147 views
5 votes
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.

User Koobz
by
7.9k points

1 Answer

4 votes

Answer:

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

User Windchime
by
8.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories