Question

A research group conducted an extensive survey of 2930 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?" 1463 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 = 2930

x = 1463

Point estimate = sample proportion = = x / n = 1463/2930=0.499

1 -   = 1-0.499 =0.501

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.499*0.501) /2930 )

= 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 :lower limit=0.484, upper limit=0.514