Slide 1 - 60 Chapter 9 Hypothesis Tests for One Population Mean
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Slide 1 - 60 Chapter 9 Hypothesis Tests for One Population Mean Slide 2 Slide 2 - 60 Definition 9.1 Slide 3 Slide 3 - 60 Example 9.4 A company that produces snack foods uses a machine to package 454 g bags of pretzels. We assume that the net weights are normally distributed and that the population standard deviation of all such weights is 7.8 g. A simple random sample of 25 bags of pretzels has the net weights, in grams, displayed in Table 9.1. Do the data provide sufficient evidence to conclude that the packaging machine is not working properly? We use the following steps to answer the question. Table 9.1 Slide 4 Slide 4 - 60 Example 9.4 a.State the null and alternative hypotheses for the hypothesis test. b.Discuss the logic of this hypothesis test. c.Identify the distribution of the variable, that is, the sampling distribution of the sample mean for samples of size 25. d.Obtain a precise criterion for deciding whether to reject the null hypothesis in favor of the alternative hypothesis. e.Apply the criterion in part (d) to the sample data and state the conclusion. Slide 5 Slide 5 - 60 Solution Example 9.4 a.The null and alternative hypotheses, as stated in Example 9.1, are H 0 : = 454 g (the packaging machine is working properly) H a : (the packaging machine is not working properly). b.If the null hypothesis is true, then the mean weight,, of the sample of 25 bags of pretzels should approximately equal 454 g. However, if the sample mean weight differs too much from 454 g, we would be inclined to reject the null hypothesis and conclude that the alternative hypothesis is true. As we show in part (d), we can use our knowledge of the sampling distribution of the sample mean to decide how much difference is too much. Slide 6 Slide 6 - 60 Solution Example 9.4 c.n = 25, = 7.8, weights are normally distributed, = (which we dont know), = and is normally distributed. In other words, for samples of size 25, the variable is normally distributed with mean and standard deviation 1.56 g. d.The 95.44 part of the 68.2695.4499.74 rule states that, for a normally distributed variable, 95.44% of all possible observations lie within two standard deviations to either side of the mean. Applying this part of the rule to the variable and referring to part (c), we see that 95.44% of all samples of 25 bags of pretzels have mean weights within 21.56 = 3.12 g of . Slide 7 Slide 7 - 60 Solution Example 9.4 d.Equivalently, only 4.56% of all samples of 25 bags of pretzels have mean weights that are not within 3.12 g of , as illustrated in Fig. 9.1. Figure 9.1 Slide 8 Slide 8 - 60 Solution Example 9.4 d.If the mean weight,, of the 25 bags of pretzels sampled is more than two standard deviations (3.12 g) from 454 g, reject the null hypothesis and conclude that the alternative hypothesis is true. Otherwise, do not reject the null hypothesis. Figure 9.2 Slide 9 Slide 9 - 60 Solution Example 9.4 e.The mean weight,, of the sample of 25 bags of pretzels whose weights are given in Table 9.1 is 450g. So, z =( 454) / 1.56 = (450 454) /1.56 = 2.56. That is, the sample mean of 450 g is 2.56 standard deviations below the null-hypothesis population mean of 454 g, as shown in Fig. 9.3. Figure 9.3 Slide 10 Slide 10 - 60 Solution Example 9.4 e.Because the mean weight of the 25 bags of pretzels sampled is more than two standard deviations from 454 g, we reject the null hypothesis ( = 454 g) and conclude that the alternative hypothesis ( ) is true. The data provide sufficient evidence to conclude that the packaging machine is not working properly. Slide 11 Slide 11 - 60 Figure 9.5 The set of values for the test statistic that leads us to reject the null hypothesis is called the rejection region. In this case, the rejection region consists of all z-scores that lie either to the left of 2 or to the right of 2; that part of the horizontal axis under the shaded areas in Fig. 9.5. Slide 12 Slide 12 - 60 Definition 9.2 Slide 13 Slide 13 - 60 For a two-tailed test, the null hypothesis is rejected when the test statistic is either too small or too large. The rejection region consists of two parts: one on the left and one on the right, (Fig. 9.6(a)). For a left-tailed test, the null hypothesis is rejected only when the test statistic is too small. The rejection region consists of only one part, on the left, (Fig.9.6(b)). For a right-tailed test, the null hypothesis is rejected only when the test statistic is too large. The rejection region consists of only one part, on the right, (Fig.9.6(c)). Figure 9.6 Slide 14 Slide 14 - 60 Definition 9.3 Slide 15 Slide 15 - 60 Definition 9.4 Slide 16 Slide 16 - 60 Key Fact 9.1 Slide 17 Slide 17 - 60 Key Fact 9.2 Slide 18 Slide 18 - 60 Key Fact 9.3 Slide 19 Slide 19 - 60 Procedure 9.1 Slide 20 Slide 20 - 60 Procedure 9.1 (cont.) Slide 21 Slide 21 - 60 Key Fact 9.4 Slide 22 Slide 22 - 60 Example 9.10 The manufacturer of