A major department store chain is interested in estimating the mean amount its credit card customers spent on their first visit

Question

2 months ago

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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: = $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. A) $50.50 ± $9.09 B) $50.50 ± $10.12 C) $50.50 ± $11.00 D) $50.50 ± $11.08

Mathematics

1 answer:

Inessa [12.5K] · Answer 1 · 2026-03-24T23:09:47+03:00

Answer: D) $\$50\pm\$11.08$

Step-by-step explanation:

Based on the provided information, we have

Sample size: n= 15

sample mean: $\overline{x}=\$50.50$

Sample standard deviation: s= $20

Since the population standard deviation is not known, we utilize a t-test.

For a significance level of 95% confidence: $\alpha=1-0.95=0.05$

Critical t-value : $t_{n-1, \alpha/2}=t_{014,0.025}=2.145$ [Using the Student's t-value table]

The required 95% confidence interval yields:-

$\overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}\\\\ =\$50.50\pm(2.145)\dfrac{\$20}{\sqrt{15}}\\\\\approx \$50\pm\$11.08$

Thus, the sought-after 95% confidence interval for the mean amount spent by credit card customers during their initial visit to the new store in the mall, assuming normal distribution of the spending amounts, is:

$\$50\pm\$11.08$