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A mail-order company has two warehouses, one on the West Coast and the second on the East Coast. The company's policy is to mail all orders placed with it within 72 hours. The company's quality control department checks quite often whether or not this policy is maintained at the two warehouses. A recently taken sample of 400 orders placed with the warehouse on the West Coast showed that 364 of them were mailed within 72 hours. Another sample of 300 orders placed with the warehouse on the East Coast showed that 279 of them were mailed within 72 hours. a. Construct a \(97 \%\) confidence interval for the difference between the proportions of all orders placed at the two warehouses that are mailed within 72 hours. b. Using a \(2.5 \%\) significance level, can you conclude that the proportion of all orders placed at the warehouse on the West Coast that are mailed within 72 hours is lower than the corresponding proportion for the warehouse on the East Coast?

Step by step solution

01

Compute the Proportions

Calculate the proportion of orders mailed within 72 hours for each warehouse. This is done by dividing the number of orders mailed within 72 hours by the total number of orders. For the West Coast warehouse: \(p_1 = \frac{364}{400} = 0.91\). For the East Coast warehouse: \(p_2 = \frac{279}{300} = 0.93\).

02

Compute the Standard Error

The standard error for the difference of proportions is calculated by the formula \(\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}\), where \(n_1\) and \(n_2\) are the sample sizes of the two warehouses. Substitute the values to get the standard error: \(\sqrt{\frac{0.91(1-0.91)}{400} + \frac{0.93(1-0.93)}{300}} = 0.02368\).

03

Construct the Confidence Interval

The 97% confidence interval for the difference in proportions is calculated by the formula \((p_1-p_2)\pm z*SE\), where \(z\) is the z-value corresponding to the desired confidence level (from the standard normal distribution table, for a 97% confidence interval, \(z = 2.170\)). Calculate the confidence interval: \((0.91 - 0.93) \pm 2.170 * 0.02368 = (-0.05124, 0.03124)\).

04

Perform the Hypothesis Test

In the hypothesis test, H0: \(p_1 - p_2 = 0\) (no difference in proportions), and H1: \(p_1 - p_2 < 0\) (the proportion for the West Coast is less than that for the East Coast). The test statistic (Z-score) is calculated by \(\frac{(p_1-p_2) - 0}{SE}\), which equals \(\frac{-0.02}{0.02}\), that is, \(z = -1\). Looking up in the standard normal distribution table, the P-value is greater than the significance level of 0.025. Hence, we do not reject the null hypothesis.

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