Answer:
d. 0.459 < p < 0.603
Step-by-step explanation:
We have to calculate a 90% confidence interval for the proportion.
The sample proportion is p=0.531.
[tex]p=X/n=69/130=0.531[/tex]
The standard error of the proportion is:
[tex]\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.531*0.469}{130}}\\\\\\ \sigma_p=\sqrt{0.001916}=0.044[/tex]
The critical z-value for a 90% confidence interval is z=1.645.
The margin of error (MOE) can be calculated as:
[tex]MOE=z\cdot \sigma_p=1.645 \cdot 0.044=0.072[/tex]
Then, the lower and upper bounds of the confidence interval are:
[tex]LL=p-z \cdot \sigma_p = 0.531-0.072=0.459\\\\UL=p+z \cdot \sigma_p = 0.531+0.072=0.603[/tex]
The 90% confidence interval for the population proportion is (0.459, 0.603).
