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

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

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=(175)/(250)=0.70`

Therefore by:

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

The lower bound and upper bound would be:

`\begin{gathered} 0.70\pm Z_(0.005)\cdot\sqrt[]{(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}`