The Television Networks Often Compete On The Evening

The Television Networks Often Compete On The Evening Of The Day Of An

The television networks often compete on the evening of the day of an election to be the first to identify the winner of the election correctly. One technique used is the random sampling of voters as they exit polling booths. Suppose that, in a two-candidate race, 55% of a random sample of 500 voters indicate that they voted Republican. Can we conclude at the 5% level of significance that the Republican candidate will win?

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Election night coverage has long been a competitive arena for television networks aiming to project the winning candidate as early as possible. A common method employed by these networks involves exit polling, where a random sample of voters is surveyed immediately after they leave polling stations. This technique provides preliminary insights into the likely election outcomes, though it comes with inherent uncertainties due to sampling variability.

In the scenario presented, a sample of 500 voters was surveyed, with 55% indicating they voted for the Republican candidate. This proportion suggests a potential lead for the Republican candidate, but statistical analysis is necessary to determine whether this observed result truly reflects the broader voting population or is simply due to random chance within the sample. The critical question here is whether the data provides sufficient evidence to conclude, at a 5% significance level, that the Republican candidate will indeed win the election.

To analyze this, we conduct a hypothesis test for a population proportion. The null hypothesis (H₀) posits that the true proportion of voters favoring the Republican candidate is 50%, indicating no advantage. The alternative hypothesis (H₁) suggests that the true proportion exceeds 50%, implying a lead for the Republican candidate:

  • H₀: p = 0.50
  • H₁: p > 0.50

The sample proportion (\(\hat{p}\)) is 0.55, and the sample size (n) is 500. Using these values, we calculate the test statistic based on a standard normal distribution:

z = (\(\hat{p}\) - p₀) / √(p₀(1 - p₀) / n)

where p₀ is the hypothesized proportion (0.50). Substituting the values:

z = (0.55 - 0.50) / √(0.50 * 0.50 / 500) = 0.05 / √(0.25 / 500) = 0.05 / √0.0005 ≈ 0.05 / 0.02236 ≈ 2.236

The critical z-value at the 5% significance level for a one-tailed test is approximately 1.645. Since our calculated z-value of 2.236 exceeds this threshold, we have sufficient evidence to reject the null hypothesis. In simple terms, this means that based on the sampled data, there is a statistically significant indication that the Republican candidate has more than a 50% chance of winning.

However, it is important to recognize the limitations of this analysis. Exit polls, although useful, are estimates and subject to sampling error. A sample proportion of 55% does not guarantee the actual election result; it only suggests a tendency within the sampled population. Furthermore, exit polls can be affected by various biases, such as non-response bias or misreporting. Therefore, while the statistical test provides evidence favoring the Republican candidate, it should not be considered an absolute prediction.

In conclusion, at the 5% significance level, the data presents enough evidence to support the claim that the Republican candidate is likely to win the election, based on the exit poll results. Nonetheless, definitive results should await official counts. The collaborative use of exit polling and statistical analysis helps media outlets and the public gauge the electoral landscape early, but it must always be interpreted within the context of its limitations.

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