185323959-Business-Stats-Ken-Black-Case-Answers.pdf

However due to the relatively small samples the error

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However, due to the relatively small samples, the error of the interval is 18.45% which is greater than the point estimate. Combining the error of the interval with the point estimate results in the confidence interval shown above. Note that zero is in the interval indicating that there is a possibility that there is no difference in the quality ratings of tractors produced at the two plants. If this were a hypothesis testing problem, then the decision would be to fail to reject the null hypothesis based on the confidence interval’s inclusion of zero.
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Case Notes 26 3. 2000 vs. 2001: t = -1.83 with a p -value of .07. This is not significant at α = .05. There is no significant difference in the mean ratings between 2000 and 2001. This is underscored by the confidence interval that includes zero. However, if α = .10 were used, there would be a significant difference. Examining the means reveals that the mean score for 2002 was higher. The sample sizes were 75 for 2000 and 93 for 2001. 4. Comparison of variances for week 1 and week 5: H 0 : σ 1 2 = σ 2 2 H a : σ 1 2 σ 2 2 Let α = .05 α /2 = .025 df 1 = n 1 - 1 = 5 df 2 = n 2 - 1 = 6 F .025,5,6 = 5.99 F .975,6,5 = 1/5.99 = 0.167 Week 1: n 1 = 6 s 1 = 1.17 Week 5: n 2 = 7 s 2 = 1.68 2 2 2 2 2 1 ) 68 . 1 ( ) 17 . 1 ( = = s s F = 0.485 Since the observed value of F = 0.485 is > the left tail critical value of F = 0.167, the decision is to fail to reject the null hypothesis. The variances of product being produced these two weeks are not significantly different. Management would probably like this because this indicative of consistent production patterns. Wide swings in variance would be of concern because it would indicate that some weeks the variability is more out-of-control than others and a less consistent product is being produced.
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Case Notes 27 Chapter 11 J. R. Clarkson Company 1. The two by three factorial design is analyzed using a two-way ANOVA. There are two independent variables, temperature and supplier. Temperature has three treatment levels: 70 o , 110 o , and 150 o . Supplier has two classifications levels: supplier A and supplier B. The dependent variable is strength of the valve as measured in psi. Shown below is Excel output for this analysis. ANOVA Source of Variation SS df MS F P-value F crit Supplier 20.056 1 20.056 1.11 0.313385 4.75 Temperature 800.111 2 400.056 22.09 0.000095 3.89 Interaction 84.778 2 42.389 2.34 0.138598 3.89 Within 217.333 12 18.111 Total 1122.278 17 First, we examine the observed F for interaction which is 2.34 with a p -value of .1386. Since interaction is not significant at any commonly used alpha, we proceed to examine main effects. There is no significant difference between the two suppliers ( F = 1.11, p -value = .31339). There is a significant difference in the strength of the valves by temperature at α = .0001. The mean psi for 70 o is 159.83, for 110 o is 158.33, and for 150 o is 145 psi. It appears that at 150 o the valves are not as strong. Using MINITAB, Tukey’s multiple comparison tests were done to determine if there were any significant differences in valve strength by temperature. The results are: Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0203 Critical value = 3.67 Intervals for (column level mean) - (row level mean) 1 2 2 -5.444 8.444 3 7.890 6.390 21.777 20.277
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