribution .

Because this is a left tailed test, we need to find the P-value of a Z-value smaller than -2.543.

We can find the P-value using a Z-table , or with a programming language function:

With Python use the Scipy Stats library norm.cdf() function find the P-value of a Z-value smaller than -2.543:

With R use the built-in pnorm() function find the P-value of a Z-value smaller than -2.543:

Using either method we can find that the P-value is \(\approx \underline{0.0055}\)

This tells us that the significance level (\(\alpha\)) would need to be bigger than 0.0055, or 0.55%, to reject the null hypothesis.

This P-value is smaller than any of the common significance levels (10%, 5%, 1%).

So the null hypothesis is rejected at all of these significance levels.

The sample data supports the claim that "less than 45% of Nobel Prize winners were born in the US" at a 10%, 5%, and 1% significance level .

Calculating a P-Value for a Hypothesis Test with Programming

Many programming languages can calculate the P-value to decide outcome of a hypothesis test.

Using software and programming to calculate statistics is more common for bigger sets of data, as calculating manually becomes difficult.

The P-value calculated here will tell us the lowest possible significance level where the null-hypothesis can be rejected.

With Python use the scipy and math libraries to calculate the P-value for a left tailed hypothesis test for a proportion.

Here, the sample size is 40, the occurrences are 10, and the test is for a proportion smaller than 0.45.

With R use the built-in prop.test() function find the P-value for a left tailed hypothesis test for a proportion.

Here, the sample size is 40, the occurrences are 10, and the test is for a proportion bigger than 0.45.

Note: The conf.level in the R code is the reverse of the significance level.

Here, the significance level is 0.01, or 1%, so the conf.level is 1-0.01 = 0.99, or 99%.

Left-Tailed and Two-Tailed Tests

This was an example of a left tailed test, where the alternative hypothesis claimed that parameter is smaller than the null hypothesis claim.

You can check out an equivalent step-by-step guide for other types here:

• Right-Tailed Test
• Two-Tailed Test

COLOR PICKER

Contact Sales

If you want to use W3Schools services as an educational institution, team or enterprise, send us an e-mail: [email protected]

Report Error

If you want to report an error, or if you want to make a suggestion, send us an e-mail: [email protected]

Top Tutorials

Top references, top examples, get certified.

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

AP®︎/College Statistics

Course: ap®︎/college statistics   >   unit 10.

• Idea behind hypothesis testing

Examples of null and alternative hypotheses

• Writing null and alternative hypotheses
• P-values and significance tests
• Comparing P-values to different significance levels
• Estimating a P-value from a simulation
• Estimating P-values from simulations
• Using P-values to make conclusions

Want to join the conversation?

• Upvote Button navigates to signup page
• Downvote Button navigates to signup page
• Flag Button navigates to signup page

Video transcript

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

10.6: Additional Information and Full Hypothesis Test Examples

• Last updated
• Save as PDF
• Page ID 155553

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

• In a hypothesis test problem, you may see words such as "the level of significance is 1%." The "1%" is the preconceived or preset \(\alpha\).
• The statistician setting up the hypothesis test selects the value of α to use before collecting the sample data.
• If no level of significance is given, a common standard to use is \(\alpha = 0.05\).
• When you calculate the \(p\)-value and draw the picture, the \(p\)-value is the area in the left tail, the right tail, or split evenly between the two tails. For this reason, we call the hypothesis test left, right, or two tailed.
• The alternative hypothesis, \(H_{a}\), tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test.
• \(H_{a}\) never has a symbol that contains an equal sign.
• Thinking about the meaning of the \(p\)-value: A data analyst (and anyone else) should have more confidence that he made the correct decision to reject the null hypothesis with a smaller \(p\)-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p -value such as 0.4, as opposed to a \(p\)-value of 0.056 (\(\alpha = 0.05\) is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules.

The following examples illustrate a left-, right-, and two-tailed test.

