Section 9.3 Hypothesis Tests for a Population Mean, Standard Deviation Unknown 441 Step 2: Choose a significance level �� and find the critical value or values. Use n − 1 degrees of freedom, where n is the sample size. Step 3: Compute the test statistic t = ̄x − ��0 √ n s∕ . Step 4: Determine whether to reject H0, as follows: Left-tailed: H1: �� < ��0 Reject if t ≤ −t��. Right-tailed: H1: �� > ��0 Reject if t ≥ t��. Two-tailed: H1: �� ≠ ��0 Reject if t ≥ t��∕2 or t ≤ −t��∕2. Step 5: State a conclusion. EXAMPLE 9.18 Test a hypothesis using the critical value method A computer software vendor claims that a new version of its operating system will crash fewer 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 large (n > 30) random sample, so the assumptions are satisfied. Step 1: State the null and alternate hypotheses. To conclude that the vendor’s claim is false, we must conclude that �� > 6. This is H1. The hypotheses are H0 : �� = 6 versus H1: �� > 6. Step 2: Choose a significance level �� and find the critical value. We will use a significance level of �� = 0.05. We use Table A.3. The number of degrees of freedom is 41−1 = 40. This is a right-tailed test, so the critical value is the t-value with area 0.05 above it in the right tail. Thus, the critical value is t�� = 1.684. Step 3: Compute the test statistic. We have ̄x = 7.1, ��0 = 6, s = 3.6, and n = 41. The test statistic is t = 7.1 − 6 √ 41 3.6∕ = 1.957 Step 4: Determine whether to reject H0. Because this is a right-tailed test, we reject H0 if t ≥ t��. Because t = 1.957 and t�� = 1.684, we reject H0. Figure 9.15 illustrates the critical region and the test statistic. Critical region tα = 1.684 t = 1.957 Figure 9.15 Step 5: State a conclusion. We conclude that the mean number of crashes is greater than six per year.
navidi_monk_elementary_statistics_2e_ch7-9
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