Question

35.4% of employees in a Canadian company had an income equal to higher than 80K in 2014.
In a simple random sample of 100, what is the probability that the proportion of employees with income higher than or equal to 80k is 44%? How large should be your sample for the standard deviation of employees with income 80k or higher income to be less than or equal to sd=0.5

Answer 1

Point estimate of the population proportion with income equal to or higher than 80k in 2014, $\hat{p}$ = 0.354

Sample size, n = 100

Thus, standard deviation of sample proportion is = 0.0478

p(1-P) V n

The probability that the proportion of employees with income higher than or equal to 80k is 44%

= P(p = 0.44)

The corresponding z value = = 1.8

0.44 -0.354 10.0478

Thus, the required probability = P(Z > 1.8) = 0.0359

Let the required sample size be m for sd to be less than or equal to 0.5

Thus,

P(1-P) 50.5 V m

->

m> P(1-P) 0.52

->

m > 0.915

Thus, any sample size will result in a standard deviation less than or equal to 0.5