Answer:
The test statistic for this hypothesis test for a proportion is z=1.90.
Step-by-step explanation:
This is a hypothesis test for a proportion.
The claim is that the unemployment rate is less than 5%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.05\\\\H_a:\pi<0.05[/tex]
The sample has a size n=1500.
The sample proportion is p=0.061.
[tex]p=X/n=92/1500=0.061[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.05*0.95}{1500}}\\\\\\ \sigma_p=\sqrt{0.000032}=0.0056[/tex]
Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.061-0.05-0.5/1500}{0.0056}=\dfrac{0.0107}{0.0056}=1.90[/tex]
