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.)
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Answer:
[tex]\begin{gathered} \text{ Lower bound= }0.625 \\ \text{ Upper bound= }0.775 \end{gathered}[/tex]Step-by-step explanation:
Confidence interval is given as.
[tex]\text{ Mean}\pm\text{ margin of error }[/tex]x is the number of successes and n is the sample size, use them to calculate the sample proportion:
[tex]p-hat=\frac{175}{250}=0.70[/tex]Therefore by:
[tex]p\pm Z_{\frac{\alpha}{2}}\cdot\sqrt[]{\frac{p(1-p)}{n}}[/tex]The lower bound and upper bound would be:
[tex]\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}[/tex]