According to Consumer Reports National Research Center, the average out-of-pocket expenditure in the United States for a dental tooth filling procedure if a person is insured is $141. Suppose a consumer analyst believes that such a procedure is more expensive now, and she wants to conduct a hypothesis test to confirm her belief. She sets up a study in which a random sample of 32 recent tooth fillings is taken from across the United States. The resulting out-of-pocket expenditures for such fillings are given below. Previous studies reveal that the population standard deviation for such data is $30. The value of alpha for her study is set at .05. Use these data and the eight-step approach for testing her hypotheses on these data. $147 152 119 142 163 151 123 118 145 166 149 130 158 175 142 160 175 125 147 210 132 156 139 122 163 175 120 135 207 141 170 110 Solution Step 1 Establish the hypotheses. Because the analyst wants to prove that this procedure is more expensive, she is conducting a one-tailed test in the upper tail. Her alternative hypothesis is that the mean expense is more than $141. The null hypothesis is that the mean still equals $141. The statistical hypotheses are Step 2 Determine the appropriate statistical test and sampling distribution. Because the population standard deviation is known ($30), the sample size is greater than 30, and the analyst is using the mean as the statistic, the z test in Formula 9.1 is the appropriate test statistic. Step 3 Specify the Type I error rate. Alpha is .05 for this problem. Step 4 Establish the decision rule. Because this test is one-tailed, alpha = .05 is in the upper tail of the distribution. Thus, the rejection region is in the upper tail of the distribution and contains 5% of the area of the distribution. There is .4500 of the area between the mean and the critical value that separates the tail of the distribution (the rejection region) from the nonrejection region. By using this .4500 and Table A.5, the critical z value can be obtained. Shown in Figure 9.8 is a graph of the problem with the rejection region and the critical value of z. The decision rule is that if the data gathered produce an observed z value greater than 1.645, the test statistic is in the rejection region and the decision is to reject the null hypothesis. If the observed z value is less than +1.645, the decision is to fail to reject the null hypothesis because the observed z value is in the nonrejection region. Graph has bell-shaped curve with nonrejection region centered at z = 0 point 0, or mu = 141. Shaded area under right tail of curve for rejection region with alpha = point 0 5. Critical value is x bar subscript c = 149 point 7 2 or z = 1 point 6 4 5. FIGURE 9.8 Graph of the Dental Problem Rejection and Nonrejection Regions Step 5 Gather the data. The 32 observed expense values are given. The sample mean of these data is $148.97. Step 6 The value of the test statistic is calculated by using , n = 32 and \sigma = $30, and a hypothesized \mu = $141. Step 7 Statistical conclusion. Because this test statistic, z = 1.50, is less than the critical value of z (+1.645) in the upper tail of the distribution, the statistical conclusion is to fail to reject the null hypothesis. The same conclusion can be reached using the p-value method. The observed test statistic is z = 1.50. From Table A.5, the probability of getting a z value at least this extreme when the null hypothesis is true is .5000 − .4332 = .0668. Since the p-value is greater than \alpha = .05, the decision is to fail to reject the null hypothesis. Remember, the p-value, in this case .0668, is the smallest value of alpha for which the null hypothesis can be rejected. If alpha had been .10, then the decision based on this p-value would have been to reject the null hypothesis. Note that we reach the same conclusion whether we use the critical value or the p-value. Step 8 Managerial implications. The consumer analyst does not have the statistical evidence to declare that the average out-of-pocket expense of getting a tooth filling is higher than $141. One might use the results of this study to argue that dentists are holding the line on expenses and/or that there is little or no inflation in the area of dental work. Using the critical value method For what sample mean (or more extreme) value would the null hypothesis be rejected? This critical sample mean can be determined by using the critical z value associated with alpha, z.05 = 1.645. The decision rule is that a sample mean more than 149.72 would be necessary to reject the null hypothesis. Because the mean obtained from the sample data is 148.97, the analysts fail to reject the null hypothesis. Figure 9.8 includes a scale with the critical sample mean and the rejection region for the critical value method.