1. Hypothesis Tests for a Population Mean, 𝜎 Unkown
Section 9.3

2. Objectives Test a hypothesis about a mean using the P-value method
Test a hypothesis about a mean using the critical value method

3. Test a hypothesis about a mean using the P-value method (Tables)
Objective 1
Test a hypothesis about a mean using the P-value method
(Tables)

4. Example – Performing a Hypothesis Test
We begin this section with an example. Suppose that in a recent medical study, 76 subjects were placed on a low-fat diet. After 12 months, their sample mean weight loss was 𝑥 = 2.2 kilograms, with a sample standard deviation of 𝑠 = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0?
If we knew the population standard deviation 𝜎, we would be able to compute the 𝑧-score of the sample mean to be 𝑧= 𝑥 −𝜇 𝜎/ 𝑛 , and use this test statistic to perform a hypothesis test. In this example, as is usually the case, we do not know the population standard deviation. To proceed, we replace 𝜎 with the sample standard deviation 𝑠, and use the 𝑡 test statistic instead: 𝑡= 𝑥 −𝜇 𝑠/ 𝑛 . When the null hypothesis is true, the 𝑡 statistic has a Student’s 𝑡 distribution with 𝑛−1 degrees of freedom.
The assumptions for performing a hypothesis test for 𝜇 when the population standard deviation 𝜎 is unknown are as follows:
We have a simple random sample.
The sample size is large (𝑛>30), or the population is approximately normal.

5. Example – Performing a Hypothesis Test
Suppose that in a recent medical study, 76 subjects were placed on a low-fat diet. After 12 months, their sample mean weight loss was 𝑥 = 2.2 kilograms, with a sample standard deviation of 𝑠 = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0? Since we have a simple random sample and the sample size is large, we may proceed with the test. The issue is whether the mean weight loss 𝜇 is greater than 0. So the null and alternate hypotheses are 𝐻 0 :𝜇=0 versus 𝐻 1 :𝜇>0. The test statistic is 𝑡= 𝑥 −𝜇 𝑠/ 𝑛 = 2.2−0 6.1/ 76 = When 𝐻 0 is true, the test statistic 𝑡 has the Student’s 𝑡 distribution with 𝑛 − 1=76−1=75 degrees of freedom. This is a right tail test, so the P-value is the area under the Student’s 𝑡 curve to the right of 𝑡 = Using technology, we find the exact P-value to be P = Since P < 0.05, we reject 𝐻 0 at the 𝛼=0.05 level. We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0.

6. Computing P-Values
The P-value of the test statistic 𝑡 is the probability, assuming 𝐻 0 is true, of observing a value for the test statistic that disagrees as strongly as or more strongly with 𝐻 0 than the value actually observed. The P-value is an area under the Student’s 𝑡 curve with 𝑛−1 degrees of freedom. The area is in the left tail, the right tail, or in both tails, depending on the type of alternate hypothesis.

7. Estimating the P-value From a Table
When using a 𝑡 table, we cannot find the P-value exactly. Instead, we can only specify that P is between two values. In the last example, there are 75 degrees of freedom. We consult Table A.3 and find that the number 75 does not appear in the degrees of freedom column. We therefore use the next smallest number, which is 60. Now look across the row for two numbers that bracket the observed value These are and The upper-tail probabilities are for and for The P-value must therefore be between and We can conclude that the P-value is small enough to reject 𝐻 0 at the 𝛼 = 0.05 level of significance.

8. P-value From a Table for a Two-tailed Test
If the alternate hypothesis were 𝐻 1 :𝜇≠0, the P-value would be the sum of the areas in two tails. If using Table A.3, we can only specify that P is between two values. We know, from Table A.3, that the area in one tail is between and Therefore, the area in both tails is between 2(0.001) = and 2(0.0025) =

9. Example – Perform a Hypothesis Test
Generic drugs are lower-cost substitutes for brand-name drugs. Before a generic drug can be sold in the United States, it must be tested and found to perform equivalently to the brand name product. The U.S. Food and Drug Administration is now supervising the testing of a new generic antifungal ointment. The brand-name ointment is known to deliver a mean of 3.5 micrograms of active ingredient to each square centimeter of skin. As part of the testing, seven subjects apply the ointment. Six hours later, the amount of drug that has been absorbed into the skin is measured. The amounts, in micrograms, are How strong is the evidence that the mean amount absorbed differs from 3.5 micrograms? Use the 𝛼 = 0.01 level of significance. Solution: We first check the assumptions. Because the sample is small, the population must be approximately normal. We check this with a dotplot of the data. There is no evidence of strong skewness, and no outliers. We may proceed. The null and alternate hypotheses are: 𝐻 0 : 𝜇=3.5 versus 𝐻 1 : 𝜇≠3.5. We compute 𝑥 and 𝑠 from the sample. The values are 𝑥 = and 𝑠=

