A major department store chain is interested in estimating t

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 variance of 400. A 95% confidence interval for the average amount the credit card customers spent on their first visit to the chain\'s new store in the mall is:

$50.50 ± $9.09.

$50.50 ± $10.12.

$50.50 ± $11.08.

None of these choices.

Solution

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=50.5
Standard deviation( sd )=20
Sample Size(n)=15
Confidence Interval = [ 50.5 ± t a/2 ( 20/ Sqrt ( 15) ) ]
= [ 50.5 ± 2.145 * (5.164) ]
= [ 50.5 ± 11.07678 ]
[ANSWER] = $50.50 ± $11.08