Question

Critical Thinking. In Exercises 17–28, use the data and confidence level to construct a confidence interval estimate of p, then address the given question.

Nonvoters Who Say They Voted In a survey of 1002 people, 70% said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records show that 61% of eligible voters actually did vote.

a. Find a 98% confidence interval estimate of the proportion of people who say that they voted.

b. Are the survey results consistent with the actual voter turnout of 61%? Why or why not?

Short answer

a. The 98% confidence interval is equal to (0.666,0.734).

b. No, the survey results are not consistent with the actual voter turnout of 61%.

Step-by-step solution

Given Information

The information about the voting percentage in an election is given. In a sample of 1002 people, 70% said that they voted in the election. 61% of the eligible voters actually did vote.

Calculation of the sample proportion

The sample size (n) is equal to 1002.

The sample proportion of voters who said they voted in the election is given below:

$$\begin{gathered} \hat{p}=70\% \\ =\frac{70}{100} \\ =0.70 \end{gathered}$$

The value of the sample of proportion is equal to 0.70.

The sample proportion of voters who did not vote in the election is given below:

$$\begin{gathered} \hat{q}=1-\hat{p} \\ =1-0.70 \\ =0.30 \end{gathered}$$

Calculation of the margin of error

The given level of significance is 0.02.

Therefore, the value of $z_{\frac{\alpha }{2}}$form the standard normal table is equal tois equal to 2.3263.

The margin of error is equal to

$$\begin{gathered} \text{E}=z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\hat{q}}{n}} \\ =2.3263\times \sqrt{\frac{0.70\times 0.30}{1002}} \\ =0.0337 \end{gathered}$$

Therefore, the margin of error is equal to 0.0337.

Calculation of the confidence interval

a.

The 98% confidence interval has the following value:

$$\begin{gathered} \hat{p}-E<p<\hat{p}+E \\ 0.70-0.0337<p<0.70+0.0337 \\ 0.666<p<0.734 \end{gathered}$$

Thus, the 98% confidence interval is equal to (0.666,0.734).

Conclusion

b.

The actual voter turnout is equal to 61% or 0.61.

The confidence interval does not contain the value of 0.61 and contains all values greater than 0.61.

This means that people have lied about voting in the election.

Therefore, the survey results are not consistent with the actual voter turnout of 61%.