Question

A research group conducted an extensive survey of 3077 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?" 1631 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

 

n = 3077

x = 1631

Point estimate = sample proportion = = x / n = 1631/3077=0.530

1 -   = 1-0.530 =0.470

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 *√((0.530*0.470) /3077 )

E = 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.530-0.015 < p <0.530+ 0.015

0.515< p <0.545

The 90% confidence interval for the population proportion p is :lower limit= 0.515,upper limit=0.545