Question

A research group conducted an extensive survey of 3058 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?" 1605 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.) the lower limit upper limit lower limit upper limit

Answer 1

 

n = 3058

x = 1605

Point estimate = sample proportion = = x / n = 1605/3058=0525

1 -   = 1- 0.525 =0.475

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

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

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

= 1.645 *√((0.525*0.475) / 3058)

E = 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.525-0.015 < p < 0.525+0.015

0.510< p < 0.540

lower limit=0.510

upper limit=0.540