Question

In a certain city, there are about one million eligible voters. A simple random sample of size 10,000 was chosen to study the relationship between gender and participation in the last election. The results were: Men Women Voted 2744 3733 Didn't Vote 1599 1924 If we are testing for a relationship between gender and participation in the last election, what is the p-value and decision at the 5% significance level? Select the [p-value, Decision to Reject (RH0) or Failure to Reject (FRH0)]

Answer 1

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).

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:

`H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\\eq 0`

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.

`p_1=X_1/n_1=2744/4343=0.6318`

The sample 2 (women), of size n2=(3733+1924)=5657 has a proportion of p2=0.6599.

`p_2=X_2/n_2=3733/5657=0.6599`

The difference between proportions is (p1-p2)=-0.0281.

`p_d=p_1-p_2=0.6318-0.6599=-0.0281`

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

`p=(X_1+X_2)/(n_1+n_2)=(2744+3733)/(4343+5657)=(6477)/(10000)=0.6477`

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

`s_(p1-p2)=\sqrt{(p(1-p))/(n_1)+(p(1-p))/(n_2)}=\sqrt{(0.6477*0.3523)/(4343)+(0.6477*0.3523)/(5657)}\\\\\\s_(p1-p2)=√(0.00005+0.00004)=√(0.00009)=0.0096`

Then, we can calculate the z-statistic as:

`z=(p_d-(\pi_1-\pi_2))/(s_(p1-p2))=(-0.0281-0)/(0.0096)=(-0.0281)/(0.0096)=-2.913`

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<-2.913)=0.0036`

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.