Statistics Homework Solutions 163

Statistics Homework - 163 SUPPLEMENTARY EXERCISES FOR CHAPTER 6 Supplementary Exercises for Chapter 6 1 This requires a test for the difference

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Unformatted text preview: 163 SUPPLEMENTARY EXERCISES FOR CHAPTER 6 Supplementary Exercises for Chapter 6 1. This requires a test for the difference between two means. The data are unpaired. Let µ 1 represent the population mean annual cost for cars using regular fuel, and let µ2 represent the population mean annual cost for cars using premium fuel. Then the appropriate null and alternate hypotheses are H0 : µ1 µ2 0 versus H1 : µ1 µ2 0. The test statistic is the difference between the sample mean costs between the two groups. The z table should be used to find the P-value. 3. This requires a test for a population proportion. Let p represent the population proportion of defective parts under the new program. The appropriate null and alternate hypotheses are H0 : p 0 10 versus H1 : p 0 10. The test statistic is the sample proportion of defective parts. The z table should be used to find the P-value. § ¡ ¡ ¡ § 16 2 739. ¨ ¡ ¤£ ¨ 0 011. ¡ 0 025. A computer package gives P 2 739. ¨ ¨ P ¡ ¨ ¡ ¤ ©¥ £ ¨ ¡ ¤ ¡ ¡ From the t table, 0 01 µ0 , the P-value is the area to the left of t ¡ ¡ (c) Since the alternate hypothesis is of the form µ ¡ ¡¥ 10 16. ¨ ¨ 0 13047 ¡ ¡ 16 9 degrees of freedom. 16 versus H1 : µ § § 15 887 1 ¡ 10. There are 10 ¡ ¡ 0 13047, n The null and alternate hypotheses are H0 : µ t ¨ 15 887, s ¡ (b) X 16 versus H1 : µ ¨ 5. (a) H0 : µ ¨ ¡ We conclude that the mean fill weight is less than 16 oz. 7 ¡¥ ¡ ¨ ¡ ¡ ¤£ 2008 147 7. There are 7 0. 1 6 degrees of freedom. The null and alternate hypothe- 2 119. ¨ ¨ ¨ ¤ ¥ £ 0 ¡ 1608 143 ¡ ¡ ¨ ¡ ¤ ¡ t ¡ (b) D 1608 143, sD 2008 147, n ses are H0 : µD 0 versus H1 : µD 0 ¡ µ2 ¡ 0 versus H1 : µ1 ¡ µ2 ¡ 7. (a) H0 : µ1 ¡ ¨ ¡ 0 010. A computer package gives P 0 078. ¨ ¨ ¡ P ¡ ¡ ¡ ¨ The null hypothesis is suspect, but one would most likely not firmly conclude that it is false. 4 61. ¡ ¨ ¤ ¥ £ 0. 5. µ0 , the P-value is the area to the right of z ¡ ¤£ ¡ ¨ ¤ ¡ ¡ Since the alternate hypothesis is of the form µ Thus P 5 versus H1 : µ 4 61. ¨ 85 ¡¥ 12 85. The null and alternate hypotheses are H0 : µ ¨ ¨ ¡ 5 ¨ 56 1 2, n ¡ z 5 6, s ¨ X ¡ 9. ¡ From the t table, 0 05 µ0 , the P-value is the sum of the areas to the right of t 2 119 ¨ (c) Since the alternate hypothesis is of the form µD and to the left of t 2 119. ...
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This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

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