Question

Your mail-order company advertises that it ships 90% of its orders within three working days. You select an SRS of 100 of the 5000 orders recieved in the past week for an audit. The audit reveals that 86 of these orders were shipped on time.

(a) If the company really ships 90% of its orders on time, what is the probability that 86 or fewer in an SRS of 100 orders are shipped on time?

(b) A critic says "Aha! You claim 90%, but in your sample the on-time percent is only 86%. So the 90% claim is wrong." Explain in simple language why your probability calcualtion in part (a) shows that the result of the sample does not refute the 90% claim.

Answer 1

a) We can use the normal approximation , but Rule of Thumb 2 is just barely satisfied : n(1 - p) =10

The standard deviation of the sample proportion is σ = √ 0.9 *0.1 /100 =0.03

The probability is P(p ≤ 0.86) = P( Z ≤ 0.86 -0.9 /0.03 ) = P(Z ≤ -1.33) =0.0918

b) If the claim is correct , then we can expect to observe 86% or fewer orders shipped on time in about 12.5% of the samples of this size.

Gettting a sample proportion at or below 0.86 is not an unlikely event