438 Chapter 9 Hypothesis Testing Step 2: Choose a significance level ��. The significance level is �� = 0.01. Step 3: Compute the value of the test statistic t. To compute t, we need to know the sample mean ̄x, the sample standard deviation s, the null hypothesis mean ��0, and the sample size n. We compute ̄x and s from the sample. The values are ̄x = 2.9429 s = 0.4995 The null hypothesis mean is ��0 = 3.5. The sample size is n = 7. The value of the t statistic is t = ̄x − ��0 s∕ √ n = 2.9429 − 3.5 √ 0.4995∕ 7 = −2.951 Step 4: Compute the P-value. The number of degrees of freedom is n − 1 = 7 − 1 = 6. The alternate hypothesis 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 2.951, in a t distribution with 6 degrees of freedom. We can use technology to find that P = 0.0256. The following TI-84 Plus display presents the results. Step-by-step instructions for performing hypothesis tests with technology are given in the Using Technology section on page 442. The P-value is given on the third line of the display. Rounding off to four decimal places, we see that P = 0.0256. Alternatively, we can use Table A.3 to specify that the P-value is between two numbers. In the row corresponding to 6 degrees of freedom, the two values closest to 2.951 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 is between 2(0.01) = 0.02 and 2(0.025) = 0.05. See Figure 9.14. Area = 0.025 Area = 0.025 −3.143 −2.447 2.447 3.143 −2.951 2.951 Area = 0.01 Area = 0.01 Figure 9.14 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 0.01 and 0.025. The sum of the areas in both tails is therefore between 2(0.01) = 0.02 and 2(0.025) = 0.05. Step 5: Interpret the P-value. The P-value of 0.0256 tells us that if H0 is true, the probability of observing a value of the test statistic as extreme as or more extreme than the value of −2.951 that we observed is 0.0256. The P-value is small enough to give us doubt about the truth of H0. However, because P > 0.01, we do not reject H0 at the 0.01 level. Step 6: State a conclusion. 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.
navidi_monk_elementary_statistics_2e_ch7-9
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