Question

Where people turn for news is different for various age groups. A study was conducted on the use of cellphones for accessing news. The study reported that 26% of users from a certain region under age 50 and 17% of users from the region age 50 and over accessed news on their cellphones. Suppose that the survey consisted of 1,100 users under age 50, of whom 286 accessed news on their cellphones, and 900 users age 50 and over, of whom 153 accessed news on their cellphones. Is there evidence of a significant difference in the proportion of users under age 50 and users age 50 years and older that accessed the news on their cellphones. The null and alternate hypotheses are: H0: π1 = π2 H1: π1 ≠ π2 Determine the value of the test statistic ZSTAT

Answer 1

Answer:

There is  enough evidence to support the claim that there is significant difference in the proportion of users under age 50 and users age 50 years and older that accessed the news on their cellphones (P-value=0.0000014).

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

The claim is that there is significant difference in the proportion of users under age 50 and users age 50 years and older that accessed the news on their cellphones.

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=1100 has a proportion of p1=0.26.

$p_1=X_1/n_1=286/1100=0.26$

The sample 2, of size n2=900 has a proportion of p2=0.17.

$p_2=X_2/n_2=153/900=0.17$

The difference between proportions is (p1-p2)=0.09.

$p_d=p_1-p_2=0.26-0.17=0.09$

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

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{286+153}{1100+900}=\dfrac{439}{2000}=0.2195$

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

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.2195*0.7805}{1100}+\dfrac{0.2195*0.7805}{900}}\\\\\\s_{p1-p2}=\sqrt{0.00016+0.00019}=\sqrt{0.00035}=0.0186$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.09-0}{0.0186}=\dfrac{0.09}{0.0186}=4.838$

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

$P-value=2\cdot P(t>4.838)=0.0000014$

As the P-value (0.0000014) 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 there is significant difference in the proportion of users under age 50 and users age 50 years and older that accessed the news on their cellphones.