Question

A political candidate has asked you to conduct a poll to determine what percentage of people supporther. If the candidate only wants a 2% margin of error at a 99% confidence level, what size of sample is needed? Give your answer in whole people

Answer 1

To determine the sample size needed to conduct a poll with a 2% margin of error at a 99% confidence level, use the formula n = (Z^2 * p * (1-p)) / E^2, plugging in the values gives a sample size of approximately 4193 people.

Step-by-step explanation:

To determine the sample size needed, we can use the formula:

n = (Z^2 * p * (1-p)) / E^2

Where:

• n is the sample size
• Z is the z-score corresponding to the desired confidence level
• p is the estimated proportion (0.5 for maximum sample size)
• E is the margin of error

In this case, the candidate wants a 2% margin of error at a 99% confidence level. The z-score corresponding to a 99% confidence level is approximately 2.576.

Plugging in the values, we get:

n = (2.576^2 * 0.5 * (1-0.5)) / 0.02^2

n ≈ 4192.96

Therefore, a sample size of 4193 people is needed to achieve a 2% margin of error at a 99% confidence level.

Answer 2

As per the values asample size of 666 people is needed.

Step-by-step explanation:

To determine the sample size needed to estimate the proportion of people who support a political candidate with a 2% margin of error at a 99% confidence level, we can use the formula:

Sample size = (Zα/2)2 * p * (1-p) / E2

Where Zα/2 is the critical value for a 99% confidence level (2.58), p is the estimated proportion, and E is the desired margin of error in decimal form.

Since the margin of error is 2% or 0.02, we can plug in the values:

Sample size = (2.58)2 * 0.5 * (1-0.5) / 0.022 = 665.64

Rounding up to the nearest whole number, a sample size of 666 people is needed.