Question

Maureen McIlvoy, owner and CEO of a mail order business for wind surfing equipment and supplies, is reviewing the order filling operations at her warehouses. Her goal is 100% of orders shipped within 24 hours. In previous years, neither warehouse has achieved the goal, but the East Coast warehouse has consistently out-performed the West Coast warehouse. Her staff randomly selected 200 orders from the West Coast warehouse (population 1) and 400 orders from the East Coast warehouse (population 2), and reports that 190 of the West Coast orders were shipped within 24 hours, and the East Coast warehouse shipped 356 orders within 24 hours. Assuming a = 0.05, the appropriate decision is ___________________.
a) reject the null hypothesis p1 – p2 = 0
b) reject the null hypothesis m1minusm2 < 0
c) do not reject the null hypothesis m1minusm2 = 0
d) do not reject the null hypothesis p1 – p2 = 0
e) do not reject the null hypothesis p1 – p2 ? 0

Answer 1

Answer:

a) reject the null hypothesis p1 – p2 = 0

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the East Coast warehouse proportion of orders delivered in 24 hours is significantly higher than the West Coast warehouse proportion.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2> 0$

The significance level is 0.05.

The sample 1 (East Coast), of size n1=200 has a proportion of p1=0.95.

$p_1=X_1/n_1=190/200=0.95$

The sample 2 (West coast), of size n2=400 has a proportion of p2=0.89.

$p_2=X_2/n_2=356/400=0.89$

The difference between proportions is (p1-p2)=0.06.

$p_d=p_1-p_2=0.95-0.89=0.06$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{190+356}{200+400}=\dfrac{546}{600}=0.91$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.91*0.09}{200}+\dfrac{0.91*0.09}{400}}\\\\\\s_{p1-p2}=\sqrt{0.00041+0.000205}=\sqrt{0.000614}=0.025$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.06-0}{0.025}=\dfrac{0.06}{0.025}=2.4209$

This test is a right-tailed test, so the P-value for this test is calculated as (using a z-table):

$\text{P-value}=P(z>2.4209)=0.0079$

As the P-value (0.0079) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the East Coast warehouse proportion of orders delivered in 24 hours is significantly higher than the West Coast warehouse proportion.