Question

A research group conducted an extensive survey of 3031 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?" 1524 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 = 3031

x = 1524

= x / n = 1524 / 3031 = 0.503

1 - = 1 - 0.503 = 0.497

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 * (((0.503 * 0.497) / 3031) = 0.015

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.503- 0.015 < p < 0.503 + 0.015

0.488 < p < 0.518

The 90% confidence interval for the population proportion p is : (0.488 ,  0.518)