99% Upper | |||||
N | Mean | SE Mean | Bound | Z | P |
227 | 19.600 | 0.485 | 20.727 | -7.02 | 0.000 |
Excel does not offer 1-sample hypothesis testing.
Frequently, the population standard deviation (σ) is not known. We can estimate the population standard deviation (σ) with the sample standard deviation (s). However, the test statistic will no longer follow the standard normal distribution. We must rely on the student’s t-distribution with n-1 degrees of freedom. Because we use the sample standard deviation (s), the test statistic will change from a Z-score to a t-score.
Steps for a hypothesis test are the same that we covered in Section 2.
Just as with the hypothesis test from the previous section, the data for this test must be from a random sample and requires either that the population from which the sample was drawn be normal or that the sample size is sufficiently large (n≥30). A t-test is robust, so small departures from normality will not adversely affect the results of the test. That being said, if the sample size is smaller than 30, it is always good to verify the assumption of normality through a normal probability plot.
We will still have the same three pairs of null and alternative hypotheses and we can still use either the classical approach or the p-value approach.
Selecting the correct critical value from the student’s t-distribution table depends on three factors: the type of test (one-sided or two-sided alternative hypothesis), the sample size, and the level of significance.
For a two-sided test (“not equal” alternative hypothesis), the critical value (t α /2 ), is determined by alpha ( α ), the level of significance, divided by two, to deal with the possibility that the result could be less than OR greater than the known value.
For a one-sided test (“a less than” or “greater than” alternative hypothesis), the critical value (t α ) , is determined by alpha ( α ), the level of significance, being all in the one side.
Find the critical value you would use to test the claim that μ ≠ 112 with a sample size of 18 and a 5% level of significance.
In this case, the critical value (t α /2 ) would be 2.110. This is a two-sided question (≠) so you would divide alpha by 2 (0.05/2 = 0.025) and go down the 0.025 column to 17 degrees of freedom.
What would the critical value be if you wanted to test that μ < 112 for the same data?
In this case, the critical value would be 1.740. This is a one-sided question (<) so alpha would be divided by 1 (0.05/1 = 0.05). You would go down the 0.05 column with 17 degrees of freedom to get the correct critical value.
In 2005, the mean pH level of rain in a county in northern New York was 5.41. A biologist believes that the rain acidity has changed. He takes a random sample of 11 rain dates in 2010 and obtains the following data. Use a 1% level of significance to test his claim.
4.70, 5.63, 5.02, 5.78, 4.99, 5.91, 5.76, 5.54, 5.25, 5.18, 5.01
The sample size is small and we don’t know anything about the distribution of the population, so we examine a normal probability plot. The distribution looks normal so we will continue with our test.
Figure 14. A normal probability plot for Example 9.
The sample mean is 5.343 with a sample standard deviation of 0.397.
Figure 15. The rejection zones for a two-sided test.
Figure 16. The critical values for a two-sided test when α = 0.01.
We will fail to reject the null hypothesis. We do not have enough evidence to support the claim that the mean rain pH has changed.
Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The government has set safety limits for cadmium in dry vegetables at 0.5 ppm. Biologists believe that the mean level of cadmium in mushrooms growing near strip mines is greater than the recommended limit of 0.5 ppm, negatively impacting the animals that live in this ecosystem. A random sample of 51 mushrooms gave a sample mean of 0.59 ppm with a sample standard deviation of 0.29 ppm. Use a 5% level of significance to test the claim that the mean cadmium level is greater than the acceptable limit of 0.5 ppm.
The sample size is greater than 30 so we are assured of a normal distribution of the means.
Figure 17. Rejection zone for a right-sided test.
Step 4) State a Conclusion.
Figure 18. Critical value for a right-sided test when α = 0.05.
The test statistic falls in the rejection zone. We will reject the null hypothesis. We have enough evidence to support the claim that the mean cadmium level is greater than the acceptable safe limit.
BUT, what happens if the significance level changes to 1%?
The critical value is now found by going down the 0.01 column with 50 degrees of freedom. The critical value is 2.403. The test statistic is now LESS THAN the critical value. The test statistic does not fall in the rejection zone. The conclusion will change. We do NOT have enough evidence to support the claim that the mean cadmium level is greater than the acceptable safe limit of 0.5 ppm.
