Essential Statistics 2E - PowerPoint Presentation

Essential Statistics 2E William Navidi and Barry Monk

Hypothesis Tests for a Population Mean, Unkown   Section 8.3

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

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

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 = 2.2 kilograms, with a sample standard deviation of = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0? If we knew the population standard deviation , we would be able to compute the -score of the sample mean to be , 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 with the sample standard deviation , and use the test statistic instead: . When the null hypothesis is true, the statistic has a Student’s distribution with degrees of freedom.  

Assumptions The assumptions for performing a hypothesis test for when the population standard deviation is unknown are as follows: We have a simple random sample. The sample size is large ( ), or the population is approximately normal.  

Performing a Hypothesis Test (Continued) Remember   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 is greater than 0. .   The test statistic is . When is true, the test statistic has the Student’s distribution with degrees of freedom. This is a right tail test, so the P- value is the area under the Student’s curve to the right of = 3.144.   Using technology, we find the exact P-value to be P = 0.0012. Since P < 0.05, we reject at the level. We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0.  

Computing P -Values The P-value of the test statistic is the probability, assuming is true, of observing a value for the test statistic that disagrees as strongly as or more strongly with than the value actually observed. The P -value is an area under the Student’s curve with degrees of freedom. The area is in the left tail, the right tail, or in both tails, depending on the type of test. Left-tailed: The P-value is the area to the left of .Right-tailed: The P-value is the area to the right of .Two-tailed: The P-value is the sum of the area to the left of and to the right of .  

Estimating the P -value From a Table When using a 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 3.144. These are 2.915 and 3.232. The upper-tail probabilities are 0.0025 for 2.915 and 0.001 for 3.232. The P-value must therefore be between 0.001 and 0.0025. We can conclude that the P-value is small enough to reject at the = 0.05 level of significance. 

P-value From a Table for a Two-tailed Test If the alternate hypothesis were , 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 0.001 and 0.0025. Therefore, the area in both tails is between 2(0 . 001) = 0 . 002 and 2(0.0025) = 0.005. 

Example: Hypothesis Test ( P -value, Table) 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 2.6 3.2 2.1 3.0 3.1 2.9 3.7 How strong is the evidence that the mean amount absorbed differs from 3.5 micrograms? Use the = 0.01 level of significance.  

Example: Hypothesis Test ( P -value, Table, Continued 1)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: : versus : . We compute and from the sample. The values are and .  

Example: Hypothesis Test ( P -value, Table, Continued 2) Remember   The test statistic is = . The number of degrees of freedom is = 6. The alternate is two-tailed, so the P -value is the sum of the area to the left of the observed statistic –2.951 and the area to the right of 2.951. Using Table A.3, the two values closest to 2.951 in the row corresponding to 6 degrees of freedom are 2.447 and 3.143. The area to the right of 2.447 is 0.025, and the area to the right of 3.143 is 0.01. Therefore, the area in the right tail is between 0.01 and 0.025. The P -value is twice the area in the right tail, so we conclude that P -value is between 0.02 and 0.05.  

Example: Hypothesis Test ( P -value, Table, Continued 3) Remember 0.02 < P -value < 0.05  The P-value is small enough to give us doubt about the truth of . However, because P > 0.01, we do not reject 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. 

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

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 = 2.2 kilograms, with a sample standard deviation of = 6.1 kilograms. Can we conclude that the mean weight loss is greater than 0? If we knew the population standard deviation , we would be able to compute the -score of the sample mean to be , 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 with the sample standard deviation , and use the test statistic instead: . When the null hypothesis is true, the statistic has a Student’s distribution with degrees of freedom.  

Assumptions ** The assumptions for performing a hypothesis test for when the population standard deviation is unknown are as follows: We have a simple random sample. The sample size is large ( ), or the population is approximately normal.  

Performing a Hypothesis Test (Continued) ** Remember   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 is greater than 0. .   The test statistic is . When is true, the test statistic has the Student’s distribution with degrees of freedom. This is a right tail test, so the P-value is the area under the Student’s curve to the right of = 3.144.   Using technology, we find the exact P-value to be P = 0.0012. Since P < 0.05, we reject at the level. We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0.  

Hypothesis Testing on the TI-84 PLUS The T-Test will perform a hypothesis test for a population mean when the population standard deviation 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. If the raw sample data are given, the Data option should be selected.  

Example 1: Hypothesis Test ( P -value, 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 = 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? 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 , 2.2 for the sample mean , 6.1 for the sample standard deviation s, and 76 for the sample size n. Since we have a right-tailed test, select the option. Select Calculate . The P -value is 0.0012. Since P < 0.05, we reject . We conclude that the mean weight loss of people who adhered to this diet for 12 months is greater than 0. 

Example 2: Hypothesis Test ( P -value, TI-84 PLUS) 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 2.6 3.2 2.1 3.0 3.1 2.9 3.7 How strong is the evidence that the mean amount absorbed differs from 3.5 micrograms? Use the = 0.01 level of significance.  

Example 2: Hypothesis Test ( P -value, TI-84 PLUS, Continued 1)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: : :  

Example 2: Hypothesis Test ( P -value, TI-84 PLUS, Continued 2) 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 field. Enter L1 as the List option and 1 as the Freq option. Since we have a two-tailed test, select the option. Select Calculate.   The P -value is 0.0256. The P -value is small enough to give us doubt about the truth of . However, because P > 0.01, we do not reject 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. 

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

Critical values for the -Statistic   To test a hypothesis about a population mean when is unknown, we use the Student’s distribution. The critical values for the Student’s distribution can be found in Table A.3 or with technology. Left-tailed : The critical value is , which has an area of to its left. Reject if .Right-tailed: The critical value is , which has an area of to its right. Reject if . Two-tailed : The critical values are , which has an area of to its right and , which has an area of to its left. Reject if or .  

Example: Hypothesis Test (Critical Value) 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 = 0.05 significance level. Solution: We first check the assumptions. We have a simple random sample and the sample size is large ( ), so the assumptions are satisfied. We have: : = 6 : > 6. 

Example: Hypothesis Test (Critical Value, Continued 1) We use a significance level of = 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 -value with area 0.05 above it in the right tail. The critical value is = 1.684. We have = 7.1, = 6, = 3.6, and = 41. The test statistic is .  

Example: Hypothesis Test (Critical Value, Continued 2) Because this is a right-tailed test, we reject if t ≥ . Since = 1.957 and = 1.684, we reject .  We conclude that the mean number of crashes is greater than six per year.

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