Question

A research group conducted an extensive survey of 3071 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?" 1477 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 :

Given that,

n = 3071

x = 1477

Point estimate = sample proportion = = x / n = 1477 / 3071 = 0.481

1 - = 1 - 0.481 = 0.519

At 90% confidence level the z is,

= 1 - 0.90 = 0.10

/2 = 0.05

Z_/2 = Z0.05 = 1.645

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

= 1.645 * (((0.481 * 0.519) / 3071)

= 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.481 - 0.015 < p < 0.481 + 0.015

0.466 < p < 0.496

The 90% confidence interval for the population proportion p is :lower limit = 0.466, upper limit = 0.496.