2. Slide 2 - 60 Copyright © 2008 Pearson Education, Inc. Chapter 9 Hypothesis Tests for One Population Mean

3. Slide 3 - 60 Copyright © 2008 Pearson Education, Inc. Definition 9.1

4. Slide 4 - 60 Copyright © 2008 Pearson Education, Inc. 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

5. Slide 5 - 60 Copyright © 2008 Pearson Education, Inc. 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.

6. Slide 6 - 60 Copyright © 2008 Pearson Education, Inc. Solution Example 9.4 a.The null and alternative hypotheses, as stated in Example 9.1, are H 0 : μ = 454 g (the packaging machine is working properly) H a : μ = 454 g (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.”

7. Slide 7 - 60 Copyright © 2008 Pearson Education, Inc. Solution Example 9.4 c.n = 25, σ = 7.8, weights are normally distributed, ･ μ = μ (which we don’t know), ･ σ = and ･ is normally distributed. In other words, for samples of size 25, the variable is normally distributed with mean μ and standard deviation 1.56 g. d.The “95.44” part of the 68.26–95.44–99.74 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 μ.

8. Slide 8 - 60 Copyright © 2008 Pearson Education, Inc. 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 μ, as illustrated in Fig. 9.1. Figure 9.1

9. Slide 9 - 60 Copyright © 2008 Pearson Education, Inc. 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

10. Slide 10 - 60 Copyright © 2008 Pearson Education, Inc. 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

11. Slide 11 - 60 Copyright © 2008 Pearson Education, Inc. 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 (μ = 454 g) and conclude that the alternative hypothesis (μ = 454 g) is true. The data provide sufficient evidence to conclude that the packaging machine is not working properly.

12. Slide 12 - 60 Copyright © 2008 Pearson Education, Inc. Definition 9.2

13. Slide 13 - 60 Copyright © 2008 Pearson Education, Inc. 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

14. Slide 14 - 60 Copyright © 2008 Pearson Education, Inc. Definition 9.3

15. Slide 15 - 60 Copyright © 2008 Pearson Education, Inc. Definition 9.4

16. Slide 16 - 60 Copyright © 2008 Pearson Education, Inc. Key Fact 9.1

17. Slide 17 - 60 Copyright © 2008 Pearson Education, Inc. Key Fact 9.2

18. Slide 18 - 60 Copyright © 2008 Pearson Education, Inc. Key Fact 9.3

19. Slide 19 - 60 Copyright © 2008 Pearson Education, Inc. Procedure 9.1

20. Slide 20 - 60 Copyright © 2008 Pearson Education, Inc. Procedure 9.1 (cont.)

21. Slide 21 - 60 Copyright © 2008 Pearson Education, Inc. Key Fact 9.4

22. Slide 22 - 60 Copyright © 2008 Pearson Education, Inc. Definition 9.5

23. Slide 23 - 60 Copyright © 2008 Pearson Education, Inc. Figure 9.14 If the null hypothesis is true, this test statistic has the standard normal distribution, and its probabilities equal areas under the standard normal curve. If we let z 0 denote the observed value of the test statistic z, we obtain the P-value as follows: Two-tailed test: The P-value is the probability of observing a value of the test statistic z that is at least as large in magnitude as the value actually observed, which is the area under the standard normal curve that lies outside the interval from − |z 0 | to |z 0 |, as in Fig. 9.14(a).

24. Slide 24 - 60 Copyright © 2008 Pearson Education, Inc. Figure 9.14 Left-tailed test: The P-value is the probability of observing a value of the test statistic z that is as small as or smaller than the value actually observed, which is the area under the standard normal curve that lies to the left of z 0, as in Fig. 9.14(b). Right-tailed test: The P-value is the probability of observing a value of the test statistic z that is as large as or larger than the value actually observed, which is the area under the standard normal curve that lies to the right of z 0, as in Fig. 9.14(c).

25. Slide 25 - 60 Copyright © 2008 Pearson Education, Inc. Figure 9.17 hypothesis would be rejected for a test at the 0.05 significance level but would not be rejected for a test at the 0.01 significance level. In fact, the P-value is precisely the smallest significance level at which the null hypothesis would be rejected. The P-value can be interpreted as the observed significance level of a hypothesis test. Suppose that the value of the test statistic for a right-tailed z-test turns out to be 1.88. Then the P-value of the hypothesis test is 0.03 (actually 0.0301), the shaded area in Fig. 9.17. The null

26. Slide 26 - 60 Copyright © 2008 Pearson Education, Inc. Procedure 9.2

27. Slide 27 - 60 Copyright © 2008 Pearson Education, Inc. Procedure 9.2 (cont.)

28. Slide 28 - 60 Copyright © 2008 Pearson Education, Inc. Table 9.9

29. Slide 29 - 60 Copyright © 2008 Pearson Education, Inc. Table 9.10

30. Slide 30 - 60 Copyright © 2008 Pearson Education, Inc. Example 9.14 Use Table IV to estimate the P-value of each one-mean t- test. a. Left-tailed test, n = 12, and t = −1.938 b. Two-tailed test, n = 25, and t = −0.895 Solution a. Because the test is left-tailed, the P-value is the area under the t-curve with df = 12−1 = 11 that lies to the left of −1.938, as shown in Fig. 9.20(a). Figure 9.20

31. Slide 31 - 60 Copyright © 2008 Pearson Education, Inc. Solution Example 9.14 A t-curve is symmetric about 0, so the area to the left of −1.938 equals the area to the right of 1.938, which we can estimate by using Table IV. In the df = 11 row of Table IV, the two t-values that straddle 1.938 are t 0.05 = 1.796 and t 0.025 = 2.201. Therefore the area under the t-curve that lies to the right of 1.938 is between 0.025 and 0.05, as shown in Fig. 9.20(b). Consequently, the area under the t-curve that lies to the left of −1.938 is also between 0.025 and 0.05, so 0.025 < P < 0.05. Hence we can reject H 0 at any significance level of 0.05 or larger, and we cannot reject H 0 at any significance level of 0.025 or smaller. For significance levels between 0.025 and 0.05, Table IV is not sufficiently detailed to help us to decide whether to reject H 0.

32. Slide 32 - 60 Copyright © 2008 Pearson Education, Inc. Solution Example 9.14 b. Because the test is two tailed, the P-value is the area under the t-curve with df = 25 − 1 = 24 that lies either to the left of −0.895 or to the right of 0.895, as shown in Fig. 9.21(a). Figure 9.21

33. Slide 33 - 60 Copyright © 2008 Pearson Education, Inc. Solution Example 9.14 Because a t-curve is symmetric about 0, the areas to the left of −0.895 and to the right of 0.895 are equal. In the df = 24 row of Table IV, 0.895 is smaller than any other t-value, the smallest being t 0.10 = 1.318. The area under the t-curve that lies to the right of 0.895, therefore, is greater than 0.10, as shown in Fig. 9.21(b). Consequently, the area under the t-curve that lies either to the left of −0.895 or to the right of 0.895 is greater than 0.20, so P > 0.20. Hence we cannot reject H 0 at any significance level of 0.20 or smaller. For significance levels larger than 0.20, Table IV is not sufficiently detailed to help us to decide whether to reject H 0.

34. Slide 34 - 60 Copyright © 2008 Pearson Education, Inc. Procedure 9.3

35. Slide 35 - 60 Copyright © 2008 Pearson Education, Inc. Procedure 9.3 (cont.)