Respuesta :
Answer:
The null hypothesis is rejected.
There is enough evidence to support the claim that the proportion of women that vote is differs from the proportion of men that vote.
P-value=0.0036 (two tailed test).
Step-by-step explanation:
This is a hypothesis test for the difference between proportions.
The claim is that the proportion of women that vote is differs from the proportion of men that vote.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0[/tex]
Being π1: proportion of men that vote, and π2: proportion of women that vote.
The significance level is 0.05.
The sample 1 (men), of size n1=(2744+1599)=4343 has a proportion of p1=0.6318.
[tex]p_1=X_1/n_1=2744/4343=0.6318[/tex]
The sample 2 (women), of size n2=(3733+1924)=5657 has a proportion of p2=0.6599.
[tex]p_2=X_2/n_2=3733/5657=0.6599[/tex]
The difference between proportions is (p1-p2)=-0.0281.
[tex]p_d=p_1-p_2=0.6318-0.6599=-0.0281[/tex]
The pooled proportion, needed to calculate the standard error, is:
[tex]p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{2744+3733}{4343+5657}=\dfrac{6477}{10000}=0.6477[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.6477*0.3523}{4343}+\dfrac{0.6477*0.3523}{5657}}\\\\\\s_{p1-p2}=\sqrt{0.00005+0.00004}=\sqrt{0.00009}=0.0096[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{-0.0281-0}{0.0096}=\dfrac{-0.0281}{0.0096}=-2.913[/tex]
This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):
[tex]P-value=2\cdot P(t<-2.913)=0.0036[/tex]
As the P-value (0.0036) 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 proportion of women that vote is differs from the proportion of men that vote.