Example \(\PageIndex{1}\)

\(H_{0}: \mu = 5, H_{a}: \mu < 5\)

Test of a single population mean. \(H_{a}\) tells you the test is left-tailed. The picture of the \(p\)-value is as follows:

Exercise \(\PageIndex{1}\)

\(H_{0}: \mu = 10, H_{a}: \mu < 10\)

Assume the \(p\)-value is 0.0935. What type of test is this? Draw the picture of the \(p\)-value.

left-tailed test

Example \(\PageIndex{2}\)

\(H_{0}: \mu \leq 0.2, H_{a}: \mu > 0.2\)

This is a test of a single population proportion. \(H_{a}\) tells you the test is right-tailed . The picture of the p -value is as follows:

Exercise \(\PageIndex{2}\)

\(H_{0}: \mu \leq 1, H_{a}: \mu > 1\)

Assume the \(p\)-value is 0.1243. What type of test is this? Draw the picture of the \(p\)-value.

right-tailed test

Example \(\PageIndex{3}\)

\(H_{0}: \mu = 50, H_{a}: \mu \neq 50\)

This is a test of a single population mean. \(H_{a}\) tells you the test is two-tailed . The picture of the \(p\)-value is as follows.

Exercise \(\PageIndex{3}\)

\(H_{0}: \mu = 0.5, H_{a}: \mu \neq 0.5\)

Assume the p -value is 0.2564. What type of test is this? Draw the picture of the \(p\)-value.

two-tailed test

Full Hypothesis Test Examples

Example \(\pageindex{4}\).

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

\(H_{0}: \mu = 16.43, H_{a}: \mu < 16.43\)

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "\(<\)" tells you this is left-tailed.

Determine the distribution needed:

Random variable: \(\bar{X} =\) the mean time to swim the 25-yard freestyle.

Distribution for the test: \(\bar{X}\) is normal (population standard deviation is known: \(\sigma = 0.8\))

\(\bar{X} - N \left(\mu, \frac{\sigma_{x}}{\sqrt{n}}\right)\) Therefore, \(\bar{X} - N\left(16.43, \frac{0.8}{\sqrt{15}}\right)\)

\(\mu = 16.43\) comes from \(H_{0}\) and not the data. \(\sigma = 0.8\), and \(n = 15\).

Calculate the \(p-\text{value}\) using the normal distribution for a mean:

\(p\text{-value} = P(\bar{x} < 16) = 0.0187\) where the sample mean in the problem is given as 16.

\(p\text{-value} = 0.0187\) (This is called the actual level of significance .) The \(p-\text{value}\) is the area to the left of the sample mean is given as 16.

\(\mu = 16.43\) comes from \(H_{0}\). Our assumption is \(\mu = 16.43\).

Interpretation of the \(p-\text{value}\): If \(H_{0}\) is true , there is a 0.0187 probability (1.87%) that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare \(\alpha\) and the \(p-\text{value}\):

\(\alpha = 0.05 p\text{-value} = 0.0187 \alpha > p\text{-value}\)

Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\).

This means that you reject \(\mu = 16.43\). In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but faster with the new goggles.

Conclusion: At the 5% significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds.

The p -value can easily be calculated.

Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Stats and press ENTER . Arrow down and enter 16.43 for \(\mu_{0}\) (null hypothesis), .8 for σ , 16 for the sample mean, and 15 for n . Arrow down to \(\mu\) : (alternate hypothesis) and arrow over to \(< \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the p -value (\(p = 0.0187\)) but it also calculates the test statistic ( z -score) for the sample mean. \(\mu < 16.43\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(z = -2.08\) (test statistic) and \(p = 0.0187\) (\(p-\text{value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

When the calculator does a \(Z\)-Test, the Z-Test function finds the p -value by doing a normal probability calculation using the central limit theorem:

\(P(\bar{X} < 16)\) 2nd DISTR normcdf (\((−10^{99},16,16.43,\frac{0.8}{\sqrt{15}})\).

The Type I and Type II errors for this problem are as follows:

The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard free-style, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{4}\)

The mean throwing distance of a football for a Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset \(\alpha = 0.05\). Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Press STAT and arrow over to TESTS. Press 1: \(Z\)-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for \(\mu_{0}\) (null hypothesis), 2 for \(\sigma\), 45 for the sample mean, and 20 for \(n\). Arrow down to \(\mu\): (alternative hypothesis) and set it either as \(<\), \(\neq\), or \(>\). Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p -value but it also calculates the test statistic ( z -score) for the sample mean. Select \(<\), \(\neq\), or \(>\) for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and \(p\)-value. Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Since the problem is about a mean, this is a test of a single population mean.

• \(H_{0}: \mu = 40\)
• \(H_{a}: \mu > 40\)
• \(p = 0.0062\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance.

