Chapter 9: Introduction to Hypothesis Testing

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386 CHAPTER 9 • INTRODUCTION TO HYPOTHESIS TESTING FIGURE 9.5 Providence Heart Institute Hypothesis Test σ x = n σ = 5 36 Rejection region α = 0.05 0.45 0.50 −z 0.05 = –1.645 0 μ = 42 z value is z 0.05 1.645. Therefore, the decision rule becomes If z 1.645, reject H 0 ; otherwise, do not reject H 0 . Step 5 Compute the test statistic. For this example we will use z. Assume the sample mean, computed using x x ∑ , is 40.2 days. Then, n Step 6 Reach a decision. (See Figure 9.5.) The decision rule is If z 1.645, reject H 0 Otherwise, do not reject. Because 2.16 1.645, reject H 0 . x z 40. 2 42 216 . 5 n Step 7 Draw a conclusion. There is sufficient evidence to conclude that the new bypass procedure does result in a shorter average recovery period. 36 EXAMPLE 9-4 Hypothesis Test for , Known TRY PROBLEM 9.5 Employee Commute Time A study in southern California claimed that the mean commute time for all employees working in Orange County exceeds 40 minutes. This figure is higher than what has been assumed in the past. The plan is to test this claim using an alpha level equal to 0.05 and a sample size of n 100 commuters. Based on previous studies, suppose that the population standard deviation is known to be σ 8 minutes. The hypothesis test can be conducted using the following steps: Step 1 Specify the population parameter of interest. The population parameter of interest is the mean commute time, μ.
CHAPTER 9 • INTRODUCTION TO HYPOTHESIS TESTING 387 Step 2 Formulate the null and alternative hypotheses. The new claim is that 40. Because this is different from past studies, it will become the alternative hypothesis. Thus, the null and alternative hypotheses are H 0 : 40 minutes H A : 40 minutes (claim) Step 3 Specify the significance level. The alpha level is specified to be 0.05. Step 4 Construct the rejection region. Alpha is the area under the standard normal distribution to the right of the critical value. Because the population standard deviation is known, the critical z-value, z α , is found by locating the z-value that corresponds to an area equal to 0.50 0.05 0.45. The critical z-value from the standard normal table is 1.645. Step 5 Compute the critical value. Suppose that the sample of 100 commuters showed a sample mean of 43.5 minutes. We can calculate using Equation 9.1 as follows: x x x z n ⎛ 40 1. 645 ⎝ ⎜ 41. 32 005 . 005 . x x 005 . 8 ⎞ 100 ⎠ ⎟ Step 6 Reach a decision. The decision rule is If x 41.32, reject H 0 ; Otherwise, do not reject. Because x 43.5 41.32, we reject H 0 . Step 7 Draw a conclusion. There is sufficient evidence to conclude that the mean commute time exceeds 40 minutes. p-value The probability (assuming the null hypothesis is true) of obtaining a test statistic at least as extreme as the test statistic we calculated from the sample. The p-value is also known as the observed significance level. p-value Approach In addition to the two hypothesis-testing approaches discussed previously, a third approach for conducting hypothesis tests also exists. This third approach uses a p-value instead of a critical value. If the calculated p-value is smaller than the probability in the rejection region (), then the null hypothesis is rejected. If the calculated p-value is greater than or equal to , then the hypothesis will not be rejected. The p-value approach is popular today because p-values are usually computed by statistical software packages, including Excel and Minitab. The advantage to reporting test results using a p-value is that it provides more information than simply stating whether the null hypothesis is rejected. The decision maker is presented with a measure of the degree of significance of the result (i.e., the p-value). This allows the reader the opportunity to evaluate the extent to which the data disagree with the null hypothesis, not just whether they disagree.
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Chapter 9: Introduction to Hypothesis Testing

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