Section 9.3 Hypothesis Tests for a Population Mean, Standard Deviation Unknown 437 Area = 0.0025 −3.232 2.915 3.232 3.144 −3.144 Area = 0.0025 Area = 0.001 Area = 0.001 −2.915 Figure 9.13 The P-value for a two-tailed test is the sum of the areas in the two tails. The area in each tail is between 0.001 and 0.0025. The sum of the areas in both tails is therefore between 2(0.001) = 0.002 and 2(0.0025) = 0.005. Check Your Understanding 1. Find the P-value for the following values of the test statistic t, sample size n, and alternate hypothesis H1. If you use Table A.3, you may specify that P is between two values. a. t = 2.584, n = 12, H1: �� > ��0 0.0127 b. t = −1.741, n = 21, H1: �� < ��0 0.0485 c. t = 3.031, n = 14, H1: �� ≠ ��0 0.0096 d. t = −2.584, n = 31, H1: �� ≠ ��0 0.0148 2. In Example 9.16, the sample size was n = 76, and we observed ̄x = 2.2 and s = 6.1. We tested H0 : �� = 0 versus H1: �� > 0, and the P-value was 0.0012. Assume that the sample size was 41 instead of 76, but that the values of ̄x and s were the same. a. Find the value of the test statistic t. 2.309 b. How many degrees of freedom are there? 40 c. Find the P-value. 0.0131 d. Is the evidence against H0 stronger or weaker than the evidence from the sample of 76? Explain. Weaker Answers are on page 447. EXAMPLE 9.17 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 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. 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. 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 There is no evidence of strong skewness, and no outliers. Therefore, we can proceed. Step 1: State the null and alternate hypotheses. The issue is whether the mean �� differs from 3.5. Therefore, the null and alternate hypotheses are H0 : �� = 3.5 H1: �� ≠ 3.5
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