Question

The mayor of a town has proposed a plan for the construction of an adjoining bridge. A political study took a sample of 900 voters in the town and found that 60% of the residents favored construction. Using the data, a political strategist wants to test the claim that the percentage of residents who favor construction is above 56%. Determine the P-value of the test statistic. Round your answer to four decimal places.

Answer 1

Answer:

Test statistic z = 2.3839.

P-value = 0.0086.

At a signficance level of 0.05, there is enough evidence to support the claim that the percentage of residents who favor construction is above 56%.

Step-by-step explanation:

This is a hypothesis test for a proportion.

The claim is that the percentage of residents who favor construction is above 56%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.56\\\\H_a:\pi>0.56$

The significance level is 0.05.

The sample has a size n=900.

The sample proportion is p=0.6.

 

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.56*0.44}{900}}\\\\\\ \sigma_p=\sqrt{0.000274}=0.017$

Then, we can calculate the z-statistic as:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.6-0.56-0.5/900}{0.017}=\dfrac{0.039}{0.017}=2.3839$

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

$\text{P-value}=P(z>2.3839)=0.0086$

As the P-value (0.0086) 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 the percentage of residents who favor construction is above 56%.