Exit polling is a popular technique used to determine the outcome of an election prior to results being tallied. Suppose a referendum to increase funding for education is on the ballot in a large town (voting population over 100,000). An exit poll of 400 voters finds that 204 voted for the referendum. How likely are the results of your sample if the population proportion of voters in the town in favor of the referendum is 0.49 ? Based on your result, comment on the dangers of using exit polling to call elections. Comment on the dangers of using exit polling to call elections. Choose the correct answer below. A. The result is unusual because the probability that $\hat{p}$ is equal to or more extreme than the sample proportion is greater than $5 \%$. Thus, it is not unusual for a wrong call to be made in an election if exit polling alone is considered. B. The result is not unusual thicause the probability that $\hat{p}$ is equal to or more extreme than the sample proportion is less than $5 \%$. Thus, it is unusual for a wrong call to be made in an election if exit polling alone is considered. C. The result is not unusual because the probability that $\hat{p}$ is equal to or more extreme than the sample proportion is greater than $5 \%$. Thus, it is not unusual for a wrong call to be made in an election if exit polling alone is considered. D. The result is unusual because the probability that $\hat{p}$ is equal to or more extreme than the sample proportion is less than $5 \%$. Thus, it is unusual for a wrong call to be made in an election if exit polling alone is considered.
Solution
Step 1 ：First, we calculate the sample proportion, denoted as \(\hat{p}\), which is the number of voters who voted for the referendum divided by the total number of voters in the sample. In this case, \(\hat{p} = \frac{204}{400} = 0.51\).
Step 2 ：Next, we calculate the standard error. The standard error is calculated using the formula \(\sqrt{p(1-p)/n}\), where p is the population proportion and n is the sample size. In this case, p = 0.49 and n = 400, so the standard error is \(\sqrt{0.49(1-0.49)/400} = 0.024994999499899976\).
Step 3 ：Then, we calculate the z-score. The z-score is calculated using the formula \((\hat{p} - p) / \text{standard error}\). In this case, the z-score is \((0.51 - 0.49) / 0.024994999499899976 = 0.8001600480160063\).
Step 4 ：After that, we calculate the probability that \(\hat{p}\) is equal to or more extreme than the sample proportion. This is done by calculating the area under the standard normal curve that is more extreme than the calculated z-score. The probability is calculated as \(2(1 - \text{cdf}(\text{abs}(z-score)))\), where cdf is the cumulative distribution function of the standard normal distribution. In this case, the probability is \(2(1 - \text{cdf}(0.8001600480160063)) = 0.4236180739868338\), or approximately 42.36%.
Step 5 ：Finally, we compare the calculated probability with 5%. If the probability is greater than 5%, the result is not considered unusual. In this case, the probability is 42.36%, which is greater than 5%, so the result is not unusual.
Step 6 ：\(\boxed{\text{Final Answer: The correct answer is C. The result is not unusual because the probability that }\hat{p}\text{ is equal to or more extreme than the sample proportion is greater than 5%. Thus, it is not unusual for a wrong call to be made in an election if exit polling alone is considered.}}\)
