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Chapter 9 Hypothesis Tests for One Population Mean Slide 1 - 60 Definition 9.1 Slide 2 - 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 3 - 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 xx , that is, the sampling distribution of the sample mean for samples of size 25. d. Obtain a precise criterionx 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 4 - 60 Solution Example 9.4 a. The null and alternative hypotheses, as stated in Example 9.1, are H0 : μ = 454 g (the packaging machine is working properly) Ha : m ¹ 454g (the packaging machine is not working properly). b. If the null hypothesis is true, then the mean weight, x , 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 5 - 60 Solution Example 9.4 c. n = 25, σ = 7.8, weights are normally distributed, ・ μ x = μ (which we don’t know), ・ σ x = s n = 7.8 25 = 1.56 and ・ x 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 2•1.56 = 3.12 g of μ. Slide 6 - 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 μ, as illustrated in Fig. 9.1. Figure 9.1 Slide 7 - 60 Solution Example 9.4 d. If the mean weight, x , 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 8 - 60 Solution Example 9.4 e. The mean weight, x , of the sample of 25 bags of pretzels whose weights are given in Table 9.1 is 450g. So, z =( x − 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 9 - 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 (μ = 454 g) and conclude that the alternative hypothesis ( m ¹ 454g) is true. The data provide sufficient evidence to conclude that the packaging machine is not working properly. Slide 10 - 60 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. Figure 9.5 Slide 11 - 60 Definition 9.2 Slide 12 - 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 13 - 60 Definition 9.3 Slide 14 - 60 Definition 9.4 Slide 15 - 60 Key Fact 9.1 Slide 16 - 60 Key Fact 9.2 Slide 17 - 60 Key Fact 9.3 Slide 18 - 60 Procedure 9.1 Slide 19 - 60 Procedure 9.1 (cont.) Slide 20 - 60 Key Fact 9.4 Slide 21 - 60 Definition 9.6 Slide 38 - 60 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 z0 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 − |z0 | to |z0 |, as in Fig. 9.21(a). Figure 9.21 Slide 39 - 60 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 z0, as in Fig. 9.21(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 z0, as in Fig. 9.21(c). Figure 9.21 Slide 40 - 60 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. 9.24. The null 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. Figure 9.24 Slide 41 - 60 Procedure 9.2 Slide 42 - 60 Procedure 9.2 (cont.) Slide 43 - 60 Table 9.11 Slide 44 - 60 Table 9.12 Slide 45 - 60 Example 9.16 Use Table IV to estimate the P-value of each one-mean ttest. 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. 9.27(a). Figure 9.27 Slide 46 - 60 Solution Example 9.16 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 t0.05 = 1.796 and t0.025 = 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. 9.27(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 H0 at any significance level of 0.05 or larger, and we cannot reject H0 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 H0. Slide 47 - 60 Solution Example 9.16 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. 9.28(a). Figure 9.28 Slide 48 - 60 Solution Example 9.16 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 t0.10 = 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. 9.28(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 H0 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 H0. Slide 49 - 60 Procedure 9.3 Slide 50 - 60 Procedure 9.3 (cont.) Slide 51 - 60 Flowchart for choosing the correct hypothesis testing procedure for a population mean. Figure 9.34 Slide 59 - 60 Homework When able, do the following by hand and check your answer with minitab. 9.54, 55, 57, 59, 62 9.117, 120 9.141, 142, 143 Slide 60 - 60