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.Provide a step-by-step solution including formulae.

Answer 1

Answer:

The 95% confidence interval is $39.42 to $61.58

Step-by-step explanation:

15 credit card accounts were sampled. So, sample size = n = 15

Sample mean = x = $ 50.5

Sample standard deviation = s = 20

The population is normally distributed. We have to construct a 95% confidence interval for the mean amount. Since, the value of population standard deviation is not known, and we have the value of sample standard deviation, we will use t-test to solve this problem.

Degrees of freedom = df = n - 1 = 15 - 1 = 14

The formula to calculate the confidence interval is:

$(x-t_{critical} \times \frac{s}{\sqrt{n}}, x+t_{critical} \times \frac{s}{\sqrt{n}})$

In order to find the value of t critical we check the value again 95% confidence level and 14 degrees of freedom from the t-table. This value comes out to be: 2.145

Using the values in above expression we get:

$(50.5-2.145 \times \frac{20}{\sqrt{15}}, 50.5-2.145 \times \frac{20}{\sqrt{15}})\\\\ =(39.42,61.58)$

Therefore, 95% confidence interval for the mean amount credit card customers spent on their first visit to the chain's new store in the mall is (39.42, 61.58)