Question

A research group conducted an extensive survey of 2822 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?" 1520 responded, "Personal satisfaction from doing a good job." Let pbe 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 :

Given that,

n = 2822

x = 1520

Point estimate = sample proportion = = x / n = 1520 / 2822 =

1 - = 1 - 0.539 = 0.461

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 * (((0.539 * 0.461) / 2822)

= 0.016

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.539 - 0.016 < p < 0.539 + 0.016

0.523 < p < 0.554

The 90% confidence interval for the population proportion p is : (0.523 , 0.554)

Lower limit = 0.523

Upper limit = 0.554