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. ยก ยคยฃ 01 ยจ 100 and standard deviation ฯX ยจ ยก 11. (a) The null distribution of X is normal with mean ยต ยก ยต0 , the rejection region will consist of both the upper and ยก 1 96 0 01 1 96, respectively. 99 9804. ยกยฅ ยจ ยค ยจ ยก ยจ ยข 99 9804. ยจ 100 0196 and 100 ยจ 100 0196 or if X ยกยฅ ยจ ยค ยจ ยจ 100 ยคยฃ 01 0 01. ยก ยจ 100 and standard deviation ฯX ยก ยง ยจ ยจ (b) The null distribution of X is normal with mean ยต ยจ Reject H0 if X 1 96 0 01 ยก Therefore the boundaries are 100 1 96 and z ยก The z-scores corresponding to the boundaries of upper and lower 2.5% are z ยจ ยก Since the alternate hypothesis is of the form ยต lower 2.5% of the null distribution. ยก ยต0 , the rejection region will consist of both the upper and 99 98355. ยกยฅ 1 645 0 01 1 645, respec- ยจ ยก ยจ ยค ยจ ยก ยจ ยกยฅ ยจ ยค ยจ 99 98355. ยจ ยข 100 01645 or if X 100 01645 and 100 ยจ ยจ ยจ ยง Reject H0 if X 1 645 0 01 ยก Therefore the boundaries are 100 1 645 and z ยก The z-scores corresponding to the boundaries of upper and lower 5% are z tively. ยจ ยก Since the alternate hypothesis is of the form ยต 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. ยจ ยก ยจ ยฃ ยฅ ยจ ยค ยจ 1 5. 0 1336, or 13.36%. ยจ ยฅ ยจ ยก ยก ยจ ยก ยค ยจ ยฃ ยฅ 0 0668 100 0 01 ยจ ยก 1 5. The z-score of 99.985 is 99 985 ยก ยค 100 and standard deviation ฯX 100 0 01 ยจ ยข ยก ยจ ยค The level of the test is therefore 0 0668 ยง The z-score of 100.015 is 100 015 ยจ The null distribution is normal with mean ยต 100 015 , computed under the null distribution. ยก 99 985 + P X ยฅ The level of the test is the sum P X ยจ 13. (a) The null hypothesis speci๏ฌes a single value for the mean: ยต 3. The level, which is 5%, is therefore the probability that the null hypothesis will be rejected when ยต 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 ยต 3 is 0.05. ยก ยก ยก (b) First ๏ฌnd the rejection region. ยคยฃ 0 10 50 0 014142. ยก ยจ ยก 3 and standard deviation ฯX ยจ ยก The null distribution of X is normal with mean ยต ยต0 , the rejection region will consist of both the upper and ยกยฅ 1 96 0 014142 2 9723. ยจ ยค ยจ ยก ยจ ยกยฅ ยจ 2 9723. ยจ ยง ยจ ยค ยจ ยก ยฅ ยจ 2 67. ยจ ยก ยจ ยก ยก ยฃ ยฉยฅ ยจ 3 01 0 014142 ยง ยข ยจ ยฃ ยฉยฅ ยจ ยจ ยก ยก ยจ ยจ ยค ยค The z-score of 2.9723 is 2 9723 1 25. ยจ ยจ 3 01 0 014142 3 0277 ยขยฅ ยจ ยค The probability that the equipment will be recalibrated is therefore equal to P X P X 2 9723 , computed under the assumption that ยต 3 01. The z-score of 3.0277 is 3 0277 1 96, respec- ยจ ยก 3 0277 or if X 3 0277 and 3 ยจ H0 will be rejected if X 1 96 0 014142 ยก Therefore the boundaries are 3 1 96 and z ยก The z-scores corresponding to the boundaries of upper and lower 2.5% are z tively. ยจ ยก Since the alternate hypothesis is of the form ยต lower 2.5% of the null distribution. ยจ ยค ...
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