Question

A research group conducted an extensive survey of 2849 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?" 1607 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:'

Solution :

Given that,

n = 2849

x = 1607

Point estimate = sample proportion = = x / n = 1607 / 2849 =0.564

1 - = 1-0.564=0.436

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.1

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

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

= 1.645(((0.564*0.436) /2849 )

= 0.015

A 90% confidence interval for p is ,

- E < p < + E

0.564-0.015< p < 0.564+0.015

0.549< p < 0.579

LOWE LIMIT 0.549

UPPER LIMIT 0.579