Credit card company wants to estimate the average balance of its customers. A random sample of 20 customers...

A credit card company wants to estimate the average balance of its customers. A random sample of 20 customers shows a sample standard deviation of $43. Construct a 99% confidence interval for the standard deviation of the balance.





a. (6.84398, 38.5822)


b. (910.5494, 5133.1243)


c. (30.1753, 71.6458)


d. (31.1563, 67.8430)

Credit card company wants to estimate the average balance of its customers. A random sample of 20 customers...
Since answers a and b do not bracket the sample SD, they are almost certainly not correct. (However, it turns out they are not just random choices, as you will see.)





Assuming the average balances are normally distributed, then





L1 = (n-1) S^2 / chisq_(1 - a/2) and


L2 = (n-1) S^2 / chisq_(a/2)





form a 100(1-a)% confidence interval for the variance of the balance, where n is the sample size, S^2 is the sample variance, and chisq_(1 - a/2) and chisq(a/2) are, respectively, the 100(1 - a/2) and 100 a/2 chi-square percentiles of the xhi-square distribution with (n-1) degrees of freedom. (This should be in your book or class notes.)





For a=0.01,





chisq(a/2) = chisq(.005) = 6.84398 and


chisq(1 - a/2) = chisq(.995) = 38.5822





(Compare these values with answer a.)





If you use these values to compute L1 and L2, you will get the range given by answer b. But this is the confidence interval for the variance, and you've been asked for a confidence interval for the standard deviation. So if





L1 %26lt; sigma_squared %26lt; L2 is a 99% confidence interval for the variance, then it would seem reasonable to me that





sqrt(L1) %26lt; sigma %26lt; sqrt(L2) would be a 99% confidence interval for the standard deviation. If so, the answer is c.
Reply:lol
Reply:No