Voters Lying? In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records showed that \({\bf{61\% }}\) of eligible voters actually did vote.

a. Given that \({\bf{61\% }}\)of eligible voters actually did vote, find the probability that among 1002 randomly selected eligible voters, at least 701 actually did vote.

b. What does the result suggest?

The percentage of eligible voters who did not vote is equal to 61%.

A sample of 1002 eligible voters is taken. Out of the given sample, 701 voters did not vote.

Let X (number of successes) denote the number of eligible voters who did not vote.

The probability of success is given to be equal to:

\(\begin{aligned}{c}p = 61\% \\ = \frac{{61}}{{100}}\\ = 0.61\end{aligned}\)

The number of trials (n) is equal to 1002.

Here, the sample is a result of 1002 independent trials with a probability of success at each trial equal to 0.61.

Also,

\(\begin{aligned}{c}np = 1002 \times 0.61\\ = 611.22\\ > 5\end{aligned}\)

\(\begin{aligned}{c}nq = n(1 - p)\\ = 1002 \times (1 - 0.61)\\ = 390.78\\ > 5\end{aligned}\)

Since the above two requirements are met, the normal distribution can be used for approximating the binomial distribution.

It is required to compute the probability of at least 701 eligible voters who did vote.

Thus, the interval of continuity correction is computed below:

\(\begin{aligned}{c}\left( {x - 0.5,x + 0.5} \right) = \left( {701 - 0.5,701 + 0.5} \right)\\ = \left( {700.5,701.5} \right)\end{aligned}\)

In terms of the bound of the continuity correction interval, the following probability needs to be computed:

\(P\left( {x > {\rm{lower}}\;{\rm{bound}}} \right) = P\left( {x > 700.5} \right)\)

The sample value equal to x=701 is converted to a z-score as follows:

\(\begin{aligned}{c}z = \frac{{x - np}}{{\sqrt {np(1 - p)} }}\\ = \frac{{700.5 - 1002(0.61)}}{{\sqrt {1002(0.61)(1 - 0.61)} }}\\ = \frac{{89.28}}{{15.44}}\\ = 5.78\end{aligned}\)

a.

The probability of at least 701 eligible voters who did vote is computed using the standard normal table as follows:

\(\begin{aligned}{c}P\left( {x > 700.5} \right) = P\left( {z > 5.78} \right)\\ = 1 - P\left( {z < 5.78} \right)\\ = 1 - 1.000\\ \approx 0.000\end{aligned}\)

Thus, the probability of at least 701 eligible voters who did vote is approximately equal to 0.000.

b.

It can be said that the result of at least 701 voters who did vote is significantly low as the probability value is almost negligible or too small (less than 0.05).

This suggests that there is enough evidence that the true percentage of eligible voters who did actually vote is not equal to 61%. Therefore, the voters lied about voting in the elections.