Question

A pollster wants to find a 95% confidence interval for the proportion of registered voters who support a certain candidate. If her confidence interval has to have a margin of error of at most 5.0%, then what’s the smallest random sample of registered voters she can poll?

Answer 1

When prior estimate for proportion is not specified then p = 0.50

Sample size = Z²/2 * p( 1 -p ) / E²

= 1.96² * 0.50 * 0.50 / 0.05²

= 384.16

Sample size = 385 (Rounded up to nearest integer)

Answer 2

To find the minimum sample size, we need to use the formula:

n = (z^2 * p * q) / E^2

where:

• n is the sample size

• z is the z-score associated with the desired confidence level (95% in this case), which is 1.96

• p is the estimated proportion of registered voters who support the candidate (we don't know this yet)

• q is the complement of p, which is 1 - p

• E is the desired margin of error, which is 5.0% or 0.05

We can rearrange the formula to solve for p:

p = (z^2 * E^2) / (n * 4) + 0.5

Substituting the given values:

p = (1.96^2 * 0.05^2) / (n * 4) + 0.5 p = 0.002401 / n + 0.5

To find the minimum value of n, we need to assume that p is 0.5 (which gives us the largest possible value of pq), and solve for n:

0.05 = z * sqrt((p * q) / n) 0.05 = 1.96 * sqrt((0.5 * 0.5) / n) n = (1.96^2 * 0.5 * 0.5) / 0.05^2 n = 384.16

Rounding up to the nearest integer, we get a minimum sample size of 385 registered voters.