Chapter 9: Introduction to Hypothesis Testing

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388 CHAPTER 9 • INTRODUCTION TO HYPOTHESIS TESTING EXAMPLE 9-5 Hypothesis Test Using p-values, σ Known TRY PROBLEM 9.9 OmniFitness Club The manager at the Omni Fitness Club in Muskegon, Michigan, believes that the recent remodeling project has greatly improved the club’s appeal for members and that they now stay longer at the club per visit than before the remodeling. Studies show that the previous mean time per visit was 36 minutes, with a standard deviation equal to 11 minutes. A simple random sample of n 200 visits is selected, and the current sample mean is 36.8 minutes. To test the manager’s claim, and partially justify the remodeling project, using an alpha 0.05 level, the following steps can be used: Step 1 Specify the population parameter of interest. The manager is interested in the mean time per visit, μ. Step 2 Formulate the null and alternative hypotheses. Based on the manager’s claim that the current mean stay is longer than before the remodeling, the null and alternative hypotheses are H 0 : 36 minutes H A : 36 minutes (claim) Step 3 Specify the significance level. The alpha level specified for this test is α 0.05. Step 4 Compute the test statistic. Because the sample size is large and the population standard deviation is assumed known, the test statistic will be a z-value, which is computed as follows: x z 36. 836 1. 0285 1. 03 11 n 200 Step 5 Calculate the p-value. In this example, the p-value is the probability of a z-value from the standard normal distribution being at least as large as 1.03. This is stated as p-value P (z 1.03) From the standard normal distribution table in Appendix D, p(z 1.03) 0.5000 0.3485 0.1515 Step 6 Reach a decision. The decision rule is If p-value 0.05, reject H 0 ; Otherwise, do not reject H 0 . Because the p-value 0.1515 0.05, do not reject the null hypothesis. Step 7 Draw a conclusion. The difference between the sample mean and the hypothesized population mean is not large enough to attribute the difference to anything but sampling error. Why do we need three methods to test the same hypothesis when they all give the same result? The answer is that we don’t. However, you need to be aware of all three methods because you will encounter each in business situations. The p-value approach is especially important because many statistical software packages provide a p-value that you can use to test a hypothesis quite easily, and a p-value provides a measure of the degree of significance
CHAPTER 9 • INTRODUCTION TO HYPOTHESIS TESTING 389 associated with the hypothesis test. This text will use both test-statistic approaches, as well as the p-value approach to hypothesis testing. One-tailed Test A hypothesis test in which the entire rejection region is located in one tail of the sampling distribution. In a one-tailed test, the entire alpha level is located in one tail of the distribution. Two-tailed Test A hypothesis test in which the entire rejection region is split into the two tails of the sampling distribution. In a two-tailed test, the alpha level is split evenly between the two tails. Business Application Types of Hypothesis Tests Hypothesis tests are formulated as either one-tailed tests or two-tailed tests depending on how the null and alternative hypotheses are presented. For instance, in the Morgan Lane application, the null and alternative hypotheses are Null hypothesis Alternative hypothesis H 0 : 25 days H A : 25 days This hypothesis test is one-tailed because the entire rejection region is located in the upper tail and the null hypothesis will be rejected only when the sample mean falls in the extreme upper tail of the sampling distribution (see Figure 9.4). In this application, it will take a sample mean substantially larger than 25 days in order to reject the null hypothesis. In Example 9.2 involving the Boulder Potato Company, the null and alternative hypotheses involving the mean fill of potato chip sacks is H 0 : μ 20 ounces (status quo) H A : μ 20 ounces In this two-tailed hypotheses test, the null hypothesis will be rejected if the sample mean is extremely large (upper tail) or extremely small (lower tail). The alpha level would be split evenly between the two tails. p-Value for Two-Tailed Tests In the previous p-value example, the rejection region was located in one tail of the sampling distribution. In those cases, the null hypothesis was of the or format. However, sometimes the null hypothesis will be stated as a direct equality. The following application involving the Golden Peanut Company shows how to use the p-value approach for a two-tailed test. GOLDEN PEANUT COMPANY Consider the Golden Peanut Company in Alpharetta, Georgia which packages salted and unsalted, unshelled peanuts in 16-ounce sacks. The company’s filling process strives for an average fill amount equal to 16 ounces. Therefore, Golden would test the following null and alternative hypotheses: H 0 : 16 ounces (status quo) H A : 16 ounces The null hypothesis will be rejected if the test statistic falls in either tail of the sampling distribution. The size of the rejection region is determined by α. Each tail has an area equal to α/2. The p-value for the two-tailed test is computed in a manner similar to that for a one-tailed test. First, determine the z-test statistic as follows: Suppose for this situation, Golden managers calculated a z 3.32. Next, find P(z 3.32) using either the standard normal table in Appendix D, Excel’s NORMSDIST function, or Minitab’s Calc Probability Distributions command. In this case, because z 3.32 exceeds the table values, we will use Excel or Minitab to obtain Then x z P(z 3.32) 0.9995 P(z 3.32) 1 0.9995 0.0005 n
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