Question

A random sample of n = 64 observations has a mean x = 29. 1, and a standard deviation of s = 3.9. Find a 90 % confidence interval for the population mean.. 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: = $50.50 and = 400. Find 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.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 29.1

sample standard deviation = s = 3.9

sample size = n = 64

Degrees of freedom = df = n - 1 = 64 - 1 = 63

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,63 = 1.699

Margin of error = E = t_/2,df * (s /n)

= 1.699 * (3.9 / 64)

= 0.8

The 90% confidence interval estimate of the population mean is,

- E < < + E

29.1 - 0.8 < < 29.1 + 0.8

28.3 < < 29.9

(28.3 , 29.9)