Statistics Homework Solutions 164

Statistics Homework Solutions 164 - 164 CHAPTER 6 100 0 01....

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Unformatted text preview: 164 CHAPTER 6 100 0 01. ¡ ¤£ 01 ¨ 100 and standard deviation σX ¨ ¡ 11. (a) The null distribution of X is normal with mean µ ¡ µ0 , the rejection region will consist of both the upper and ¡ 1 96 0 01 1 96, respectively. 99 9804. ¡¥ ¨ ¤ ¨ ¡ ¨ ¢ 99 9804. ¨ 100 0196 and 100 ¨ 100 0196 or if X ¡¥ ¨ ¤ ¨ ¨ 100 ¤£ 01 0 01. ¡ ¨ 100 and standard deviation σX ¡ § ¨ ¨ (b) The null distribution of X is normal with mean µ ¨ Reject H0 if X 1 96 0 01 ¡ Therefore the boundaries are 100 1 96 and z ¡ The z-scores corresponding to the boundaries of upper and lower 2.5% are z ¨ ¡ Since the alternate hypothesis is of the form µ lower 2.5% of the null distribution. ¡ µ0 , the rejection region will consist of both the upper and 99 98355. ¡¥ 1 645 0 01 1 645, respec- ¨ ¡ ¨ ¤ ¨ ¡ ¨ ¡¥ ¨ ¤ ¨ 99 98355. ¨ ¢ 100 01645 or if X 100 01645 and 100 ¨ ¨ ¨ § Reject H0 if X 1 645 0 01 ¡ Therefore the boundaries are 100 1 645 and z ¡ The z-scores corresponding to the boundaries of upper and lower 5% are z tively. ¨ ¡ Since the alternate hypothesis is of the form µ lower 5% of the null distribution. (c) Yes (d) No (e) Since this is a two-tailed test, there are two critical points, equidistant from the null mean of 100. Since one critical point is 100.015, the other is 99.985. 0 01. ¨ ¡ ¨ £ ¥ ¨ ¤ ¨ 1 5. 0 1336, or 13.36%. ¨ ¥ ¨ ¡ ¡ ¨ ¡ ¤ ¨ £ ¥ 0 0668 100 0 01 ¨ ¡ 1 5. The z-score of 99.985 is 99 985 ¡ ¤ 100 and standard deviation σX 100 0 01 ¨ ¢ ¡ ¨ ¤ The level of the test is therefore 0 0668 § The z-score of 100.015 is 100 015 ¨ The null distribution is normal with mean µ 100 015 , computed under the null distribution. ¡ 99 985 + P X ¥ The level of the test is the sum P X ¨ 13. (a) The null hypothesis specifies a single value for the mean: µ 3. The level, which is 5%, is therefore the probability that the null hypothesis will be rejected when µ 3. The machine is shut down if H0 is rejected at the 5% level. Therefore the probability that the machine will be shut down when µ 3 is 0.05. ¡ ¡ ¡ (b) First find the rejection region. ¤£ 0 10 50 0 014142. ¡ ¨ ¡ 3 and standard deviation σX ¨ ¡ The null distribution of X is normal with mean µ µ0 , the rejection region will consist of both the upper and ¡¥ 1 96 0 014142 2 9723. ¨ ¤ ¨ ¡ ¨ ¡¥ ¨ 2 9723. ¨ § ¨ ¤ ¨ ¡ ¥ ¨ 2 67. ¨ ¡ ¨ ¡ ¡ £ ©¥ ¨ 3 01 0 014142 § ¢ ¨ £ ©¥ ¨ ¨ ¡ ¡ ¨ ¨ ¤ ¤ The z-score of 2.9723 is 2 9723 1 25. ¨ ¨ 3 01 0 014142 3 0277 ¢¥ ¨ ¤ The probability that the equipment will be recalibrated is therefore equal to P X P X 2 9723 , computed under the assumption that µ 3 01. The z-score of 3.0277 is 3 0277 1 96, respec- ¨ ¡ 3 0277 or if X 3 0277 and 3 ¨ H0 will be rejected if X 1 96 0 014142 ¡ Therefore the boundaries are 3 1 96 and z ¡ The z-scores corresponding to the boundaries of upper and lower 2.5% are z tively. ¨ ¡ Since the alternate hypothesis is of the form µ lower 2.5% of the null distribution. ¨ ¤ ...
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