Historical Note

The traditional way to compare the two probabilities, \(\alpha\) and the \(p-\text{value}\), is to compare the critical value (\(z\)-score from \(\alpha\)) to the test statistic (\(z\)-score from data). The calculated test statistic for the \(p\)-value is –2.08. (From the Central Limit Theorem, the test statistic formula is \(z = \frac{\bar{x}-\mu_{x}}{\left(\frac{\sigma_{x}}{\sqrt{n}}\right)}\). For this problem, \(\bar{x} = 16\), \(\mu_{x} = 16.43\) from the null hypotheses is, \(\sigma_{x} = 0.8\), and \(n = 15\).) You can find the critical value for \(\alpha = 0.05\) in the normal table (see 15.Tables in the Table of Contents). The \(z\)-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The \(z\)-score is –1.645. Since –1.645 > –2.08 (which demonstrates that \(\alpha > p-\text{value}\)), reject \(H_{0}\). Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities \(\alpha\) and the \(p\)-value is very common. For this problem, the \(p-\text{value}\), 0.0187 is considerably smaller than \(\alpha = 0.05\). You can be confident about your decision to reject. The graph shows \(\alpha\), the \(p-\text{value}\), and the test statistics and the critical value.

Example \(\PageIndex{5}\)

A college football coach thought that his players could bench press a mean weight of 275 pounds . It is known that the standard deviation is 55 pounds . Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1).

Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.

Since the problem is about a mean weight, this is a test of a single population mean.

• \(H_{0}: \mu = 275\)
• \(H_{a}: \mu > 275\)

This is a right-tailed test.

Calculating the distribution needed:

Random variable: \(\bar{X} =\) the mean weight, in pounds, lifted by the football players.

Distribution for the test: It is normal because \(\sigma\) is known.

• \(\bar{X} - N\left(275, \frac{55}{\sqrt{30}}\right)\)
• \(\bar{x} = 286.2\) pounds (from the data).
• \(\sigma = 55\) pounds (Always use \(\sigma\) if you know it.) We assume \(\mu = 275\) pounds unless our data shows us otherwise.

Calculate the p -value using the normal distribution for a mean and using the sample mean as input (see [link] for using the data as input):

\[p\text{-value} = P(\bar{x} > 286.2) = 0.1323.\nonumber \]

Interpretation of the p -value: If \(H_{0}\) is true, then there is a 0.1331 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.

\(\alpha = 0.025 p-value = 0.1323\)

Make a decision: Since \(\alpha < p\text{-value}\), do not reject \(H_{0}\).

Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.

The \(p-\text{value}\) can easily be calculated.

Put the data and frequencies into lists. Press STAT and arrow over to TESTS . Press 1:Z-Test . Arrow over to Data and press ENTER . Arrow down and enter 275 for \(\mu_{0}\), 55 for \(\sigma\), the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to \(\mu\) : and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p-\text{value}\) (\(p = 0.1331\)), a little different from the previous calculation - in it we used the sample mean rounded to one decimal place instead of the data) but it also calculates the test statistic ( z -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 275\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(z = 1.112\) (test statistic) and \(p = 0.1331\) (\(p-\text{value})\). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Example \(\PageIndex{6}\)

Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples ten statistics students and obtains the scores 65 65 70 67 66 63 63 68 72 71. He performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.

Set up the hypothesis test:

A 5% level of significance means that \(\alpha = 0.05\). This is a test of a single population mean .

\(H_{0}: \mu = 65  H_{a}: \mu > 65\)

Since the instructor thinks the average score is higher, use a "\(>\)". The "\(>\)" means the test is right-tailed.

Random variable: \(\bar{X} =\) average score on the first statistics test.

Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given . You are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a student's \(t\).

Use \(t_{df}\). Therefore, the distribution for the test is \(t_{9}\) where \(n = 10\) and \(df = 10 - 1 = 9\).

Calculate the \(p\)-value using the Student's \(t\)-distribution:

\(p\text{-value} = P(\bar{x} > 67) = 0.0396\) where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data.

Interpretation of the p -value: If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 65 or more.

Since \(α = 0.05\) and \(p\text{-value} = 0.0396\). \(\alpha > p\text{-value}\).

This means you reject \(\mu = 65\). In other words, you believe the average test score is more than 65.

Conclusion: At a 5% level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks.

The \(p\text{-value}\) can easily be calculated.

