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**Unformatted text preview: **11.4 Hypothesis Testing 415 Figure 11.5 Critical Region for One-Tailed Tests ation of 120 hours. If the president of the company claims that the mean lifetime E [ X ] of all the lightbulbs produced by the company is 1600 hours, test the hypoth- esis that E [ X ] is not equal to 1600 hours using a level of significance of (a) 0.05 and (b) 0.01. Solution The null hypothesis is H : E [ X ] = 1600 hours Similarly, the alternative hypothesis is H 1 : E [ X ] negationslash= 1600 hours Since E [ X ] negationslash= 1600 includes numbers that are both greater than and less than 1600, this is a two-tailed test. From the available data, the normalized value of the sample mean is z = overbar X − E [ overbar X ] σ X / √ n = 1570 − 1600 120 / √ 100 = − 30 12 = − 2 . 50 (a) At a level of significance of 0.05, z c = ± 1 . 96 for a two-tailed test. Thus, our acceptance region is [− 1 . 96 , 19 . 6 ] of the standard normal distribution. The rejection and acceptance regions are illustrated in Figure 11.6. Since z = − 2 . 50 lies outside the range [− 1 . 96 , 1 . 96 ] (that is, it is in the rejection region), we reject H at the 0.05 level of significance, which means that the difference in mean lifetimes is statistically significant. 416 Chapter 11 Introduction to Statistics Figure 11.6 Critical Region for Problem 11.9(a) Figure 11.7 Critical Region for Problem 11.9(b) (b) At the 0.01 level of significance, z c = ± 2 . 58. The acceptance and rejection regions are shown in Figure 11.7. Since z = − 2 . 50 lies within the range [− 2 . 58 , 2 . 58 ] , which is the acceptance region, we accept H at the 0.01 level of significance, which means that the difference in mean lifetimes is not sta- tistically significant. trianglesolid Example 11.10 For Example 11.9, test the hypothesis that E [ X ] is less than 1600 hours using a level of significance of (a) 0.05 and (b) 0.01. Solution Here we define the null hypothesis and alternative hypothesis as fol- lows: 11.4 Hypothesis Testing 417 Figure 11.8 Critical Region for Problem 11.10(a) H : E [ X ] = 1600 hours H 1 : E [ X ] < 1600 hours This is a one-tailed test; or more precisely, a left-tail test. Since the z-score is the same as in Example 11.9, we only need to find the confidence limits for the two cases. (a) Since H 1 is concerned with values that are less than E [ X ] , we have a left- tail test, which means that we choose the rejection region that is below the acceptance region. Therefore, we choose z c = − 1 . 645 for the 0.05 level of significance in Table 11.3. Since z = − 2 . 50 lies in the rejection region (i.e., − 2 . 50 < − 1 . 645), as illustrated in Figure 11.8, we reject H at the 0.05 level of significance and thus accept H 1 . This implies that the difference in mean lifetimes is statistically significant....

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