Question

A research group conducted an extensive survey of 3134 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?" 1565 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.)

Answer 1

Solution :

Given that,

n = 3134

x = 1565

= x / n = 1565 /3134 = 0.499

1 - = 1 - 0.499 = 0.501

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 * (((0.499 * 0.501) / 3134) = 0.015

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.499 - 0.015 < p < 0.499 + 0.015

0.484 < p < 0.514

The 90% confidence interval for the population proportion p is : ( 0.484 , 0.514)