Put the data into a list. Press STAT and arrow over to TESTS . Press 2:T-Test . Arrow over to Data and press ENTER . Arrow down and enter 65 for \(\mu_{0}\), the name of the list where you put the data, and 1 for Freq: . Arrow down to \(\mu\): and arrow over to \(> \mu_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator not only calculates the \(p\text{-value}\) (p = 0.0396) but it also calculates the test statistic ( t -score) for the sample mean, the sample mean, and the sample standard deviation. \(\mu > 65\) is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(t = 1.9781\) (test statistic) and \(p = 0.0396\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

Exercise \(\PageIndex{6}\)

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: \mu = 5\)
• \(H_{a}: \mu < 5\)
• \(p = 0.0082\)

Because \(p < \alpha\), we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week.

• Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false).

Example \(\PageIndex{7}\)

Joon believes that 50% of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50% . Joon samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1% level of significance.

The 1% level of significance means that α = 0.01. This is a test of a single population proportion .

\(H_{0}: p = 0.50\)  \(H_{a}: p \neq 0.50\)

The words "is the same or different from" tell you this is a two-tailed test.

Calculate the distribution needed:

Random variable: \(P′ =\) the percent of of first-time brides who are younger than their grooms.

Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′ , the estimated proportion.

\[P' - N\left(p, \sqrt{\frac{p-q}{n}}\right)\nonumber \]

\[P' - N\left(0.5, \sqrt{\frac{0.5-0.5}{100}}\right)\nonumber \]

where \(p = 0.50, q = 1−p = 0.50\), and \(n = 100\)

Calculate the p -value using the normal distribution for proportions:

\[p\text{-value} = P(p′ < 0.47 \space or \space p′ > 0.53) = 0.5485\nonumber \]

where \[x = 53, p' = \frac{x}{n} = \frac{53}{100} = 0.53\nonumber \].

Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability (54.85%) that the sample (estimated) proportion \(p'\) is 0.53 or more OR 0.47 or less (see the graph in Figure).

\(\mu = p = 0.50\) comes from \(H_{0}\), the null hypothesis.

\(p′ = 0.53\). Since the curve is symmetrical and the test is two-tailed, the \(p′\) for the left tail is equal to \(0.50 – 0.03 = 0.47\) where \(\mu = p = 0.50\). (0.03 is the difference between 0.53 and 0.50.)

Compare \(\alpha\) and the \(p\text{-value}\):

Since \(\alpha = 0.01\) and \(p\text{-value} = 0.5485\). \(\alpha < p\text{-value}\).

Make a decision: Since \(\alpha < p\text{-value}\), you cannot reject \(H_{0}\).

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.

Press STAT and arrow over to TESTS . Press 5:1-PropZTest . Enter .5 for \(p_{0}\), 53 for \(x\) and 100 for \(n\). Arrow down to Prop and arrow to not equals \(p_{0}\). Press ENTER . Arrow down to Calculate and press ENTER . The calculator calculates the \(p\text{-value}\) (\(p = 0.5485\)) and the test statistic (\(z\)-score). Prop not equals .5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate ). Press ENTER . A shaded graph appears with \(z = 0.6\) (test statistic) and \(p = 0.5485\) (\(p\text{-value}\)). Make sure when you use Draw that no other equations are highlighted in \(Y =\) and the plots are turned off.

The Type I and Type II errors are as follows:

The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true).

The Type II error is there is not enough evidence to conclude that the proportion of first time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. (Do not reject the null hypothesis when the null hypothesis is false.)

Exercise \(\PageIndex{7}\)

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

First, determine what type of test this is, set up the hypothesis test, find the \(p\text{-value}\), sketch the graph, and state your conclusion.

Since the problem is about percentages, this is a test of single population proportions.

• \(H_{0} : p = 0.85\)
• \(H_{a}: p \neq 0.85\)
• \(p = 0.7554\)

Because \(p > \alpha\), we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%.

Example \(\PageIndex{8}\)

Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.

\(H_{0}: p = 0.30, H_{a}: p \neq 0.30\)

The random variable is \(P′ =\) proportion of households that have three cell phones.

The distribution for the hypothesis test is \(P' - N\left(0.30, \sqrt{\frac{(0.30 \cdot 0.70)}{150}}\right)\)

Exercise \(\PageIndex{8}\).2

a. The value that helps determine the \(p\text{-value}\) is \(p′\). Calculate \(p′\).

a. \(p' = \frac{x}{n}\) where \(x\) is the number of successes and \(n\) is the total number in the sample.

