Answer:
P-value = 0.0367
Step-by-step explanation:
This is a hypothesis test for a proportion.
The claim is that the percentage of residents who favor construction is significantly over 42%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.42\\\\H_a:\pi>0.42[/tex]
The sample has a size n=900.
The sample proportion is p=0.45.
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.42*0.58}{900}}\\\\\\ \sigma_p=\sqrt{0.000271}=0.016[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.45-0.42-0.5/900}{0.016}=\dfrac{0.029}{0.016}=1.79[/tex]
This test is a right-tailed test, so the P-value for this test is calculated as:
[tex]\text{P-value}=P(z>1.79)=0.0367[/tex]