Question

A manager of a cafeteria wants to estimate the average time customers wait before being served. A sample of 51 customers has an average waiting time of 8.4 minutes with a standard deviation of 3.5 minutes.

a) With 90% confidence, what can the manager conclude about the possible size of his error in using 8.4 minutes to estimate the true average waiting time?

b) Find a 90% confidence interval for the true average customer waiting time.

c) Repeat part (a) using 98% confidence interval

d) Repeat part (b) using 98% confidence interval

Answer 1

here,

sample size, n = 51

mean = 8.4

standard deviation, s = 3.5

a) for 90% of CI, alpha = 0.1

degree of freedom, df = n-1 = 50

therefore t_{alpha/2 , df} = t =  1.6759

margin of error, E = t*s/sqrt(n) = 1.6759*3.5/sqrt(51) = 0.8214

b) 90% of CI = mean +/- E = 8.4 +/- 0.8214 = (7.58 , 9.22)

c) for 98% of CI, alpha = 0.02

therefore, t_{alpha/2 , df} = t = 2.403

margin of error, E = t*s/sqrt(n) = 2.403*3.5/sqrt(51) = 1.1777

d) 98% of CI = mean +/- E = 8.4 +/- 1.1777 = (7.22 , 9.58)