\(x = 43, n = 150\)

\(p′ = 43150\)

Exercise \(\PageIndex{8}\).3

b. What is a success for this problem?

b. A success is having three cell phones in a household.

Exercise \(\PageIndex{8}\).4

c. What is the level of significance?

c. The level of significance is the preset \(\alpha\). Since \(\alpha\) is not given, assume that \(\alpha = 0.05\).

Exercise \(\PageIndex{8}\).5

d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately.

Calculate the \(p\text{-value}\).

d. \(p\text{-value} = 0.7216\)

Exercise \(\PageIndex{8}\).6

e. Make a decision. _____________(Reject/Do not reject) \(H_{0}\) because____________.

e. Assuming that \(\alpha = 0.05, \alpha < p\text{-value}\). The decision is do not reject \(H_{0}\) because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.

Exercise \(\PageIndex{8}\)

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

• \(H_{0}: p = 0.92\)
• \(H_{a}: p < 0.92\)
• \(p\text{-value} = 0.0046\)

Because \(p < 0.05\), we reject the null hypothesis. There is sufficient evidence to conclude that fewer than 92% of American adults own cell phones.

• Type I Error: To conclude that fewer than 92% of American adults own cell phones when, in fact, 92% of American adults do own cell phones (reject the null hypothesis when the null hypothesis is true).
• Type II Error: To conclude that 92% of American adults own cell phones when, in fact, fewer than 92% of American adults own cell phones (do not reject the null hypothesis when the null hypothesis is false).

The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter \(p\). The distribution for the test is normal. The estimated proportion \(p′\) is the proportion of fleas killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived \(\alpha = 0.01\), for comparison, and a 95% confidence interval computation. The poem is clever and humorous, so please enjoy it!

Example \(\PageIndex{9}\)

My dog has so many fleas,

They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25% of the fleas, Unfortunately I was not pleased.

I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe.

A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas.

I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42.

At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased.

Now it is time for you to have some fun With the level of significance being .01, You must help me figure out

Use the new shampoo or go without?

\(H_{0}: p \leq 0.25\)   \(H_{a}: p > 0.25\)

In words, CLEARLY state what your random variable \(\bar{X}\) or \(P′\) represents.

\(P′ =\) The proportion of fleas that are killed by the new shampoo

State the distribution to use for the test.

\[N\left(0.25, \sqrt{\frac{(0.25){1-0.25}}{42}}\right)\nonumber \]

Test Statistic: \(z = 2.3163\)

Calculate the \(p\text{-value}\) using the normal distribution for proportions:

\[p\text{-value} = 0.0103\nonumber \]

In one to two complete sentences, explain what the p -value means for this problem.

If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 \(\left(\frac{17}{42}\right)\) or more.

Use the previous information to sketch a picture of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) corresponding to the \(p\text{-value}\).

Indicate the correct decision (“reject” or “do not reject” the null hypothesis), the reason for it, and write an appropriate conclusion, using complete sentences.

Conclusion: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25%.

Construct a 95% confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval.

Confidence Interval: (0.26,0.55) We are 95% confident that the true population proportion p of fleas that are killed by the new shampoo is between 26% and 55%.

This test result is not very definitive since the \(p\text{-value}\) is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return.

Example \(\PageIndex{10}\)

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98 1.02; .95; .95

Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.

Let’s follow a four-step process to answer this statistical question.

• \(H_{0}: \mu \leq 1\)
• \(H_{a}: \mu > 1\)
• Plan : We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations : We will input the sample data into the TI-83 as follows.

4. State the Conclusions : Since the \(p\text{-value} (p = 0.036)\) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Example \(\PageIndex{11}\)

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

We will follow the four-step process.

• \(H_{0}: p \leq 0.00034\)
• \(H_{a}: p > 0.00034\)

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with \(x = 172\) and \(n = 420,019\). The sample is sufficiently large because we have \(np = 420,019(0.00034) = 142.8\), \(nq = 420,019(0.99966) = 419,876.2\), two independent outcomes, and a fixed probability of success \(p = 0.00034\). Thus we will be able to generalize our results to the population.

Figure \(\PageIndex{11}\).

Figure \(\PageIndex{12}\).

• Since the \(p\text{-value} = 0.0073\) is greater than our alpha value \(= 0.005\), we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.

