Mathematics, 05.05.2020 17:15 s77969

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?

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Answer from: amchavez

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.

Step-by-step 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:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.5373*0.4627}{630}}\\\\\\ \sigma_p=\sqrt{0.00039}=0.0199$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.5032-0.5373+0.5/630}{0.0199}=\dfrac{-0.0333}{0.0199}=-1.6766$

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

P-value=P(z

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.

Answer from: kameahtravis

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

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

p_v =P(Z

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

Step-by-step 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

$p_{1}=\frac{223}{415}=0.537$ represent the proportion before

$p_{2}=\frac{317}{630}=0.503$ 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:

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

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{223+317}{415+630}=0.517$

Calculate the statistic

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

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

Statistical decision

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

p_v =P(Z

Answer from: Quest

lo.lll

step-by-step explanation:

Answer from: Quest

answer: idk

step-by-step explanation:

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