Question

It is thought that the front cover and the nature of the first question on mail surveys influence the response rate. An article tested this theory by experimenting with different cover designs. One cover was plain; the other used a picture of a skydiver. The researchers speculated that the return rate would be lower for the plain cover.
Cover Number Sent Number Returned
Plain 209 103
Skydiver 215 109
Does this data support the researchers' hypothesis? Test the relevant hypotheses using ? = 0.10 by first calculating a P-value.

State the relevant hypotheses. (Use p1 for the plain cover and p2 for the skydiver cover.)

H0: p1? p2 = 0
Ha: p1? p2 > 0H0: p1? p2 = 0
Ha: p1? p2 ? 0 H0: p1? p2 = 0
Ha: p1? p2? 0H0: p1? p2 = 0
Ha: p1? p2 < 0

Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z =
P-value =

State the conclusion in the problem context.

Reject H0. The data does not suggest that the front cover and nature of the first question on mail surveys influence the response rate.Fail to reject H0. The data does not suggest that the front cover and nature of the first question on mail surveys influence the response rate. Reject H0. The data suggests that the front cover and nature of the first question on mail surveys does influence the response rate.Fail to reject H0. The data suggests that the front cover and nature of the first question on mail surveys does influence the response rate.

Answer 1

Answer:

The null and alternative hypothesis are:

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

Test statistic z = -0.29

P-value = 0.7709

Fail to reject H0. The data does not suggest that the front cover and nature of the first question on mail surveys influence the response rate.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the return rate is different for the Plain cover and the Skydiver cover.

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 0.1.

The sample 1 (plain cover), of size n1=209 has a proportion of p1=0.493.

$p_1=X_1/n_1=103/209=0.493$

The sample 2 (skydiver cover), of size n2=215 has a proportion of p2=0.507.

$p_2=X_2/n_2=109/215=0.507$

The difference between proportions is (p1-p2)=-0.014.

$p_d=p_1-p_2=0.493-0.507=-0.014$

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

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{103+109}{209+215}=\dfrac{212}{424}=0.5$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.5*0.5}{209}+\dfrac{0.5*0.5}{215}}\\\\\\s_{p1-p2}=\sqrt{0.001196+0.001163}=\sqrt{0.002359}=0.049$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.014-0}{0.049}=\dfrac{-0.014}{0.049}=-0.29$

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

$\text{P-value}=2\cdot P(z$

As the P-value (0.771) is bigger than the significance level (0.1), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the return rate is different for the Plain cover and the Skydiver cover.