Question

A sample of 200 observations from the first population indicated that X1 is 170. A sam- ple of 150 observations from the second population revealed X2 to be 110. Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?

Answer 1

Answer:

a. If the P-value is smaller than the significance level, the null hypothesis is rejected.

b. Pooled proportion = 0.8

c. z = 2.7

d. As the P-value (0.0072) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is enough evidence to support the claim that the proportions differ significantly.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

We will use the P-value approach, so the decision rule is that if the P-value is lower than the significance level, the null hypothesis is rejected.

The claim is that the proportions differ significantly.

Then, the null and alternative hypothesis are:

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

The significance level is 0.05.

The sample 1, of size n1=200 has a proportion of p1=0.85.

$p_1=X_1/n_1=170/200=0.85$

The sample 2, of size n2=150 has a proportion of p2=0.7333.

$p_2=X_2/n_2=110/150=0.7333$

The difference between proportions is (p1-p2)=0.1167.

$p_d=p_1-p_2=0.85-0.7333=0.1167$

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

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{170+110}{200+150}=\dfrac{280}{350}=0.8$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.8*0.2}{200}+\dfrac{0.8*0.2}{150}}\\\\\\s_{p1-p2}=\sqrt{0.0008+0.00107}=\sqrt{0.00187}=0.0432$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.1167-0}{0.0432}=\dfrac{0.1167}{0.0432}=2.7$

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

$P-value=2\cdot P(z>2.7)=0.0072$

As the P-value (0.0072) 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 proportions differ significantly.