The level of significance is the probability that you, as the researcher, set to decide if there is enough statistical evidence to support the alternative claim. It should be set before the experiment begins.
We can also use the p-value approach for a hypothesis test about the mean when the population standard deviation ( σ ) is unknown. However, when using a student’s t-table, we can only estimate the range of the p-value, not a specific value as when using the standard normal table. The student’s t-table has area (probability) across the top row in the table, with t-scores in the body of the table.
Estimating P-value from a Student’s T-table
If your test statistic is 3.789 with 3 degrees of freedom, you would go across the 3 df row. The value 3.789 falls between the values 3.482 and 4.541 in that row. Therefore, the p-value is between 0.02 and 0.01. The p-value will be greater than 0.01 but less than 0.02 (0.01<p<0.02).
If your level of significance is 5%, you would reject the null hypothesis as the p-value (0.01-0.02) is less than alpha ( α ) of 0.05.
If your level of significance is 1%, you would fail to reject the null hypothesis as the p-value (0.01-0.02) is greater than alpha ( α ) of 0.01.
Software packages typically output p-values. It is easy to use the Decision Rule to answer your research question by the p-value method.
(referring to Ex. 12)
Test of mu = 0.5 vs. > 0.5
95% Lower | ||||||
N | Mean | StDev | SE Mean | Bound | T | P |
51 | 0.5900 | 0.2900 | 0.0406 | 0.5219 | 2.22 | 0.016 |
Additional example: www.youtube.com/watch?v=WwdSjO4VUsg .
Frequently, the parameter we are testing is the population proportion.
Recall that the best point estimate of p , the population proportion, is given by
when np (1 – p )≥10. We can use both the classical approach and the p-value approach for testing.
The steps for a hypothesis test are the same that we covered in Section 2.
The test statistic follows the standard normal distribution. Notice that the standard error (the denominator) uses p instead of p̂ , which was used when constructing a confidence interval about the population proportion. In a hypothesis test, the null hypothesis is assumed to be true, so the known proportion is used.
A botanist has produced a new variety of hybrid soy plant that is better able to withstand drought than other varieties. The botanist knows the seed germination for the parent plants is 75%, but does not know the seed germination for the new hybrid. He tests the claim that it is different from the parent plants. To test this claim, 450 seeds from the hybrid plant are tested and 321 have germinated. Use a 5% level of significance to test this claim that the germination rate is different from 75%.
This is a two-sided question so alpha is divided by 2.
Figure 19. Critical values for a two-sided test when α = 0.05.
The test statistic does not fall in the rejection zone. We fail to reject the null hypothesis. We do not have enough evidence to support the claim that the germination rate of the hybrid plant is different from the parent plants.
Let’s answer this question using the p-value approach. Remember, for a two-sided alternative hypothesis (“not equal”), the p-value is two times the area of the test statistic. The test statistic is -1.81 and we want to find the area to the left of -1.81 from the standard normal table.
Now compare the p-value to alpha. The Decision Rule states that if the p-value is less than alpha, reject the H 0 . In this case, the p-value (0.0702) is greater than alpha (0.05) so we will fail to reject H 0 . We do not have enough evidence to support the claim that the germination rate of the hybrid plant is different from the parent plants.
You are a biologist studying the wildlife habitat in the Monongahela National Forest. Cavities in older trees provide excellent habitat for a variety of birds and small mammals. A study five years ago stated that 32% of the trees in this forest had suitable cavities for this type of wildlife. You believe that the proportion of cavity trees has increased. You sample 196 trees and find that 79 trees have cavities. Does this evidence support your claim that there has been an increase in the proportion of cavity trees?
Use a 10% level of significance to test this claim.
This is a one-sided question so alpha is divided by 1.
Figure 20. Critical value for a right-sided test where α = 0.10.
Figure 21. Comparison of the test statistic and the critical value.
The test statistic is larger than the critical value (it falls in the rejection zone). We will reject the null hypothesis. We have enough evidence to support the claim that there has been an increase in the proportion of cavity trees.
