Question

In: Statistics and Probability

A major department store chain is interested in estimating the average amount its credit card customers...

A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. 15 credit card accounts randomly sampled produced a mean of $50.50 and a standard deviation of $20. Find a 95% confidence interval for the average amount the credit card customers spent on their first visit to the chain's new supply store in the mall.

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Expert Solution


Solution :

Given that,

= 50.50

s = 20

n = 15

Degrees of freedom = df = n - 1 = 15 - 1 = 14

At 95% confidence level the t is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,14 = 2.145

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

= 2.145 * (20 / 15)

= 11.08

Margin of error = 11.08

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

- E < < + E

50.50 - 11.08 < < 50.50 + 11.08

39.42 < < 61.58

(39.42, 61.58)

orchestra answered 1 year ago
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