Section 9.5 Hypothesis Tests for a Standard Deviation 459 Performing a Hypothesis Test for a Standard Deviation Check to be sure that the assumptions are satisfied. We must have a simple random sample from a normal population. Step 1: State the null and alternate hypotheses. The null hypothesis will be of the form H0 : = 0. The alternate hypothesis can be stated in one of three ways: Left-tailed: H1: < 0 Right-tailed: H1: > 0 Explain It Again Two-tailed: H1: ≠ 0 Degrees of freedom: When a hypothesis test for a standard deviation is performed, the degrees of freedom is always 1 less than the sample size. Step 2: Choose a significance level , and find the critical value, using Table A.4 with n − 1 degrees of freedom. Left-tailed: H1: < 0: The critical value is 2 1−. Right-tailed: H1: > 0: The critical value is 2 . Two-tailed: H1: ≠ 0: The critical values are 2 ∕2 and 2 1−∕2. Step 3: Compute the test statistic 2 = (n − 1) ⋅ s2 2 0 . Step 4: Determine whether to reject H0, as follows: Left-tailed: H1: < 0: Reject if 2 ≤ 2 1−. Right-tailed: H1: > 0: Reject if 2 ≥ 2 . Two-tailed: H1: ≠ 0: Reject if 2 ≥ 2 ∕2 or 2 ≤ 2 1−∕2. Step 5: State a conclusion. EXAMPLE 9.22 Performing a hypothesis test To check the reliability of a scale in a butcher shop, a test weight known to weigh 400 grams was weighed 16 times. For the scale to be considered reliable, the standard deviation of repeated measurements must be less than 1 gram. The standard deviation of the 16 measured weights was s = 0.8 gram. Assume that the measured weights are independent and follow a normal distribution. Can we conclude that the population standard deviation of the measurements is less than 1 gram? Use the = 0.05 level of significance. Solution We first check the assumptions. We have a random sample from a normal population, so the assumptions are satisfied. Step 1: State the null and alternate hypotheses. Because we are interested in determining whether < 1, the hypotheses are H0 : = 1 H1: < 1 Step 2: Choose a significance level and find the critical value. We will use a significance level of = 0.05. We have 16 − 1 = 15 degrees of freedom. This is a 1− = 2 left-tailed test, so the critical value is 2 0.95 = 7.261. Step 3: Compute the test statistic. The sample size is n = 16. The sample variance is s2 = (0.8)2 = 0.64, and the value specified by H0 is 0 = 1. The value of the test statistic is 2 = (n − 1)s2 2 0 = (16 − 1)(0.64) 12 = 9.6 Step 4: Determine whether to reject H0. The value of the test statistic is 2 = 9.6. 0.95 = 7.261. This is a left-tailed test, so we reject H0 if The critical value is 2 0.95. Because 9.6 > 7.261, we do not reject H0 at the = 0.05 level. 2 < 2 Step 5: State a conclusion. There is not enough evidence to conclude that the population standard deviation is less than 1 gram. We cannot consider the scale to be reliable.
navidi_monk_elementary_statistics_2e_ch7-9
To see the actual publication please follow the link above