Question

In a sample of credit card holders the mean monthly value of credit card purchases was $ 361 and the sample variance was 75 ($ 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 15 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 1

Here

$\overline{X}=361$               $S_{x}^{2}=75$     So $S_{x}=\sqrt{75}=8.66$

a)

Sample size(n)=15

We will use Ti83/84

Press STAT...TESTS...T-Interval

Enter values as

TInterval Inpt: Data State X:361 SX:8.66 n: 15 C-Level:.95 Calcul ate

The result is

TInterval (356.2.365.8) X=361 Sx=8.66 n=15

Confidence Interval is ( 356.20   ,    365.80)

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b)

Increasing the sample size decreases the width of confidence intervals, because it decreases the standard error.

So confidence interval would be narrower.

Sample size(n)=19

We will use Ti83/84

Press STAT...TESTS...T-Interval

Enter values as

TInterval Inpt: Data State X: 361 Sx: 8.66 n: 19 C-Level: .95 Calculate

The result is

TInterval (356.83,365.17) X=361 Sx=8.66 n=19

Confidence Interval is ( 356.83 ,    365.17)