Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

Answer 1

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.

$p=X/n=69/130=0.531$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.531*0.469}{130}}\\\\\\ \sigma_p=\sqrt{0.001916}=0.044$

The critical z-value for a 90% confidence interval is z=1.645.

The margin of error (MOE) can be calculated as:

$MOE=z\cdot \sigma_p=1.645 \cdot 0.044=0.072$

Then, the lower and upper bounds of the confidence interval are:

$LL=p-z \cdot \sigma_p = 0.531-0.072=0.459\\\\UL=p+z \cdot \sigma_p = 0.531+0.072=0.603$

The 90% confidence interval for the population proportion is (0.459, 0.603).