Now use the p-value approach to answer the question. This is a right-sided question (“greater than”), so the p-value is equal to the area to the right of the test statistic. Go to the positive side of the standard normal table and find the area associated with the Z-score of 2.49. The area is 0.9936. Remember that this table is cumulative from the left. To find the area to the right of 2.49, we subtract from one.
p-value = (1 – 0.9936) = 0.0064
The p-value is less than the level of significance (0.10), so we reject the null hypothesis. We have enough evidence to support the claim that the proportion of cavity trees has increased.
(referring to Ex. 15)
Test of p = 0.32 vs. p > 0.32
90% Lower | ||||||
Sample | X | N | Sample p | Bound | Z-Value | p-Value |
1 | 79 | 196 | 0.403061 | 0.358160 | 2.49 | 0.006 |
Using the normal approximation. |
When people think of statistical inference, they usually think of inferences involving population means or proportions. However, the particular population parameter needed to answer an experimenter’s practical questions varies from one situation to another, and sometimes a population’s variability is more important than its mean. Thus, product quality is often defined in terms of low variability.
Sample variance S 2 can be used for inferences concerning a population variance σ 2 . For a random sample of n measurements drawn from a normal population with mean μ and variance σ 2 , the value S 2 provides a point estimate for σ 2 . In addition, the quantity ( n – 1) S 2 / σ 2 follows a Chi-square ( χ 2 ) distribution, with df = n – 1.
The properties of Chi-square ( χ 2 ) distribution are:
Figure 22. The chi-square distribution.
Alternative hypothesis:
where the χ 2 critical value in the rejection region is based on degrees of freedom df = n – 1 and a specified significance level of α .
As with previous sections, if the test statistic falls in the rejection zone set by the critical value, you will reject the null hypothesis.
A forester wants to control a dense understory of striped maple that is interfering with desirable hardwood regeneration using a mist blower to apply an herbicide treatment. She wants to make sure that treatment has a consistent application rate, in other words, low variability not exceeding 0.25 gal./acre (0.06 gal. 2 ). She collects sample data (n = 11) on this type of mist blower and gets a sample variance of 0.064 gal. 2 Using a 5% level of significance, test the claim that the variance is significantly greater than 0.06 gal. 2
H 0 : σ 2 = 0.06
H 1 : σ 2 >0.06
The critical value is 18.307. Any test statistic greater than this value will cause you to reject the null hypothesis.
The test statistic is
We fail to reject the null hypothesis. The forester does NOT have enough evidence to support the claim that the variance is greater than 0.06 gal. 2 You can also estimate the p-value using the same method as for the student t-table. Go across the row for degrees of freedom until you find the two values that your test statistic falls between. In this case going across the row 10, the two table values are 4.865 and 15.987. Now go up those two columns to the top row to estimate the p-value (0.1-0.9). The p-value is greater than 0.1 and less than 0.9. Both are greater than the level of significance (0.05) causing us to fail to reject the null hypothesis.
(referring to Ex. 16)
Test and CI for One Variance
Method | ||
Null hypothesis | Sigma-squared | = 0.06 |
Alternative hypothesis | Sigma-squared | > 0.06 |
The chi-square method is only for the normal distribution.
Test | |||
Method | Statistic | DF | P-Value |
Chi-Square | 10.67 | 10 | 0.384 |
Excel does not offer 1-sample χ 2 testing.
To test a claim about μ when σ is known.
Table 4. A summary table for critical Z-scores.
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Course: ap®︎/college statistics > unit 10.
After you have completed the statistical analysis and decided to reject or fail to reject the Null hypothesis, you need to state your conclusion about the claim. To get the correct wording, you need to recall which hypothesis was the claim.
If the claim was the null, then your conclusion is about whether there was sufficient evidence to reject the claim. Remember, we can never prove the null to be true, but failing to reject it is the next best thing. So, it is not correct to say, “Accept the Null.”
If the claim is the alternative hypothesis, your conclusion can be whether there was sufficient evidence to support (prove) the alternative is true.
Use the following table to help you make a good conclusion.
The best way to state the conclusion is to include the significance level of the test and a bit about the claim itself.
