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. Fifteen credit card accounts were randomly sampled and analyzed with the following results: x = $50.50 and s2 = 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:

a. $50.50 ± $11.00
b. $50.50 ± $9.09
c. $50.50 ± $11.79
d. $50.50 ± $10.12
e. $50.50 ± $11.08

$50.50+-11.08$

10.12

To calculate the 95% confidence interval for the average amount the credit card customers spent on their first visit, we can use the t-distribution since the population standard deviation is not known.

The formula for calculating the confidence interval is:

CI = x ± t * (s / sqrt(n))

Where:
x = sample mean ($50.50 in this case)
s = sample standard deviation (sqrt(s^2), which would be sqrt(400) = 20 in this case)
n = sample size (15 in this case)
t = t-value for a 95% confidence level with n-1 degrees of freedom.

To find the t-value, we need to determine the degrees of freedom. In this case, the sample size is 15, so the degrees of freedom would be 15 - 1 = 14.

Using a t-distribution table or a statistical software, the t-value for a 95% confidence level with 14 degrees of freedom is approximately 2.145.

Plug in the values into the formula:

CI = 50.50 ± 2.145 * (20 / sqrt(15))

Calculating this, we get:

CI = 50.50 ± 2.145 * (20 / 3.87)

CI = 50.50 ± 2.145 * 5.17

CI ≈ 50.50 ± 11.08

Therefore, the 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 ± $11.08.

So the correct answer is e. $50.50 ± $11.08.

95% = mean ± 1.96SEm

SEm = s/√n

s = √(s^2)

I'll let you do the calculations.