Question

A mail-order business prides itself in its ability to fill customers' orders in six calendar days or less on the average. Periodically, the operations manager selects a random sample of customer orders and determines the number of days required to fill the orders. Based on this sample information, he decides if the desired standard is not being met. He will assume that the average number of days to fill customers' orders is six or less unless the data suggest strongly otherwise. On one occasion where a sample of 40 customers was selected, the average number of days was 6.65, with a sample standard deviation of 1.5 days. Can the operations manager conclude that his mail-order business is achieving its goal? Use a significance level of 0.025 to answer this question.

Answer 1

Answer:

No. Operations manager cannot conclude that his mail-order business is achieving its goal.

Step-by-step explanation:

We make hypothesis test about the manager's assuption:

Null hypothesis, $H_{0}$: Average number of days to fill customers' orders is six or less

Alternate Hypothesis: $H_{a}$:  Average number of days to fill customers' orders is more than six.

According to the null hypothesis we assume number of days to fill customers' orders follows a normal distribution with mean 6 and standard deviation 1.5. We would test if the sample mean is in the critical field or not in the given significance level.

One tailed critical value for the significance level 0.025 is 1.96. We'll compare this value with the z-score of the sample mean 6.65, which is calculated as:

z=$\frac{6.65-6}{\frac{1.5}{\sqrt{40} } }$ ≈ 2.74 where

• 6,65 is the sample mean
• 6 is the null hypothesis
• 1.5 is the standard deviation
• 40 is the sample size

Since 2.74>1.96, we can conclude that sample mean is in the critical region, we reject the null hypothesis.

Therefore operations manager can conclude that average number of days to fill customers' orders is more than 6 days.