Question

Construct a confidence interval of the population proportion at the given level of confidence.

Answer 1

Answer:

The lower bound of the confidence interval is 0.6922.

Step-by-step explanation:

We have to calculate a 94% confidence interval for the proportion.

The sample proportion is p=0.7167.

$p=X/n=860/1200=0.7167$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.7167*0.2833}{1200}}\\\\\\ \sigma_p=\sqrt{0.000169}=0.013$

The critical z-value for a 94% confidence interval is z=1.8808.

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

$MOE=z\cdot \sigma_p=1.8808 \cdot 0.013=0.0245$

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

$LL=p-z \cdot \sigma_p = 0.7167-0.0245=0.6922\\\\UL=p+z \cdot \sigma_p = 0.7167+0.0245=0.7412$

The 94% confidence interval for the population proportion is (0.6922, 0.7412).