For example, if the claim was the alternative that the mean score on a test was greater than 85, and your decision was to Reject then Null , then you could conclude: “ At the 5% significance level, there is sufficient evidence to support the claim that the mean score on the test was greater than 85. ”
The reason you should include the significance level is that the decision, and thus the conclusion, could be different if the significance level was not 5%.
If you are curious why we say “Fail to Reject the Null” instead of “Accept the Null,” this short video might be of interest: Here
It is concluded that the null hypothesis Ho is not rejected proportion p is greater than 0.5, at the 0.05 significance
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S.3.1 hypothesis testing (critical value approach).
The critical value approach involves determining "likely" or "unlikely" by determining whether or not the observed test statistic is more extreme than would be expected if the null hypothesis were true. That is, it entails comparing the observed test statistic to some cutoff value, called the " critical value ." If the test statistic is more extreme than the critical value, then the null hypothesis is rejected in favor of the alternative hypothesis. If the test statistic is not as extreme as the critical value, then the null hypothesis is not rejected.
Specifically, the four steps involved in using the critical value approach to conducting any hypothesis test are:
Mean gpa section .
In our example concerning the mean grade point average, suppose we take a random sample of n = 15 students majoring in mathematics. Since n = 15, our test statistic t * has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05 so that we have only a 5% chance of making a Type I error.
The critical value for conducting the right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the t -value, denoted t \(\alpha\) , n - 1 , such that the probability to the right of it is \(\alpha\). It can be shown using either statistical software or a t -table that the critical value t 0.05,14 is 1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3 if the test statistic t * is greater than 1.7613. Visually, the rejection region is shaded red in the graph.
The critical value for conducting the left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the t -value, denoted -t ( \(\alpha\) , n - 1) , such that the probability to the left of it is \(\alpha\). It can be shown using either statistical software or a t -table that the critical value -t 0.05,14 is -1.7613. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ < 3 if the test statistic t * is less than -1.7613. Visually, the rejection region is shaded red in the graph.
There are two critical values for the two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 — one for the left-tail denoted -t ( \(\alpha\) / 2, n - 1) and one for the right-tail denoted t ( \(\alpha\) / 2, n - 1) . The value - t ( \(\alpha\) /2, n - 1) is the t -value such that the probability to the left of it is \(\alpha\)/2, and the value t ( \(\alpha\) /2, n - 1) is the t -value such that the probability to the right of it is \(\alpha\)/2. It can be shown using either statistical software or a t -table that the critical value -t 0.025,14 is -2.1448 and the critical value t 0.025,14 is 2.1448. That is, we would reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ ≠ 3 if the test statistic t * is less than -2.1448 or greater than 2.1448. Visually, the rejection region is shaded red in the graph.
COMMENTS
When writing the conclusion of a hypothesis test, we typically include: Whether we reject or fail to reject the null hypothesis. The significance level. A short explanation in the context of the hypothesis test. For example, we would write: We reject the null hypothesis at the 5% significance level.
The null hypothesis varies by the type of statistic and hypothesis test. Remember that inferential statistics use samples to draw conclusions about populations. Consequently, when you write a null hypothesis, it must make a claim about the relevant population parameter. Further, that claim usually indicates that the effect does not exist in the ...
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 and alternative hypotheses offer competing answers to your research question. When the research question asks "Does the independent variable affect the dependent variable?": The null hypothesis ( H0) answers "No, there's no effect in the population.". The alternative hypothesis ( Ha) answers "Yes, there is an effect in the ...
Basic definitions. The null hypothesis and the alternative hypothesis are types of conjectures used in statistical tests to make statistical inferences, which are formal methods of reaching conclusions and separating scientific claims from statistical noise.. The statement being tested in a test of statistical significance is called the null hypothesis. . The test of significance is designed ...
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.
6. Write a null hypothesis. If your research involves statistical hypothesis testing, you will also have to write a null hypothesis. The null hypothesis is the default position that there is no association between the variables. The null hypothesis is written as H 0, while the alternative hypothesis is H 1 or H a.
The first step in hypothesis testing is to set up two competing hypotheses. The hypotheses are the most important aspect. If the hypotheses are incorrect, your conclusion will also be incorrect. The two hypotheses are named the null hypothesis and the alternative hypothesis. The null hypothesis is typically denoted as H 0.
