Question

The government of Preon (a small island nation) was voted in at the last election with 58% of the votes. That was 2 years ago, and ever since then the government has assumed that their approval rating has been the same. Some recent events have affected public opinion and the government suspects that their approval rating might have changed. They decide to run a hypothesis test for the proportion of people who would still vote for them.

The null and alternative hypotheses are:

H0: TT = 0.58

HA: Pi symbol ≠ 0.58

The level of significance used in the test is α = 0.05. A random sample of 114 people are asked whether or not they would still vote for the government. The proportion of people that would is equal to 0.684. You may find this standard normal table useful throughout this question.

a)Calculate the test statistic (z) for this hypothesis test. Give your answer to 3 decimal places.

z =

b)According to this test statistic, you should accept, reject, not reject the null hypothesis.

Answer 1

Answer:

a) $z=\frac{0.684 -0.58}{\sqrt{\frac{0.58(1-0.58)}{114}}}=2.250$  

b) $p_v =2*P(z>2.250)=0.0244$  

If we compare the p value and the significance level given we see that $p_v <\alpha$ we have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Data given and notation

n=114 represent the random sample taken

$\hat p=0.684$ estimated proportion of people that their approval rating might have changed

$p_o=0.58$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Hypothesis

We need to conduct a hypothesis in order to test the claim that true proportion of people that their approval rating might have changed is 0.58 or no.:  

Null hypothesis:$p=0.58$  

Alternative hypothesis:$p \neq 0.58$  

Part a

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

Calculate the statistic  

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

$z=\frac{0.684 -0.58}{\sqrt{\frac{0.58(1-0.58)}{114}}}=2.250$  

Part b: Statistical decision  

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

Since is a bilateral test the p value would be:  

$p_v =2*P(z>2.250)=0.0244$  

If we compare the p value and the significance level given we see that $p_v <\alpha$ we have enough evidence to reject the null hypothesis at 5% of significance.