Question

A research group conducted an extensive survey of 2942 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?" 1562 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

Given that,

n = 2942

x = 1562

Point estimate = sample proportion = = x / n =1562/ 2942=0.531

1 - = 1-0.531=0.469

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.531*0.469) / 2942)

E= 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.531-0.015 < p <0.531+ 0.015

0.516< p < 0.546

The 90% confidence interval for the population proportion p is : lower limit 0.516,upper limit 0.546