Question

A research group conducted an extensive survey of 3140 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1473 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p. (Round your answers to three decimal places.)

lower limit =
upper limit =

Answer 1

Solution:

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

x = 1473

n = 3140

P = x/n = 1473/3140 = 0.46910828

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.46910828 ± 1.6449* sqrt(0.46910828*(1 – 0.46910828)/3140)

Confidence Interval = 0.46910828 ± 1.6449* 0.0089

Confidence Interval = 0.46910828 ± 0.0146

Lower limit = 0.46910828 - 0.0146 = 0.454

Upper limit = 0.46910828 + 0.0146 = 0.484

Confidence interval = (0.454, 0.484)