SOLUTION: In a sample of credit card holders the mean monthly value of credit card purchases was $ 400 and the sample variance was 90 ($ squared). Assume that the population distribution is

Question 1193340: In a sample of credit card holders the mean monthly value of credit card purchases was $ 400 and the sample variance was 90 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate.
(a) Suppose the sample results were obtained from a random sample of 11 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.

(b) Suppose the sample results were obtained from a random sample of 19 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.

Before you answer (b) consider whether the confidence interval will be wider than or narrower than the confidence interval found for (a). Then check that your answer verifies this.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Because b has more information, its interval will be narrower. Ci also is inversely proportional to the square root of the sample size and directly proportional to the t-value which is also less for a larger n, everything else's being the same.
95% half-interval of a is t(0.975,df=10)*9.49/sqrt(11); the 9.49 is the sd, the sqrt(90), and the units are $.
=2.228*9.49/sqrt(11)=6.375 or 6.38
the interval is ($393.62, $406.38)
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For the second, t is 2.101, and half-interval is 2.101*9.49/sqrt(19)=4.57
interval is ($395.43, $404.57) It is narrower
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