Question

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-bar = $50.50 and s2 = 400. Construct a 95% confidence interval for the average amount its credit card customers spent on their first visit to the chain's new store in the mall.

Answer 1

Answer: (39.424, 61.576)

Step-by-step explanation:

When population standard deviation($\sigma$) unknown ,The confidence interval for population mean is given by :-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size

$\overline{x}$ = sample mean.

s= sample standard deviation

$t^*$ = Critical t-value (two-tailed)

Given : n= 15

Degree of freedom= 14  [df=n-1]

$\overline{x}=\ 50.50$

$s^2=400\\\\\Rightarrow\ s=\sqrt{400}=20$

Significance level = $\alpha=1-0.95=0.05$

For $\alpha=0.05$ and df = 14, the critical t-values : $t^*=\pm2.1448$

Then the 95% confidence interval for population mean will be  :

$50.50\pm (2.1448)\dfrac{20}{\sqrt{15}}\\\\=50.50\pm(2.1448)(5.1640)\\\\\approx50.50\pm11.076\\\\=(50.50-11.076,\ 50.50+11.076)\\\\=(39.424,\ 61.576)$

Hence, a 95% confidence interval for the average amount its credit card customers spent on their first visit to the chain's new store in the mall. : (39.424, 61.576)