help me! 1. You work for a consumeradvocate agency and want to find the mean rep
ID: 2954958 • Letter: H
Question
help me! 1. You work for a consumeradvocate agency and want to find the mean repair cost of a washingmachine. As part of your study, you randomly select 50 repair costsand find the mean to be $139.00. The sample atandard deviation is$17.10. Complete part (a) and (b) 2. Construct the confidence interval for theppoulation mean . C=0.95,-/x = 16.4, s=7.0 and n=60 A 95% confidence interval for is= (Round to one decimal placeneeded.) 3. Construct the indicated confidenceinterval for the population mean using (a) at-distribution,(b) if you had incorrectly used a normaldistribution, Which interval would be wider? C=0.95, -/x=14.4, s=20,n=9 4. Find margin of error for the given values ofC, S, and N. C=0.90, S=2.1,N=100 E= (round to three decimals places.)help me! 1. You work for a consumeradvocate agency and want to find the mean repair cost of a washingmachine. As part of your study, you randomly select 50 repair costsand find the mean to be $139.00. The sample atandard deviation is$17.10. Complete part (a) and (b) 2. Construct the confidence interval for theppoulation mean . C=0.95,-/x = 16.4, s=7.0 and n=60 A 95% confidence interval for is= (Round to one decimal placeneeded.) 3. Construct the indicated confidenceinterval for the population mean using (a) at-distribution,(b) if you had incorrectly used a normaldistribution, Which interval would be wider? C=0.95, -/x=14.4, s=20,n=9 4. Find margin of error for the given values ofC, S, and N. C=0.90, S=2.1,N=100 E= (round to three decimals places.)
Explanation / Answer
(2) Given =0.05, Z(0.025)=1.96 (check normal table) The 95% CI is xbar ± Z*s/n --> 16.4 ± 1.96*7/sqrt(60) --> ( 14.62, 18.17)
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