Example \(\PageIndex{12}\)

According to the US Census there are approximately 268,608,618 residents aged 12 and older. Statistics from the Rape, Abuse, and Incest National Network indicate that, on average, 207,754 rapes occur each year (male and female) for persons aged 12 and older. This translates into a percentage of sexual assaults of 0.078%. In Daviess County, KY, there were reported 11 rapes for a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local sexual assault percentage and the national sexual assault percentage. Use a significance level of 0.01.

We will follow the four-step plan.

• We need to test whether the proportion of sexual assaults in Daviess County, KY is significantly different from the national average.
• \(H_{0}: p = 0.00078\)
• \(H_{a}: p \neq 0.00078\)

Figure \(\PageIndex{13}\).

Figure \(\PageIndex{14}\).

• Since the \(p\text{-value}\), \(p = 0.00063\), is less than the alpha level of 0.01, the sample data indicates that we should reject the null hypothesis. In conclusion, the sample data support the claim that the proportion of sexual assaults in Daviess County, Kentucky is different from the national average proportion.

The hypothesis test itself has an established process. This can be summarized as follows:

• Determine \(H_{0}\) and \(H_{a}\). Remember, they are contradictory.
• Determine the random variable.
• Determine the distribution for the test.
• Draw a graph, calculate the test statistic, and use the test statistic to calculate the \(p\text{-value}\). (A z -score and a t -score are examples of test statistics.)
• Compare the preconceived α with the p -value, make a decision (reject or do not reject H 0 ), and write a clear conclusion using English sentences.

Notice that in performing the hypothesis test, you use \(\alpha\) and not \(\beta\). \(\beta\) is needed to help determine the sample size of the data that is used in calculating the \(p\text{-value}\). Remember that the quantity \(1 – \beta\) is called the Power of the Test . A high power is desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same.If the power is low, the null hypothesis might not be rejected when it should be.

Assume \(H_{0}: \mu = 9\) and \(H_{a}: \mu < 9\). Is this a left-tailed, right-tailed, or two-tailed test?

This is a left-tailed test.

Exercise \(\PageIndex{9}\)

Assume \(H_{0}: \mu \leq 6\) and \(H_{a}: \mu > 6\). Is this a left-tailed, right-tailed, or two-tailed test?

Exercise \(\PageIndex{10}\)

Assume \(H_{0}: p = 0.25\) and \(H_{a}: p \neq 0.25\). Is this a left-tailed, right-tailed, or two-tailed test?

This is a two-tailed test.

Exercise \(\PageIndex{11}\)

Draw the general graph of a left-tailed test.

Exercise \(\PageIndex{12}\)

Draw the graph of a two-tailed test.

Exercise \(\PageIndex{13}\)

A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?

Exercise \(\PageIndex{14}\)

Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?

a right-tailed test

Exercise \(\PageIndex{15}\)

A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?

Exercise \(\PageIndex{16}\)

You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?

a left-tailed test

Exercise \(\PageIndex{17}\)

If the alternative hypothesis has a not equals ( \(\neq\) ) symbol, you know to use which type of test?

Exercise \(\PageIndex{18}\)

Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?

Exercise \(\PageIndex{19}\)

Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?

Exercise \(\PageIndex{20}\)

Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?

• Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC.
• Data from Bloomberg Businessweek . Available online at www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html.
• Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013).
• Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013).
• Data from Growing by Degrees by Allen and Seaman.
• Data from La Leche League International. Available online at www.lalecheleague.org/Law/BAFeb01.html.
• Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013).
• Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013).
• Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm .
• Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013)
• Data from the U.S. Census Bureau, available online at quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013).
• Data from the United States Census Bureau. Available online at www.census.gov/hhes/socdemo/language/.
• Data from Toastmasters International. Available online at http://toastmasters.org/artisan/deta...eID=429&Page=1 .
• Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013).
• Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013).
• “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at research.fhda.edu/factbook/DA...t_da_2006w.pdf.
• Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013).
• Rape, Abuse & Incest National Network. “How often does sexual assault occur?” RAINN, 2009. Available online at www.rainn.org/get-information...sexual-assault (accessed June 27, 2013).

Six Sigma Study Guide

Study notes and guides for Six Sigma certification tests

Tailed Hypothesis Tests

Posted by Ted Hessing

A tailed hypothesis tests is an assumption about a population parameter. The assumption may or may not be true.  Hypothesis testing is a key procedure in inferential statistics used to make statistical decisions using experimental data.  It is basically an assumption that we make about the population parameter.

Null and alternative hypothesis

Null hypothesis (H 0 ): A statistical hypothesis assumes that the observation is due to the chance factor. In other words, the null hypothesis states that there is no (statistical significance) difference or effect.

