Question

Construct a 99% confidence interval of the population proportion using the given information. x = 175n = 250The lower bound is __The upper bound is __(Round to three decimal places as needed.)

Answer 1

Answer:

$\begin{gathered} \text{ Lower bound= }0.625 \\ \text{ Upper bound= }0.775 \end{gathered}$

Step-by-step explanation:

Confidence interval is given as.

$\text{ Mean}\pm\text{ margin of error }$

x is the number of successes and n is the sample size, use them to calculate the sample proportion:

$p-hat=\frac{175}{250}=0.70$

Therefore by:

$p\pm Z_{\frac{\alpha}{2}}\cdot\sqrt[]{\frac{p(1-p)}{n}}$

The lower bound and upper bound would be:

$\begin{gathered} 0.70\pm Z_{0.005}\cdot\sqrt[]{\frac{0.7*0.3}{250}} \\ 0.70\pm\mleft(-2.576\mright)*\sqrt[]{0.00084} \\ \text{ Lower bound= }0.625 \\ \text{ Upper bound= }0.775 \end{gathered}$