Question

Two individuals are running for mayor of Shinbone. You conduct an election survey a week before the election and find that \(51 \%\) of the respondents prefer candidate A. Can you predict a winner? Use the \(99 \%\) level. (HINT: In a two-candidate race, what percentage of the vote would the winner need? Does the confidence interval indicate that candidate \(A\) has a sure margin of victory? Remember that while the population parameter is probably \(\alpha=0.01\) in the confidence interval, it may be anywhere in the interval.) $$ \begin{aligned} &P_{s}=0.51 \\ &N=578 \end{aligned} $$

Short answer

No, the confidence interval suggests that candidate A does not have a sure margin of victory.

Step-by-step solution

Determine the proportion and sample size

From the problem, the sample proportion (p_s) is given as 0.51, and the sample size (N) is 578.

Find the critical value for the 99% confidence level

For a 99% confidence level, the critical value (z) is approximately 2.576.

Calculate the standard error

The standard error (SE) is calculated using the formula: \[ SE = \sqrt{\frac{p_s(1 - p_s)}{N}} = \sqrt{\frac{0.51 \times 0.49}{578}} \approx 0.0208 \]

Calculate the margin of error

The margin of error (MOE) is given by: \[ MOE = z \times SE = 2.576 \times 0.0208 \approx 0.0536 \]

Construct the confidence interval

The confidence interval (CI) is calculated as follows: \[ CI = p_s \pm MOE = 0.51 \pm 0.0536 \] This results in a confidence interval of: \[ CI = (0.4564, 0.5636) \]

Interpret the confidence interval

Since the confidence interval (0.4564, 0.5636) includes values below 0.50, candidate A does not have a sure margin of victory. There is a possibility that the true proportion of voters who prefer candidate A could be less than 50%.

Key concepts

Confidence Interval

A confidence interval gives an estimated range of values likely to include an unknown population parameter. In this scenario, the population parameter is the true proportion of voters who prefer candidate A. By using a confidence interval, we can say with 99% confidence that this range contains the true proportion.
The confidence interval is calculated using the sample proportion and the margin of error. The sample proportion in this case is 0.51, and the margin of error helps adjust this estimate to account for sampling variability.
For a 99% confidence level, the calculated interval was (0.4564, 0.5636). This interval means that if we repeated the survey many times, 99% of the intervals calculated from those samples would contain the true proportion of voters for candidate A.

Margin of Error

The margin of error (MOE) reflects the range of uncertainty around the sample proportion and is crucial in constructing the confidence interval. It considers the variability within the data and the confidence level we desire.
In our example, the margin of error is determined by multiplying the standard error with the critical value (z) for the desired confidence level. The critical value for a 99% confidence level is approximately 2.576.
The formula used here is:

Standard Error

The standard error (SE) measures the variability of the sample proportion. It determines how much the sample proportion is expected to fluctuate due to random sampling variation.
The standard error is calculated using the following formula:

Sample Proportion

The sample proportion (p_s) represents the proportion of respondents in the sample who prefer a certain candidate. It's an estimate of the true population proportion.
In our election survey, the sample proportion is given as: