Question

According to an exit poll for an​ election, 55.6​% of the sample size of 836 reported voting for a specific candidate. Is this enough evidence to predict who​ won? Test that the population proportion who voted for this candidate was 0.50 against the alternative that it differed from 0.50.

Report the test statistic and P-value and interpret the latter.

Answer 1

`z=\frac{0.556 -0.5}{\sqrt{(0.5(1-0.5))/(836)}}=3.238`

`p_v =2*P(z>3.238)=0.0012`

The p value is a reference value and is useful in order to take a decision for the null hypothesis is this p value is lower than a significance level given we reject the null hypothesis and otherwise we have enough evidence to fail to reject the null hypothesis.

Explanation:

Data given and notation

n=836 represent the random sample taken

`\hat p=0.556` estimated proportion of interest

`p_o=0.5` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that ture proportion is equal to 0.5 or no.:

Null hypothesis:
`p=0.5`

Alternative hypothesis:
`p \\eq 0.5`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.556 -0.5}{\sqrt{(0.5(1-0.5))/(836)}}=3.238`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z>3.238)=0.0012`

The p value is a reference value and is useful in order to take a decision for the null hypothesis is this p value is lower than a significance level given we reject the null hypothesis and otherwise we have enough evidence to fail to reject the null hypothesis.