here error =t*standard deviation/sqrt(n)
therefore ; critical t =10*sqrt(80)/62.35 =1.435
for above test statistic at (n-1=79) degree of freedom; level of confidence =84.72%
option D is correct
56)Consider a random sample of n 80 and S62.35. Find the level of confidence (1- a%...
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{Exercise 8.14 Algorithmic} A simple random sample with n = 56 provided a sample mean of 21.0 and a sample standard deviation of 4.5. a. Develop a 90% confidence interval for the population mean (to 2 decimals). ( 20 ®, 22 ) b. Develop a 95% confidence interval for the population mean (to 2 decimals). c. Develop a 99% confidence interval for the population mean (to 2 decimals). d. What happens to the margin of error and the confidence interval...
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