a new model car, the Orion, claims that a typical car gets 26 miles per gallon (mpg). A consumer advocacy group is skeptical of this claim and thinks that the mean gas mileage, , of all Orions may be less than 26 mpg. The group plans to perform the hypothesis test H 0 : = 26 mpg (manufacturers claim) H a : < 26 mpg (consumer groups conjecture), at the 5% significance level, using a sample of 30 Orions. Do the data provide sufficient evidence that the manufacturers claim is incorrect for: a. 25.8 mpg.b. 25.0 mpg. Assume that gas mileages of Orions are normally distributed with a standard deviation of 1.4 mpg. Slide 23 Slide 23 - 60 Solution Example 9.10 The inference under consideration is a left-tailed hypothesis test for a population mean at the 5% significance level. The test statistic is and the critical value is z = z 0.05 = 1.645. Thus the decision criterion for the hypothesis test is: If z 1.645, reject H 0 ; if z > 1.645, do not reject H 0. Computing error probabilities is somewhat simpler if the decision criterion is expressed in terms of instead of z. The decision criterion can thus be expressed in terms of as: If 25.6 mpg, reject H 0 ; if > 25.6 mpg, do not reject H 0. See Fig. 9.14. Slide 24 Slide 24 - 60 Solution Example 9.10 Figure 9.14 Slide 25 Slide 25 - 60 Solution Example 9.10 a. If = 25.8 mpg, then Thus, the variable is normally distributed with a mean of 25.8 mpg and a standard deviation of 0.26 mpg. The normal curve for is shown in Fig. 9.15. and is normally distributed. Slide 26 Slide 26 - 60 Figure 9.15 Slide 27 Slide 27 - 60 Solution Example 9.10 A Type II error occurs if we do not reject H 0, that is, if > 25.6 mpg. The probability of this happening equals the percentage of all samples whose means exceed 25.6 mpg, which we obtain in Fig. 9.15. Thus, if the true mean gas mileage of all Orions is 25.8 mpg, the probability of making a Type II error is 0.7794; that is, = 0.7794. Interpretation There is roughly a 78% chance that the consumer group will fail to reject the manufacturers claim that the mean gas mileage of all Orions is 26 mpg when in fact the true mean is 25.8 mpg. Although this result is a rather high chance of error, we probably would not expect the hypothesis test to detect such a small difference in mean gas mileage (25.8 mpg as opposed to 26 mpg) with a sample size of only 30. Slide 28 Slide 28 - 60 Solution Example 9.10 b. We proceed as we did in part (a) but this time assume that = 25.0 mpg. Fig. 9.16 shows the required computations. Figure 9.16 Slide 29 Slide 29 - 60 Solution Example 9.10 If the true mean gas mileage of all Orions is 25.0 mpg, the probability of making a Type II error is 0.0104; that is, = 0.0104. Interpretation There is only about a 1% chance that the consumer group will fail to reject the manufacturers claim that the mean gas mileage of all Orions is 26 mpg when in fact the true mean is 25.0 mpg. Slide 30 Slide 30 - 60 Figure 9.17 We combined Figs. 9.15 and 9.16 with two others. Figure 9.17 shows clearly that the farther the true mean is from the null hypothesis mean of 26 mpg, the smaller will be the probability of a Type II error. This result is hardly surprising: We would expect that a false null hypothesis is more likely to be detected when the true mean is far from the null hypothesis mean than when the true mean is close to the null hypothesis mean. Slide 31 Slide 31 - 60 Definition 9.5 Slide 32 Slide 32 - 60 In reality, the true value of the parameter in question will be unknown. Consequently, constructing a table of powers for various values of the parameter is helpful in evaluating the effectiveness of the hypothesis test. For the gas mileage illustration where the parameter in question is the mean gas mileage, , of all Orions we have already obtained the Type II error probability, , when the true mean is 25.8 mpg, 25.6 mpg, 25.3 mpg, and 25.0 mpg, as depicted in Fig. 9.17. Similar calculations yield the other probabilities shown in the second column of Table 9.8. The third column of Table 9.8 shows the power that corresponds to each value of , obtained by subtracting from 1. Slide 33 Slide 33 - 60 Table 9.8 Slide 34 Slide 34 - 60 We can use Table 9.8 to evaluate the overall effectiveness of the hypothesis test. We can also obtain from Table 9.8 a visual display of that effectiveness by plotting points of power against and then connecting the points with a smooth curve. The resulting curve is called a power curve and is shown in Fig. 9.18 on the next slide. In general, the closer a power curve is to 1 (i.e., the horizontal line 1 unit above the horizontal axis), the better the hypothesis test is at detecting a false null hypothesis. Slide 35 Slide 35 - 60 Figure 9.18 Slide 36 Slide 36 - 60 Figure 9.20 The smaller we specify the significance level, the smaller will be the power. However, by using a large sample, we can have both a small significance level and large power. Here is a graph of power for = 0.05 and samples sizes of n = 30 and n = 100.