Your mail-order company advertises that it ships 95% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that only 91 of these orders were shipped on time.

A) What is the sample proportion of orders shipped on time?

B) If the company really ships 95% of its orders on time, what is the probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller? (Use a normal approximation, but be sure to justify the standard deviation and approximation first.)

C) A critic says, "Aha! You claim 95% but in your sample the on-time percentage is lower than that. So the 95% claim is wrong." Does your probability calculation in (b) support or refute the 95% claim? Explain.

A) To find the sample proportion of orders shipped on time, we divide the number of orders shipped on time (91) by the total number of orders in the sample (100).

Sample proportion of orders shipped on time = 91/100 = 0.91

B) In order to determine the probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller, we will use a normal approximation. However, before proceeding with the normal approximation, we need to justify the standard deviation and approximation.

Since the company advertises that it ships 95% of its orders on time, the claim implies that the population proportion of orders shipped on time is 0.95. The standard deviation for the proportion can be calculated using the formula:

Standard deviation = sqrt((p * (1-p))/n)

Where p is the population proportion, and n is the sample size.

Substituting the given values:
Standard deviation = sqrt((0.95 * (1-0.95))/100)
Standard deviation = sqrt(0.95 * 0.05/100)
Standard deviation = sqrt(0.0475/100)
Standard deviation = sqrt(0.000475)
Standard deviation ≈ 0.0218

Now, we can proceed with the normal approximation. The distribution of the sample proportion can be approximated by a normal distribution with mean equal to the population proportion (0.95) and standard deviation equal to the calculated standard deviation (0.0218).

To find the probability that the proportion in the sample is as small as or smaller than the sample proportion, we need to calculate the z-score for the sample proportion. The z-score formula is:

z = (sample proportion - population proportion) / standard deviation

Substituting the given values:
z = (0.91 - 0.95) / 0.0218
z = -0.04 / 0.0218
z ≈ -1.83

Using the z-table or a statistical calculator, we find that the probability associated with a z-score of -1.83 is approximately 0.0336.

Therefore, the probability that the proportion in an SRS of 100 orders is as small as the proportion in your sample or smaller is approximately 0.0336.

C) The probability calculation in part (B) does not support or refute the 95% claim directly. It only provides the probability of observing a sample proportion as small as or smaller than the sample proportion in your audit, assuming that the true population proportion is indeed 95%.

To make a conclusion about the claim, we can interpret the probability value obtained. In this case, the probability is approximately 0.0336, which is relatively small. This suggests that it is unlikely to observe a sample proportion as small as or smaller than the one in your audit if the true population proportion is 95%.

However, we cannot definitively conclude that the claim is wrong based on this information alone. Further analysis, additional data, or confidence intervals would be needed to provide stronger evidence for or against the 95% claim.