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

Answer:

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

$p_v =2*P(z>3.238)=0.0012$  

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.

Step-by-step 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:$p \neq 0.5$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

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.