Question

A random sample of 415 potential voters was interviewed 3 weeks before the start of a state-wide campaign for governor; 223 of the 415 said they favored the new candidate over the incumbent. However, the new candidate made several unfortunate remarks one week before the election. Subsequently, a new random sample of 630 potential voters showed that 317 voters favored the new candidate.Do these data support the conclusion that there was a decrease in voter support for the new candidate after the unfortunate remarks were made?

Answer 1

`z=\frac{0.503-0.537}{\sqrt{0.517(1-0.517)((1)/(415)+(1)/(630))}}=-1.076`

Since is a left tailed test the p value would be:

`p_v =P(Z<-1.076)=0.141`

For this case if we use a significance level of 10% we see that
`p_v >\alpha` so then we don;'t have enough evidence to conclude that there was a decrease in voter support for the new candidate after the unfortunate remarks were made

Explanation:

Data given and notation

`X_(1)=223` represent the number of people who support the candidate before

`X_(2)=317` represent the number of people who support the candidate after

`n_(1)=415` sample before

`n_(2)=630` sample after

`p_(1)=(223)/(415)=0.537` represent the proportion before

`p_(2)=(317)/(630)=0.503` represent the proportion after

`\hat p` represent the pooled estimate of p

z would represent the statistic (variable of interest)

`p_v` represent the value for the test (variable of interest)

`\alpha=0.05` significance level given

Concepts and formulas to use

We need to conduct a hypothesis in order to check if is the proportion after is lower than the proportion before, the system of hypothesis would be:

Null hypothesis:
`p_(2) \geq p_(1)`

Alternative hypothesis:
`p_(2) < p_(1)`

We need to apply a z test to compare proportions, and the statistic is given by:

`z=\frac{p_(2)-p_(1)}{\sqrt{\hat p (1-\hat p)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(223+317)/(415+630)=0.517`

Calculate the statistic

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.503-0.537}{\sqrt{0.517(1-0.517)((1)/(415)+(1)/(630))}}=-1.076`

Statistical decision

Since is a left tailed test the p value would be:

`p_v =P(Z<-1.076)=0.141`

For this case if we use a significance level of 10% we see that
`p_v >\alpha` so then we don;'t have enough evidence to conclude that there was a decrease in voter support for the new candidate after the unfortunate remarks were made

Answer 2

Yes. There is enough evidence to support the claim that there was a decrease in voter support for the new candidate after the unfortunate remarks.

Explanation:

This is a hypothesis test for a proportion.

The previous proportion, to which we will test the new proportion, was:

`\pi=X_0/n_0=223/415=0.5373`

The claim is that there was a decrease in voter support for the new candidate after the unfortunate remarks.

Then, the null and alternative hypothesis are:

`H_0: \pi=0.5373\\\\H_a:\pi< 0.5373`

The significance level is 0.05.

The sample has a size n=630.

The sample proportion of the new sample is p=0.5032.

`p=X/n=317/630=0.5032`

The standard error of the proportion is:

`\sigma_p=\sqrt{(\pi(1-\pi))/(n)}=\sqrt{(0.5373*0.4627)/(630)}\\\\\\ \sigma_p=√(0.00039)=0.0199`

Then, we can calculate the z-statistic as:

`z=(p-\pi-0.5/n)/(\sigma_p)=(0.5032-0.5373+0.5/630)/(0.0199)=(-0.0333)/(0.0199)=-1.6766`

This test is a left-tailed test, so the P-value for this test is calculated as:

`P-value=P(z<-1.6766)=0.0468`

As the P-value (0.0468) 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 was a decrease in voter support for the new candidate after the unfortunate remarks.