Question

there is no prior information about the proportion of americans who support gun control in 2018. if we want to estimate 95% confidence interval for the true proportion of americans who support gun control in 2018 with a 0.36 margin of error, how many randomly selected americans must be surveyed? answer: (round up your answer to nearest whole number)

Answer 1

Explanation:

To estimate the required sample size for estimating the true proportion of Americans who support gun control in 2018 with a 95% confidence level and a margin of error of 0.36, we need to use the formula:

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

Where:

n = required sample size

Z = Z-score corresponding to the desired confidence level (for 95% confidence level, Z ≈ 1.96)

p = estimated proportion (since we have no prior information, we can use p = 0.5, which gives the maximum sample size required)

E = margin of error

Substituting the values into the formula:

n = (1.96^2 * 0.5 * (1 - 0.5)) / 0.36^2

n = (3.8416 * 0.25) / 0.1296

n ≈ 9.6042 / 0.1296

n ≈ 74.0842

Rounding up to the nearest whole number, the required sample size is approximately 75. Therefore, you would need to survey at least 75 randomly selected Americans to estimate the true proportion of Americans who support gun control in 2018 with a 95% confidence level and a margin of error of 0.36.