Question

A survey asked "Do you think the president is doing a great job?" Of the 1200 Americans surveyed, 800 responded yes. For a 95% level of confidence find the sample proportion and the margin of error associated with the poll.

Answer 1

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

$\begin{gathered} \hat{p}=\frac{800}{1200}=\frac{2}{3}\approx0.666666666667\approx0.67 \\ \\ \hat{p}=\frac{800}{1200}\approx0.67 \end{gathered}$

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:

$E=Z_c\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Where:

$\begin{gathered} Z_c\text{ is the critical value for a 95\% level of confidence} \\ \\ n\text{ is the sample size \lparen n = 1200\rparen} \\ \\ \hat{p}\text{ is the sample proportion \lparen800/1200\rparen} \\ \end{gathered}$

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:

$\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}$$\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}$

Therefore, in summary, we have that:

1. The sample proportion is:

$\hat{p}=\frac{800}{1,200}\approx0.67$

2. The margin of error associated with the poll is:

$E=0.0266722216436\approx0.027$