Question

A research group conducted an extensive survey of 2875 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?" 1519 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.)

upper limit =

lower limit =

Answer 1

 

n = 2875

x = 1519

Point estimate = sample proportion = = x / n = 1519 / 2875 = 0.528

1 - = 1 - 0.528 = 0.472

At 90% confidence level the z is ,

   = 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * √ (( * (1 - )) / n)

= 1.645 * (√((0.528 * 0.472) / 2875)

= 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.528 - 0.015 < p < 0.528 + 0.015

0.513 < p < 0.543

The 90% confidence interval for the population proportion p is : 0.513 , 0.543

upper limit = 0.513

lower limit =0.543