A researcher wishes to estimate with 90 confidence the propo

A researcher wishes to estimate, with 90% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 5% of the true proportion. a) Find the minimum sample size needed. using a prior study that found that 52% of the respondents said they have high-speed Internet access. b) No preliminary estimate is available. Find the minimum sample size needed. a) What is the minimum sample size needed using a prior study that found that 52% of the respondents said they have high-speed Internet access? n (Round up to the nearest whole number as needed-) b) What is the minimum sample size needed assuming that no preliminary estimate is available? n (Round up to the nearest whole number as needed.)

Solution

given:

z-score at 90% = 1.645
n = size of sample = ?
std = standard deviation = sqrt(p * (1-p) = sqrt(0.52 * 0.48)

standard error of mean = std * z-score/sqrt(n) = 0.05

a)

=>
1.645 * sqrt(0.52 * 0.48)/sqrt(n) = 0.05

minimum sample size n = 270.17 = 270

b)

no preliminary estimate is available so p = 0.5

=>

1.645 * sqrt(0.5 * 0.5)/sqrt(n) = 0.05

minimum sample size n = 270.60 = 271