Question

A major department store chain is interested in estimating the mean 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 S = 20.
Construct a 95% confidence interval for the mean amount its credit card customers spent on their first visit to the chain's new store in the mall assuming that the amount spent follows a normal distribution.

Answer 1

Answer:

95% Confidence interval: (39.43, 61.58)

Step-by-step explanation:

We are given the following in the question:

Sample mean, $\bar{x}$ = $50.50

Sample size, n = 15

Alpha, α = 0.05

Sample standard deviation = 20

95% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 14 and}~\alpha_{0.05} = \pm 2.1447$  

$=50.50 \pm 2.1447(\dfrac{20}{\sqrt{15}} ) \\\\= 50.50 \pm 11.0751 \\= (39.4249,61.5751)\\\approx (39.43, 61.58)$  

95% Confidence interval: (39.43, 61.58)