Step 7: Based on Steps 5 and 6, draw a conclusion about H 0. If F calculated is larger than F α, then you are in the rejection region and you can reject the null hypothesis with ( 1 − α) level of confidence. Note that modern statistical software condenses Steps 6 and 7 by providing a p -value. The p -value here is the probability of getting ...
Present the findings in your results and discussion section. Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps. Table of contents. Step 1: State your null and alternate hypothesis. Step 2: Collect data. Step 3: Perform a statistical test.
5.2 - Writing Hypotheses. The first step in conducting a hypothesis test is to write the hypothesis statements that are going to be tested. For each test you will have a null hypothesis ( H 0) and an alternative hypothesis ( H a ). When writing hypotheses there are three things that we need to know: (1) the parameter that we are testing (2) the ...
The null and alternative hypotheses are two competing claims that researchers weigh evidence for and against using a statistical test: Null hypothesis (H0): There's no effect in the population. Alternative hypothesis (HA): There's an effect in the population. The effect is usually the effect of the independent variable on the dependent ...
Null Hypothesis Examples. "Hyperactivity is unrelated to eating sugar " is an example of a null hypothesis. If the hypothesis is tested and found to be false, using statistics, then a connection between hyperactivity and sugar ingestion may be indicated. A significance test is the most common statistical test used to establish confidence in a ...
A null hypothesis is rejected if the measured data is significantly unlikely to have occurred and a null hypothesis is accepted if the observed outcome is consistent with the position held by the null hypothesis. Rejecting the null hypothesis sets the stage for further experimentation to see if a relationship between two variables exists.
The Purpose of Null Hypothesis Testing. As we have seen, psychological research typically involves measuring one or more variables for a sample and computing descriptive statistics for that sample. In general, however, the researcher's goal is not to draw conclusions about that sample but to draw conclusions about the population that the ...
Onward! We use p -values to make conclusions in significance testing. More specifically, we compare the p -value to a significance level α to make conclusions about our hypotheses. If the p -value is lower than the significance level we chose, then we reject the null hypothesis H 0 in favor of the alternative hypothesis H a .
Examples. A research hypothesis, in its plural form "hypotheses," is a specific, testable prediction about the anticipated results of a study, established at its outset. It is a key component of the scientific method. Hypotheses connect theory to data and guide the research process towards expanding scientific understanding.
Components of a Formal Hypothesis Test. The null hypothesis is a statement about the value of a population parameter, such as the population mean (µ) or the population proportion (p).It contains the condition of equality and is denoted as H 0 (H-naught).. H 0: µ = 157 or H 0: p = 0.37. The alternative hypothesis is the claim to be tested, the opposite of the null hypothesis.
Below these are summarized into six such steps to conducting a test of a hypothesis. Set up the hypotheses and check conditions: Each hypothesis test includes two hypotheses about the population. One is the null hypothesis, notated as H 0, which is a statement of a particular parameter value. This hypothesis is assumed to be true until there is ...
It is the opposite of your research hypothesis. The alternative hypothesis--that is, the research hypothesis--is the idea, phenomenon, observation that you want to prove. If you suspect that girls take longer to get ready for school than boys, then: Alternative: girls time > boys time. Null: girls time <= boys time.
The best way to state the conclusion is to include the significance level of the test and a bit about the claim itself. For example, if the claim was the alternative that the mean score on a test was greater than 85, and your decision was to Reject then Null, then you could conclude: " At the 5% significance level, there is sufficient ...
The P -value is, therefore, the area under a tn - 1 = t14 curve to the left of -2.5 and to the right of 2.5. It can be shown using statistical software that the P -value is 0.0127 + 0.0127, or 0.0254. The graph depicts this visually. Note that the P -value for a two-tailed test is always two times the P -value for either of the one-tailed tests.
The critical value for conducting the left-tailed test H0 : μ = 3 versus HA : μ < 3 is the t -value, denoted -t( α, n - 1), such that the probability to the left of it is α. It can be shown using either statistical software or a t -table that the critical value -t0.05,14 is -1.7613. That is, we would reject the null hypothesis H0 : μ = 3 ...