Question

A hair salon in Cambridge, Massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited the salon were 41, 64, 43, 28, 54, 44, and 56. It can be assumed that weekday customer visits follow a normal distribution. [You may find it useful to reference the t table.]

a. Construct the 90% confidence interval for the average number of customers who visit the salon on weekdays. (Round intermediate calculations to at least 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)

b. Construct the 99% confidence interval for the average number of customers who visit the salon on weekdays. (Round intermediate calculations to at least 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.)

c. What happens to the width of the interval as the confidence level increases?

• As the confidence level increases, the interval becomes wider and less precise.

• As the confidence level increases, the interval becomes narrower and less precise.

Answer 1

a)
sample mean, xbar = 47.14
sample standard deviation, s = 11.84
sample size, n = 7
degrees of freedom, df = n - 1 = 6

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.943

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (47.14 - 1.943 * 11.84/sqrt(7) , 47.14 + 1.943 * 11.84/sqrt(7))
CI = (38.44 , 55.84)

b)
Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 3.707

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (47.14 - 3.707 * 11.84/sqrt(7) , 47.14 + 3.707 * 11.84/sqrt(7))
CI = (30.55 , 63.73)

c)
As the confidence level increases, the interval becomes wider and less precise.