Question

A research group conducted an extensive survey of 2854 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?" 1619 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 answers to three decimal places.)

lower limit
upper limit

Answer 1

Solution :

Given that,

n = 2854

x = 1619

Point estimate = sample proportion = = x / n = 1619/2854=0.567

1 -   = 1-0.567 =0.433

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 *((0.567*0.433) /2854 )

E = 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.567-0.015 < p < 0.567+ 0.015

0.552< p < 0.582

The 90% confidence interval for the population proportion p is : lower limit =0.552,upper limit=0.582