Question

In a sample of credit card holders the mean monthly value of credit card purchases was $ 298 and the sample variance was 53 ($ 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 13 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 25 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 298

sample standard deviation = s = 7.28

A)

sample size = n = 13

Degrees of freedom = df = n - 1 = 12

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,12 =1.179

Margin of error = E = t_/2,df * (s /n)

= 1.179 * ( 7.28/ 13)

= 2.381

The 95% confidence interval estimate of the population mean is,

- E < < + E

298 - 2.391 < < 298 + 2.391

295.609 < < 300.391

(295.609 , 300.391)

B)

sample size = n = 25

Degrees of freedom = df = n - 1 = 24

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,24 = 2.064

Margin of error = E = t_/2,df * (s /n)

= 2.064 * ( 7.28/ 25)

= 3.005

The 95% confidence interval estimate of the population mean is,

- E < < + E

298 - 3.005 < < 298 + 3.005

294.995 < < 301.005

(294.995 , 301.005)