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}=(800)/(1200)=(2)/(3)\approx0.666666666667\approx0.67 \\ \\ \hat{p}=(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{((8)/(12)(1-(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{((8)/(12)(1-(8)/(12)))/(1200)} \\ \\ E=0.0266722216436\approx0.027 \end{gathered}`

Therefore, in summary, we have that:

1. The sample proportion is:

`\hat{p}=(800)/(1,200)\approx0.67`

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

`E=0.0266722216436\approx0.027`