Question

A random sample of 225 individuals working in a large city indicated that 45 are dissatisfied with their working conditions. Based upon this, compute a 90% confidence interval for the proportion of all individuals in this city who are dissatisfied with their working conditions. Then find the lower limit and upper limit of the 90% confidence interval.

Answer 1

CI 90 % = [ 0,165 ; 0,235 ]

Lower limit 0,165

upper limit 0,235

Explanation:

Sample information:

sample size n = 225

number of dissatisfied individuals x = 45

p = 45/225

p = 0,2 and q = 1 - p q = 1 - 0,2 q = 0,8

p*n = 0,2*225 = 45 and q*n = 0,8*225 = 180

p*n and q*n big enough to use the approximation of binomial distribution to normal distribution

90 % of Confidence Interval then a significance level is α = 10%

α = 0,1 in z table we get z(c) for that significance level

z(c) = 1,28

CI 90 % = [ p ± z(c)*√(p*q)/n ]

z(c) * √(p*q)/n = 1,28 * √ ( 0,2*0,8)/225

z(c) * √(p*q)/n = 1,28 * 0,027

z(c) * √(p*q)/n = 0,035

CI 90 % = [ 0,2 ± 0,035 ]

CI 90 % = [ 0,165 ; 0,235 ]

Lower limit 0,165

upper limit 0,235