91影视

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.

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:

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

The given level of significance is 0.02.

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

The margin of error is equal to

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

Therefore, the margin of error is equal to 0.0337.

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).

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%.