Question

4. A firm would like to know the proportion of its employees that expect to receive a raise within the next year. The firm surveys 150 employees and finds that 60 expect a raise. Construct a 90% confidence interval for the population proportion.

a) What is the sample proportion?

b) State the critical value:

c) Calculate the margin of error (round to the thousandths place):

d) State the lower and upper values for the confidence interval:

Answer 1

Solution :

Given that,

n = 150

x = 60

Point estimate = sample proportion = = x / n = 60/150=0.4

1 -   = 1- 0.4 =0.6

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 *((0.4*0.6) /150 )

E = 0.066

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.4-0.066 < p < 0.4+0.066

0.334< p < 0.466

lower 0.334

and upper 0.466