Question

Exercise 8- 5. Perception and Reality In a presidential election, 308 out of 611 voters surveyed said that they voted for the candidate who won (based on data from ICR Survey Research Group). Use a 0.05 significance level to test the claim that among all voters, the percentage who believe that they voted for the winning candidate is equal to 43%, which is the actual percentage of votes for the winning candidate. What does the result suggest about voter perceptions?

Short answer

There is not enough evidence to support the claim that 43% of voters favored the candidate. This suggests that the voters did not reveal the true votes.

Step-by-step solution

Given information

The summary of the results obtained from the survey is as follows.

Sample size $n=611$.

Votes in favor of winning candidate $x=308$.

Level of significance $\alpha =0.05$.

It is claimed that the proportion of voters in favor of the winning candidate is 43% or 0.43.

State the hypotheses

Null hypothesis: The votes for the winning candidate are equal to 43%.

Alternative hypothesis: The votes for the winning candidate are not equal to 43%

Let p be the true proportion of voters who vote for the candidate who won.

Mathematically,

$$\begin{gathered} H_0:p=0.43 \\ {H}_1:p\ne 0.43 \end{gathered}$$

Compute the test statistic

In testing claims for the population proportion, the z-test is used.

The sample proportion is given as follows.

$$\begin{gathered} \hat{p}=\frac{308}{611} \\ =0.5041 \end{gathered}$$

From the given information,

$$\begin{gathered} p=0.43 \\ q=1-p \\ =1-0.43 \\ =0.57 \end{gathered}$$

The test statistic is computed as follows.

$$\begin{gathered} z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}} \\ =\frac{0.5041-0.43}{\sqrt{\frac{0.43\times 0.57}{611}}} \\ =3.70 \end{gathered}$$

Compute the critical value

From the standard normal table, the critical values for a two-tailed test with a 0.05 significance level are obtained as follows.

The left-tailed critical value will have an area of 0.025, corresponding to row -1.9 and column 0.06.

The right-tailed critical value will have an area of 0.975, corresponding to row 1.9 and column 0.06.

Thus, the two critical values are -1.96 and 1.96.

State the decision

If the test statistic lies between the critical values, the null hypothesis is failed to be rejected. Otherwise, the null hypothesis is rejected.

Here, 3.70 does not fall between the critical values -1.96 and 1.96.

Thus, the null hypothesis is rejected.

Conclusion

The result suggests that there is not sufficient evidence to support the claim that the actual percentage of voters who favored the winning candidate is 43%.

Asthe proportion of voters in the survey is quite high, it suggests that the voters did not reveal the true votes in the survey.