Question

Industrialist H.E. Pennypacker wants information on the customers that patronize his bicycle stores. He surveyed 81 randomly-selected individuals who made a purchase at his Pasadena store to find out how much they spent, on average. The mean amount spent was $62 with a standard deviation of $15.

i) What is the population in this study?

ii) Construct a 95% CI for the mean amount of money an individual spends at the Pasadena store. Explain the meaning of this CI (i.e., what does it say about the parameter of interest?).

iii) Construct a 99% CI for the mean amount of money spent at the Pasadena store. Contrast this interval with the one from 4b and explain why it is different.

iv) Suppose 100 people were initially surveyed, but 19 of them actually refused to answer (leading to the final sample size of 81). If the 19 individuals who refused to answer spent considerably less money than the 81 who did respond, how would this affect the estimate of the mean and the CI?

Answer 1

i) Population: the customers that patronize his bicycle stores.

ii) x̅ = 62, s = 18, n = 81

95% Confidence interval for the mean amount of money an individual spends at the Pasadena store:

At α = 0.05 and df = n-1 = 80, two tailed critical value, t-crit = T.INV.2T(0.05, 80) = 1.990

Lower Bound = x̅ - t-crit*s/√n = 62 - 1.99 * 18/√81 = 58.0199

Upper Bound = x̅ + t-crit*s/√n = 62 + 1.99 * 18/√81 = 65.9801

58.0199 < µ < 65.9801

iii) 99% Confidence interval for the mean amount of money an individual spends at the Pasadena store:

At α = 0.01 and df = n-1 = 80, two tailed critical value, t-crit = T.INV.2T(0.01, 80) = 2.639

Lower Bound = x̅ - t-crit*s/√n = 62 - 2.639 * 18/√81 = 56.7226

Upper Bound = x̅ + t-crit*s/√n = 62 + 2.639 * 18/√81 = 67.2774

56.7226 < µ < 67.2774

This interval is wider than the interval in 4b.

iv) If the are included in the study, then the mean will decreases.

And the CI would also change.