Question

a company wants to determine if there has been gender bias in their hiring. over the past 5 years, 72% of applicants have been male and 28% female. over the duration of 5 years, they hired 110 employees (80% male, 20% female).

a) construct a 99% confidence interval to estimate the probability of a hired employee being male. are male applicants more likely to be hired at this company?

b) suppose the company wants to estimate the probability of a hired employee being male with a 95% confidence interval and margin of error no greater than 5%. what is the smallest sample size they could have to guarantee this margin of error?

Answer 1

n = 110, p̅ = 0.8

a) 99% Confidence interval to estimate the probability of a hired employee being male:

At α = 0.01, two tailed critical value, z_c = NORM.S.INV(0.01/2) = 2.576

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.8 - 2.576 *√(0.8*0.2/110) = 0.702

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.8 + 2.576 *√(0.8*0.2/110) = 0.898

0.702 < p < 0.898

As the value 0.72 is in the confidence interval, so, there is not enough evidence to conclude that male applicants are more likely to be hired at this company.

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b)

Proportion, p = 0.72

Margin of error, E = 0.05

Confidence Level, CL = 0.95

Significance level, α = 1 - CL = 0.05

Critical value, z = NORM.S.INV(0.05/2) = 1.9600

Sample size, n = (z² * p * (1-p)) / E² = (1.96² * 0.72 * 0.28)/ 0.05²

= 309.775 = 310