Question

A research group conducted an extensive survey of 3086 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?" 1633 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 = 3086

x = 1633

Point estimate = sample proportion = = x / n = 1633 / 3086=0.529

1 - = 1 - 0.529 =0.471

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 * (((0.529 * 0.471) / 3086) = 0.015

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.529 - 0.015< p < 0.529 + 0.015

0.514 < p < 0.544

The 90% confidence interval for the population proportion p is : ( lower limit = 0.514 upper limit =0.544)