Respuesta :
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:
[tex]H_0: \pi=0.56\\\\H_a:\pi>0.56[/tex]
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:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.56*0.44}{900}}\\\\\\ \sigma_p=\sqrt{0.000274}=0.017[/tex]
Then, we can calculate the z-statistic as:
[tex]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[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=P(z>2.3839)=0.0086[/tex]
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%.