10. Example – Perform a Hypothesis Test
Solution (continued): The 𝑡 test statistic is 𝑡= 𝑥 − 𝜇 0 𝑠 𝑛 = − =− The number of degrees of freedom is 𝑛−1 = 6. The alternate is two-tailed, so the P-value is the sum of the area to the left of the observed 𝑡 statistic –2.951 and the area to the right of Using Table A.3, the two values closest to in the row corresponding to 6 degrees of freedom are and The area to the right of is 0.025, and the area to the right of is Therefore, the area in the right tail is between 0.01 and The P-value is twice the area in the right tail, so we conclude that P-value is between 0.02 and The P-value is small enough to give us doubt about the truth of 𝐻 0 . However, because P > 0.01, we do not reject 𝐻 0 at the 0.01 level. There is not enough evidence to conclude that the mean amount of drug absorbed differs from 3.5 micrograms. The mean may be equal to 3.5 micrograms.
Remember 𝐻 0 :𝜇=3.5
𝐻 1 :𝜇≠3.5
𝑥 = 𝑠=0.4995
𝑛=7

11. Test a hypothesis about a mean using the P-value method (TI-84 PLUS)
Objective 1
Test a hypothesis about a mean using the P-value method
(TI-84 PLUS)

12. Example – Performing a Hypothesis Test
We begin this section with an example. Suppose that in a recent medical study, 76 subjects were placed on a low-fat diet. After 12 months, their sample mean weight loss was 𝑥 = 2.2 kilograms, with a sample standard deviation of 𝑠 = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0?
If we knew the population standard deviation 𝜎, we would be able to compute the 𝑧-score of the sample mean to be 𝑧= 𝑥 −𝜇 𝜎/ 𝑛 , and use this test statistic to perform a hypothesis test. In this example, as is usually the case, we do not know the population standard deviation. To proceed, we replace 𝜎 is the sample standard deviation 𝑠, and use the 𝑡 test statistic instead: 𝑡= 𝑥 −𝜇 𝑠/ 𝑛 . When the null hypothesis is true, the 𝑡 statistics has a Student’s 𝑡 distribution with 𝑛−1 degrees of freedom.
The assumptions for performing a hypothesis test for 𝜇 when the population standard deviation 𝜎 is unknown are as follows:
We have a simple random sample.
The sample size is large (𝑛>30), or the population is approximately normal.

13. Example – Performing a Hypothesis Test
We begin this section with an example. Suppose that in a recent medical study, 76 subjects were placed on a low-fat diet. After 12 months, their sample mean weight loss was 𝑥 = 2.2 kilograms, with a sample standard deviation of 𝑠 = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0? Since we have a simple random sample and the sample size is large, we may proceed with the test. The issue is whether the mean weight loss 𝜇 is greater than 0. So the null and alternate hypotheses are 𝐻 0 :𝜇=0 & 𝐻 1 :𝜇>0. The test statistic is 𝑡= 𝑥 −𝜇 𝑠/ 𝑛 = 2.2−0 6.1/ 76 = When 𝐻 0 is true, the test statistic 𝑡 has the Student’s 𝑡 distribution with 𝑛 − 1=76−1=75 degrees of freedom. This is a right tail test, so the P-value is the area under the Student’s 𝑡 curve to the right of 𝑡 = Using technology, we find the exact P-value to be P = Since P < 0.05, we reject 𝐻 0 at the 𝛼=0.05 level. We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0.

14. Hypothesis Testing on the TI-84 PLUS
The T-Test will perform a hypothesis test when the population standard deviation 𝜎 is not known. This command is accessed by pressing STAT and highlighting the TESTS menu. If the summary statistics are given the Stats option should be selected for the input option. If the raw sample data are given, the Data option should be selected.

