Answer
There is sufficient evidence to reject that the percentage of voters who say they voted is 61%.
Work Step by Step
$H_{0}:p=61$%=0.61. $H_{a}:p\ne0.61$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=0.7.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.7-0.61}{\sqrt{0.61(1-0.61)/1002}}=5.84.$ The P is the probability of the z-score being more than 5.84 or less than -5.84 is the sum of the probability of the z-score being less than -5.84 plus 1 minus the probability of the z-score being less than 5.84, hence:P=0.0001+1-0.9999=0.0002. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0002 is less than $\alpha=0.01$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to reject that the percentage of voters who say they voted is 61%.
