452 Chapter 9 Hypothesis Testing Step 4: Determine whether to reject H0, as follows: Critical region: −zα Area = α Left-tailed: H1: p < p0 Reject if z ≤ −z��. Critical region: zα Area = α Right-tailed: H1: p > p0 Reject if z ≥ z��. Critical region: Area = α/2 Critical region: Area = α/2 −zα/2 zα/2 Two-tailed: H1: p ≠ p0 Reject if z ≥ z��∕2 or z ≤ −z��∕2. Step 5: State a conclusion. EXAMPLE 9.20 Test a hypothesis about a population proportion using the critical value method A nationwide survey of working adults indicates that only 50% of them are satisfied with their jobs. The president of a large company believes that more than 50% of employees at his company are satisfied with their jobs. To test his belief, he surveys a random sample of 100 employees, and 54 of them report that they are satisfied with their jobs. Can he conclude that more than 50% of employees at the company are satisfied with their jobs? Use the �� = 0.05 level of significance. Solution We first check the assumptions. We have a simple random sample from the population of employees. Each employee is categorized as being satisfied or not satisfied. The sample size is n = 100 and the proportion p0 specified by H0 is 0.5. Therefore, we calculate that np0 = 100(0.5) = 50 > 10, and n(1 − p0) = 100(1 − 0.5) = 50 > 10. If the total number of employees in the company is more than 2000, as we shall assume, then the population is more than 20 times as large as the sample. All the assumptions are therefore satisfied. Step 1: State the null and alternate hypotheses. The issue is whether the proportion of employees that are satisfied with their jobs is more than 0.5. Therefore, the null and alternate hypotheses are H0 : p = 0.5 H1: p > 0.5 Step 2: Choose a significance level and find the critical value. The significance level is �� = 0.05. The alternate hypothesis is p > 0.5, so this is a right-tailed test. The critical value corresponding to �� = 0.05 is z�� = 1.645. Step 3: Compute the test statistic. The test statistic is z = ̂p − p0 √ p0(1 − p0) n The value of the sample proportion ̂p is ̂p = Number of satisfied employees Sample size = 54 100 = 0.54 The quantity p0 is the value of p specified by H0, so p0 = 0.5. The sample size is n = 100. Therefore, the value of the test statistic is z = 0.54 − 0.5 √ 0.5(1 − 0.5) 100 = 0.80
navidi_monk_elementary_statistics_2e_ch7-9
To see the actual publication please follow the link above