Question

A research group conducted an extensive survey of 2970 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?" 1584 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 :

Given that,

n = 2970

x = 1587

Point estimate = sample proportion = = x / n = 1587 /2970=0.534

1 - = 1 -0.534=0.466

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 (((0.534*0.466) / 2970)

E = 0.015

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.534 -0.015 < p <0.534+ 0.015

0.519< p < 0.549

The 90% confidence interval for the population proportion p is : lower limit 0.519 upper limit 0.549