Alternative hypothesis (H 1 ): The complementary hypothesis of the null hypothesis is an alternative hypothesis. In other words, the alternative hypothesis shows that observations are the results of a real effect.

What are the tails in a hypothesis test?

A tail in hypothesis testing refers to the tail at either end of a distribution curve. Generally, in hypothesis tests, test statistic means to obtain all of the sample data and convert it to a single value. For example, Z-test calculates Z statistics, t-test calculates t-test statistic, and F-test calculates F values etc., are the test statistics. Test statistics need to compare to an appropriate critical value. A decision can then be made to reject or not reject the null hypothesis.

In probability distribution plots, the shaded area in the plot (one side in one-tailed hypothesis and two sides in a two-tailed hypothesis) indicates the probability of a value falls within that range.

Critical region: In a hypothesis test, critical regions are ranges of the distributions where the values represent statistically significant results. If the test statistic falls in the critical region, reject the null hypothesis.

Types of tailed hypothesis tests

There are three basic types of ‘tails’ that hypothesis tests can have:

• Right-tailed test: where the alternative hypothesis includes a ‘>’ symbol.
• Left-tailed test: where the alternative hypothesis includes a ‘<’ symbol.
• Two-tailed test: where the alternative hypothesis includes a ≠.

One-tailed hypothesis tests

A test of hypothesis where the area of rejection is only in one direction. In other words, when change is expected to have occurred in one direction, i.e expecting output either increase or to decrease.

If the level of significance is 0.05, a one-tail test allots the entire alpha (α) in the one direction to test the statistical significance. Since the statistical significance in the one direction of interest, it is also known as a directional hypothesis.

Reject the null hypothesis; If the test statistic falls in the critical region, that means the test statistic has a greater value than the critical value (for the right-tailed test) and the test statistic has a lesser value than the critical value (for the left tailed test).

Generally, one-tailed tests are more powerful than two-tailed tests; because of that, one-tailed tests are preferred.

The basic disadvantage of a one-tailed test is it considers effects in one direction only. There is a chance that an important effect may miss in another direction. For example, a new material used in the production and checking whether the yield improved over the existing material. There is a possibility that new material may give less yield than the current material.

One tailed tests are further divided into

Right-tailed test

Left-tailed test.

Right tailed test is also called the upper tail test. A hypothesis test is performed if the population parameter is suspected to be greater than the assumed parameter of the null hypothesis.

• H 0 : The sampling mean (x̅) is less than are equal to µ
• H 1 : The sampling mean (x̅) is greater than µ.

Example: The average weight of an iron bar population is 90lbs. Supervisor believes that the average weight might be higher. Random samples of 5 iron bars are measured, and the average weight is 110lbs and a standard deviation of 18lbs. With a 95% confidence level, is there enough evidence to suggest the average weight is higher?

Population average score (µ) = 90

• Sample average (x̅) = 110
• number of samples (n) = 5
• Level of significance α=0.05

H 0 : The average weight is equal to 90, µ=90.

H 1 : The average score is higher than 90, µ>90

Since supervisor is keen to check the average weight is higher, it is a right-tailed test.

Compute the critical value: For 95% confidence level t value with a degrees of freedom n-1= 2.132

Critical value =2.132

Calculate the test statistics t = x̅-µ/(s/√n) = 110-90/(18/√5)=2.484.

Conclusion: Test statistic is greater than the critical value, and it is in the rejection region. Hence, we can reject the null hypothesis. So the average weight of the iron bar is may be higher than the 90lbs.

Left-tailed test is also known as a lower tail test. A hypothesis test is performed if the population parameter is suspected to be less than the assumed parameter of the null hypothesis.

• H 0 : The sampling mean (x̅) is greater than are equal to µ
• H 1 : The sampling mean (x̅) is less than µ.

Example: The average weight of an iron bar population is 90lbs. Supervisor believes that the average weight might be lower. Random samples of 6 iron bars are measured, and the average weight is 82lbs and a standard deviation of 18lbs. With a 95% confidence level, is there enough evidence to suggest the average weight is lower?

• Sample average (x̅) = 82
• number of samples (n) = 6

H 1 : The average score is less than 90, µ<90

Since the supervisor is keen to check the average weight is lower, hence it is a left-tailed test.

