两道概率题,求高手解答
a)Giventhepopulationstandarddeviation=15andMarginoferror(i.e.tolerance)=2,computether...
a)Given the population standard deviation = 15 and Margin of error(i.e. tolerance) = 2, compute the required sample size for 90% confidence level.
b) If the standard deviation is now 4 times of the original, and the accuracy is thrice as accurate as before, what would be the impact on the sample size. 展开
b) If the standard deviation is now 4 times of the original, and the accuracy is thrice as accurate as before, what would be the impact on the sample size. 展开
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a) α=0.1, so we need to find t-test(α/2,n-1) which means t-test0.05,n-1. We don't know exact number of simple n, but we could assume an initial value like 21. Normally you can check appendix of your probability textbook to find that t-test0.05, 20 = 1.725. Simple size for 90% confidence level =>
R1 = (1.725*15/2)^2 = 167.38
Noticing that 167 is greater than our assumption 21. so we need to enlarge assumption simple size to at least 167 and do this again. T-test0.05,166 = 1.654.
R2 = (1.654*15/2)^2 = 153.88
This number of simple is smaller than 167, so, the required simple size of 90% confidence level is 154.
b) Now the standard deviation is 60 and margin of error is 6. We assume initial value of n is 501. T-test0.05,500 = 1.648
R1= (1.648*60/6)^2 = 271.59
This time, required simple size is 272.
R1 = (1.725*15/2)^2 = 167.38
Noticing that 167 is greater than our assumption 21. so we need to enlarge assumption simple size to at least 167 and do this again. T-test0.05,166 = 1.654.
R2 = (1.654*15/2)^2 = 153.88
This number of simple is smaller than 167, so, the required simple size of 90% confidence level is 154.
b) Now the standard deviation is 60 and margin of error is 6. We assume initial value of n is 501. T-test0.05,500 = 1.648
R1= (1.648*60/6)^2 = 271.59
This time, required simple size is 272.
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