A city ballot includes a local initiative that would legalize gambling. The issue is hotly contested, and two groups decide to conduct polls to predict the outcome. The local newspaper finds that 53% of 1200 randomly selected voters plan to vote "yes," while a college statistics class finds that 54% of 450 randomly selected voters in support.
Once of the groups concludes the outcome is too close to call. Which group and why?
A) The newspaper since their point estimate is lower.
B) The statistics class since their point estimate is higher.
C) The statistics class since the lower bound of their confidence interval is below 50%.
D) The newspaper since the lower bound of their confidence interval is just above 50%.
E) It is impossible to determine with the given information.
Is this D? 0.53*1200=636
636= success
1200= population
90% interval
confidence interval is 50.5-55.3?
To determine which group concludes that the outcome is too close to call, we need to evaluate the information given.
The newspaper conducted a poll of 1,200 randomly selected voters and found that 53% plan to vote "yes" on the initiative. From this data, we can calculate a confidence interval to estimate the true proportion of voters in favor of the initiative. The confidence interval is a range within which the true proportion is likely to fall.
Now, let's calculate the confidence interval for the newspaper's poll. The point estimate is 53% of 1,200, which is equivalent to 0.53 * 1,200 = 636. This represents the number of voters in the sample who plan to vote "yes." The sample size is 1,200, and we want to construct a 90% confidence interval.
Using the formula for calculating a confidence interval for a proportion, we can find the margin of error:
Margin of Error = (Z-score) * (Standard Error)
Since the sample size is large (n > 30) and the population proportion is unknown, we can use the standard normal distribution for the Z-score. For a 90% confidence level, the Z-score is approximately 1.645.
To calculate the Standard Error, we use the formula:
Standard Error = sqrt((p * (1 - p)) / n)
Where "p" is the point estimate (in this case, 0.53) and "n" is the sample size (1,200).
Substituting the values, we get:
Standard Error = sqrt((0.53 * (1 - 0.53)) / 1,200)
Standard Error ≈ 0.014
Now, we can calculate the margin of error:
Margin of Error = 1.645 * 0.014 ≈ 0.023
The confidence interval for the newspaper's poll is therefore:
53% ± 2.3% or (50.7%, 55.3%)
On the other hand, the college statistics class conducted a poll of 450 randomly selected voters and found that 54% plan to vote "yes" on the initiative. However, we do not have any information about the class's confidence level or the confidence interval calculated for their poll. So we cannot determine the lower bound of their confidence interval.
Based on the given information, the only conclusion we can make is that the newspaper's lower bound of the confidence interval is just above 50%. Therefore, the correct answer is:
D) The newspaper since the lower bound of their confidence interval is just above 50%.