Compute the critical value: For 95% confidence level t value with a degrees of freedom n-1= -2.015

Critical value =-2.015

Calculate the test statistics t = x̅-µ/(s/√n) = 82-90/(18/√6)=-1.088

Conclusion: Test statistic is not in the rejection region. Hence, we failed to reject the null hypothesis. So the average weight of the iron bar is 90lbs.

Two-tailed hypothesis tests

A test of hypothesis where the area of rejection is on both sides of the sampling distribution.

If level of significance is 0.05 of a two-tailed test, it distributes the alpha (α) into two equal parts (α/2 & α/2) on both sides to test the statistical significance. 

• H 0 : The sampling mean (x̅) is equal to µ
• H 1 : The sampling mean (x̅) is not equal to µ

Two-tailed tests also known as two-sided or non-directional test, as it tests the effects on both sides. In a two-tailed test, extreme values above or below are evidence against the null hypothesis.

Reject the null hypothesis if test statistics fall on either side of the critical region.

Example: The average score for the mean population is 80, with a standard deviation of 10. With a new training method, the professor believes that the score might change. Professor tested randomly 36 students’ scores. The average score of the sample is 88. With a 95% confidence level, is there enough evidence to suggest the average score changed?

Population average score (µ) = 80

• Sample average (x̅) = 88
• number of samples (n) = 36

H 0 : The average score is equal to 80, µ=80.

H 1 : The average score is not equal to 80, µ≠80

Since the professor is keen to check the change in average score, it is a two tail test

Compute the critical value: For 95% confidence level Z value = 1.96

Critical value =±1.96

Calculate the test statistics Z = x̅-µ/(σ/√n) = 88-80/(10/√36)=4.8

Conclusion: The test statistic is greater than the critical value, which means the test statistic is in the rejection region. So, we can reject the null hypothesis.

Few more Example

For Example, consider a null hypothesis that states that cars traveling on a particular road have a mean velocity of 40 miles/hour:

• A right-tailed test would state that cars traveling on a particular road have a mean velocity greater than 40 miles/hour.
• A left-tailed test would state that cars traveling on a particular road have a mean velocity less than 40 miles/hour.
• A two-tailed test would state that cars traveling on a particular road have a mean velocity greater than or less than 40 miles/hour.

Additional Resources:

• https://www.statisticshowto.datasciencecentral.com/how-to-decide-if-a-hypothesis-test-is-a-left-tailed-test-or-a-right-tailed-test/  

Important Videos

I originally created SixSigmaStudyGuide.com to help me prepare for my own Black belt exams. Overtime I've grown the site to help tens of thousands of Six Sigma belt candidates prepare for their Green Belt & Black Belt exams. Go here to learn how to pass your Six Sigma exam the 1st time through!

Comments (8)

I believe you have the right tail and left tails mixed up in this article. It conflicts with other articles you have written and also with my separate findings. In general, the right tailed test is when the alternative hypothesis is > null hypothesis. Putting you on the right side of the bell curve, hence the “right tailed” name.

You’re absolutely correct, Montgomery. The article has been updated. Thanks for bringing this to my attention!

Good Day Ted,

Just brushing up a little while on my break at work. Has the following below been updated yet? When looking at an actual graph it’s otherwise.

-When performing a right-tailed test, we reject the null hypothesis if the test statistics are less than the critical value.

-When performing a left-tailed test, we reject the null hypothesis if the test statistics are greater than the critical value.

We should be all set now with our current re-write, Lemarcus.

Thank you for the note!

What is the difference btw z statistic and t statistic ?

I have seen on your content that, for a right tailed test

For Z statistic, reject null hypothesis if p value is less than critical value

For t statistic, reject null hypothesis if t value is more than critical value

Also, in the two-tailed test, you chose Z instead of t statistic. Why ?

Hello Rahul,

A t-test is used to compare the mean of two given samples. Like a z-test, a t-test also assumes a normal distribution of the sample. A t-test is used when the population parameters (mean and standard deviation) are not known.

Since we know the population mean and standard deviation, selected the z test instead of t test

Your explanations are better and easier to follow than those of my Statistics profs… Thanks

Thanks, Ben. That’s high praise!

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed .

Insert/edit link

Enter the destination URL

Or link to existing content

Statistics Examples

• Privacy Policy--New
• Terms ( Premium )
• DO NOT SELL MY INFO
• Mathway © 2024

Please ensure that your password is at least 8 characters and contains each of the following:

• a special character: @$#!%*?&