Question

Suppose that early in an election campaign, a telephone poll of 800 registered voters shows that 460 favor a particular candidate. Just before Election Day, a second poll shows that 520 of 1,000 registered voters now favor that candidate. At the 0.05 significance level, is there sufficient evidence that the candidate's popularity has changed?

Answer 1

Yes. At the 0.05 significance level, there is enough evidence to support the claim that the proportion that support the candidate has significantly changed.

Explanation:

This is a hypothesis test for the difference between proportions.

The claim is that the proportion that support the candidate has significantly changed.

Then, the null and alternative hypothesis are:

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

The significance level is 0.05.

The sample 1, of size n1=800 has a proportion of p1=0.58.

`p_1=X_1/n_1=460/800=0.58`

The sample 2, of size n2=1000 has a proportion of p2=0.52.

`p_2=X_2/n_2=520/1000=0.52`

The difference between proportions is (p1-p2)=0.05.

`p_d=p_1-p_2=0.58-0.52=0.05`

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

`p=(X_1+X_2)/(n_1+n_2)=(464+520)/(800+1000)=(980)/(1800)=0.54`

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{(0.54*0.46)/(800)+(0.54*0.46)/(1000)}\\\\\\s_(p1-p2)=√(0.00031+0.000248)=√(0.000558)=0.02`

Then, we can calculate the z-statistic as:

`z=(p_d-(\pi_1-\pi_2))/(s_(p1-p2))=(0.05-0)/(0.02)=(0.05)/(0.02)=2.33`

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>2.33)=0.02`

As the P-value (0.02) 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 that support the candidate has significantly changed.