Question

In a poll of 1000 randomly selected voters in a local election, 625 voters were against school bond measures. What is the sample proportion p? What is the margin of error m for the 99% confidence level?

Answer 1

p = 625 / 1000 = 0.625

m = 2.576 * sqrt((0.625 * (1 - 0.625)) / 1000) = 0.03

Step-by-step explanation:

The sample proportion p is the proportion of voters in the sample who were against school bond measures. To calculate this, divide the number of voters against the bond measures (625) by the total number of voters in the sample (1000):

p = 625 / 1000 = 0.625

The margin of error m for a sample proportion at the 99% confidence level can be calculated using the formula:

m = zsqrt(p(1-p)/n)

where z is the z-score for the desired level of confidence (for a 99% confidence level, z = 2.576), p is the sample proportion, and n is the sample size.

m = 2.576 * sqrt((0.625 * (1 - 0.625)) / 1000) = 0.03

So the margin of error for this poll at the 99% confidence level is 0.03.