Question

According to a recent poll 53% of Americans would vote for the incumbent president. If a random sample of 100 people results in 40% who would vote for the incumbent, test whether the claim that the actual percentage is different from 53% is supported or not supported.

(1) State the null hypothesis.
(2) State the alternative hypothesis.
(3) What is the test statistic used for the test (z or t)?
(4) State the significance or alpha (α) level?

Answer 1

Answer:

1) Null hypothesis:$p=0.53$  

2)Alternative hypothesis:$p \neq 0.53$  

3) $z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605$  

4) We assume that $\alpha=0.05$

$p_v =2*P(z<-2.605)=0.0092$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53.  

Step-by-step explanation:

Data given and notation  

n=100 represent the random sample taken

$\hat p=0.4$ estimated proportion of people who would vote for the incumbent

$p_o=0.53$ is the value that we want to test

$\alpha=0.05$ represent the significance level  (assumed)

Confidence=95% or 0.95  (Assumed)

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 the true proportion is 0.53 or not.:  

1) Null hypothesis:$p=0.53$  

2)Alternative hypothesis:$p \neq 0.53$  

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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.4 -0.53}{\sqrt{\frac{0.53(1-0.53)}{100}}}=-2.605$  

4) 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 significance level assumed $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z<-2.605)=0.0092$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who would vote for the incumbent is different from 0.53 .