15. Example (TI-84 PLUS)
In a recent medical study, 76 subjects were placed on a low-fat diet. After 12 months, their sample mean weight loss was 𝑥 = 2.2 kilograms, with a sample standard deviation of s = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0? Use the 𝛼 = 0.05 level of significance.
Solution:
We press STAT and highlight the TESTS menu and select T-Test.
Select Stats as the input option and enter 0 as the null hypothesis mean 𝜇 0 , 2.2 for the sample mean 𝑥 , 6.1 for the sample standard deviation s, and 76 for the sample size n. Since we have a right- tailed test, select the >𝝁 𝟎 option.
Select Calculate.
The P-value is Since P < 0.05, we reject 𝐻 We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0.

16. Example – Perform a Hypothesis Test
Generic drugs are lower-cost substitutes for brand-name drugs. Before a generic drug can be sold in the United States, it must be tested and found to perform equivalently to the brand name product. The U.S. Food and Drug Administration is now supervising the testing of a new generic antifungal ointment. The brand-name ointment is known to deliver a mean of 3.5 micrograms of active ingredient to each square centimeter of skin. As part of the testing, seven subjects apply the ointment. Six hours later, the amount of drug that has been absorbed into the skin is measured. The amounts, in micrograms, are How strong is the evidence that the mean amount absorbed differs from 3.5 micrograms? Use the 𝛼 = 0.01 level of significance. Solution: We first check the assumptions. Because the sample is small, the population must be approximately normal. We check this with a dotplot of the data. There is no evidence of strong skewness, and no outliers. We may proceed. The null and alternate hypotheses are: 𝐻 0 : 𝜇=3.5 versus 𝐻 1 : 𝜇≠3.5.

17. Example (TI-84 PLUS) Solution (continued):
We enter the data (2.6, 3.2, 2.1, 3.0, 3.1, 2.9, 3.7) into list L1.
Press STAT and highlight the TESTS menu and select T-Test.
Select Data as the input option and enter 3.5 in the 𝝁 𝟎 field. Enter L1 as the List option and 1 as the Freq option. Since we have a two-tailed test, select the ≠𝝁 𝟎 option.
Select Calculate.
The P-value is The P-value is small enough to give us doubt about the truth of 𝐻 0 . However, because P > 0.01, we do not reject 𝐻 0 at the 0.01 level.
There is not enough evidence to conclude that the mean amount of drug absorbed differs from 3.5 micrograms. The mean may be equal to 3.5 micrograms.

18. Test a hypothesis about a mean using the critical value method
Objective 2
Test a hypothesis about a mean using the critical value method

19. Critical values for the 𝑡-Statistic
The critical value method for a hypothesis test of a population mean when 𝜎 is unknown is the same as that when 𝜎 is known. The only exception is that we use the Student’s 𝑡 distribution rather than a normal distribution. The critical values for the Student’s 𝑡 distribution can be found in Table A.3 or with technology.

20. Example
A computer software vendor claims that a new version of their operating system will crash less than six times per year on average. A system administrator installs the operating system on a random sample of 41 computers. At the end of a year, the sample mean number of crashes is 7.1, with a standard deviation of 3.6. Can you conclude that the vendor’s claim is false? Use the 𝛼 = 0.05 significance level. Solution: We first check the assumptions. We have a large (𝑛 > 30) random sample, so the assumptions are satisfied. The null and alternate hypotheses are: 𝐻 0 : 𝜇 = 6 versus 𝐻 1 : 𝜇 > 6.

21. Solution
We use a significance level of 𝛼 = 0.05 and Table A.3. The number of degrees of freedom is 41 − 1 = 40. Since this is a right-tailed test, the critical value is the 𝑡-value with area 0.05 above it in the right tail. Thus, the critical value is 𝑡 𝛼 = We have 𝑥 = 7.1, 𝜇 0 = 6, 𝑠 = 3.6, and 𝑛 = 41. The test statistic is 𝑡= 𝑥 − 𝜇 0 𝑠 𝑛 = 7.1 − =1.957

22. Solution
Because this is a right-tailed test, we reject 𝐻 0 if t ≥ 𝑡 𝛼 . Since 𝑡 = and 𝑡 𝛼 = 1.684, we reject 𝐻 0 . We conclude that the mean number of crashes is greater than six per year.

23. You Should Know…
The assumptions for hypothesis tests for 𝜇 when 𝜎 is unknown
How to perform hypothesis tests for 𝜇 when 𝜎 is unknown using the P-value method
How to estimate a P-value from a table for one-tailed and two-tailed tests
How to perform hypothesis tests for 𝜇 when 𝜎 is unknown using the critical value method