Question

A research group conducted an extensive survey of 2862 wage and salaried workers on issues ranginf 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?” 1531 responded, “personal satisfaction from doing a good job.” Let p be population proportion of all wage and salaried workers who would respond the same way to the stated question. Find 90% confidence interval for p.

a) lower limit

b) upper limit

Answer 1

Solution :

Given that,

n = 2862

x = 1531

= x / n = 1531 /2862 = 0.535

1 - = 1 - 0.535 = 0.465

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.535 * 0.465) / 2862) = 0.0153

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.535 - 0.0153 < p < 0.535 + 0.0153

0.5197 < p < 0.5503

The 90% confidence interval for the population proportion p is : (lower bound = 0.5197 , upper bound = 0.5503)