Math Solutions

[See Steps] The point estimate is p#770;=(0.375+0.325)/(2)=0.350 and the margin of error is E=(0.375-0.325)/(2)=0.025 7.2, #18: The margin of error is computed

Question: #14: The point estimate is

\[\hat{p}=\frac{0.375+0.325}{2}=0.350\]

and the margin of error is

\[E=\frac{0.375-0.325}{2}=0.025\]

7.2, #18: The margin of error is computed as

\[MOE={{z}_{\alpha /2}}\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}}=\text{2}\text{.58}\sqrt{\frac{\text{0}\text{.667}\left( 1-\text{0}\text{.667} \right)}{\text{1200}}}=\text{0}\text{.035}\]

7.2, #22: Based on the information provided, the \(\text{99}%\) confidence interval for the actual population proportion is computed as

\[CI=\left( \hat{p}-{{z}_{\alpha /2}}\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}},\,\,\hat{p}+{{z}_{\alpha /2}}\sqrt{\frac{\hat{p}\left( 1-\hat{p} \right)}{n}} \right)\]

Here, we have that

\[\begin{aligned} & \hat{p}=\,\text{Sample Proportion}=\frac{x}{n}=\frac{\text{800}}{\text{1200}}=\text{0}\text{.667} \\ & {{z}_{\alpha /2}}=\text{ Critical z-value}=\text{2}\text{.58} \\ \end{aligned}\]

and n corresponds to the sample size. In this case, we have that \(n=\text{1200}\). The confidence interval is therefore

\[CI=\left( \text{0}\text{.667}-\text{2}\text{.58}\sqrt{\frac{\text{0}\text{.667}\left( 1-\text{0}\text{.667} \right)}{\text{1200}}},\,\,\text{0}\text{.667}-\text{2}\text{.58}\sqrt{\frac{\text{0}\text{.667}\left( 1-\text{0}\text{.667} \right)}{\text{1200}}} \right)=(\text{0}\text{.632},\,\,\text{0}\text{.702})\]

This means that there is a probability of \(\text{99}%\) that the actual population proportion p is contained by the interval \(\left( \text{0}\text{.632},\,\text{ 0}\text{.702} \right)\).

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