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Sagot :
We have here working with estimating a population proportion.
We have the following information from the question:
• The sample size, n, is equal to 1200 (n = 1200).
,• We have that the fraction that responded "Yes" is 800.
,• We need to find a 95% level of confidence for the margin error associated with the poll.
Now, we have the sample proportion for the sample size, n = 1200 is as follows:
Sample Proportion
[tex]\begin{gathered} \hat{p}=\frac{800}{1200}=\frac{2}{3}\approx0.666666666667\approx0.67 \\ \\ \hat{p}=\frac{800}{1200}\approx0.67 \end{gathered}[/tex]Therefore, the sample proportion for the sample size is 800/1200, which is, approximately, 0.67.
The margin of error associated with the poll
The margin of error, in this case, is given by the next formula:
[tex]E=Z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}[/tex]Where:
[tex]\begin{gathered} Z_c\text{ is the critical value for a 95\% level of confidence} \\ \\ n\text{ is the sample size \lparen n = 1200\rparen} \\ \\ \hat{p}\text{ is the sample proportion \lparen800/1200\rparen} \\ \end{gathered}[/tex]Now, we have that, for a level of confidence of 95%, the critical value is equal to z = 1.96:
Now, using all of the values at our disposal, we can use the formula to find the margin of error as follows:
[tex]\begin{gathered} E=Z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ \\ E=1.96\sqrt{\frac{\frac{8}{12}(1-\frac{8}{12})}{1200}} \\ \\ E=0.0266722216436\approx0.027 \end{gathered}[/tex][tex]\begin{gathered} E=Z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ \\ E=1.96\sqrt{\frac{\frac{8}{12}(1-\frac{8}{12})}{1200}} \\ \\ E=0.0266722216436\approx0.027 \end{gathered}[/tex]Therefore, in summary, we have that:
1. The sample proportion is:
[tex]\hat{p}=\frac{800}{1,200}\approx0.67[/tex]2. The margin of error associated with the poll is:
[tex]E=0.0266722216436\approx0.027[/tex]
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