Question

In a random sample of four mobile devices, the mean repair cost was $60.00 and the standard deviation was $14.00. Assume the population is normally distributed and use a t-distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results. The 90% confidence interval for the population mean is (DO (Round to two decimal places as needed.) The margin of error is $ (Round to two decimal places as needed.) Interpret the results. Choose the correct answer below. O A. With 90% confidence, it can be said that the population mean repair cost is between the bounds of the confidence interval. OB. With 90% confidence, it can be said that the repair cost is between the bounds of the confidence interval. O c. If a large sample of mobile devices are taken approximately 90% of them will have repair costs between the bounds of the confidence interval. OD. It can be said that 90% of mobile devices have a repair cost between the bounds of the confidence interval.

Answer 1

sample mean, xbar = 60
sample standard deviation, s = 14
sample size, n = 4
degrees of freedom, df = n - 1 = 3

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 2.353

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (60 - 2.353 * 14/sqrt(4) , 60 + 2.353 * 14/sqrt(4))
CI = (43.53 , 76.47)

ME = tc * s/sqrt(n)
ME = 2.353 * 14/sqrt(4)
ME = 16.47

Margin of error is 16.47

OPtion A) is interpretation