Question

A firm would like to know the proper portion of its employees that expect to receive a raise within the next year the firm surveys 150 employees and find 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 thousands place

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

Answer 1

Solution =   

Given,

n = 150 ....... Sample size

x = 60 .......no. of successes in the sample

a) Let denotes the sample proportion.

     = x/n   = 60 / 150 = 0.40

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

b) = 1.645 (use z table)

Now , the margin of error is given by

c) E = *  

= 1.645 * [ 0.40 *(1 - 0.40 ) / 150 ]

= 0.0658  

Now the confidence interval is given by

( - E)   ( + E)

( 0.4 - 0.0658 )   ( 0.4 + 0.0658 )

0.334   0.466  

d) Required 90% Confidence Interval is ( 0.334 , 0.466 )

Lower value 90% Confidence Interval is 0.334

Upper value 90% Confidence Interval is 0.466