A sociologist is interested in surveying workers in computer

A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such workers who have changed jobs within the past year.

a) In the absence of preliminary data, how large a sample must be taken to ensure that a 95% confidence interval will specify the proportion to within 0.05?

b) In a sample of 100 workers, 20 of them had changed jobs within the past year. Find a 95% confidence interval for the proportion of workers who have changed jobs within the past year.

c) Based on the data in part (b), estimate the sample size needed so that the 95% confidence interval will specify the proportion to within .05.  

Solution

a)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.5
ME = 0.05
n = ( 1.96 / 0.05 )^2 * 0.5*0.5
= 384.16 ~ 385

b)
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=20
Sample Size(n)=100
Sample proportion = x/n =0.2
Confidence Interval = [ 0.2 ±Z a/2 ( Sqrt ( 0.2*0.8) /100)]
= [ 0.2 - 1.96* Sqrt(0.002) , 0.2 + 1.96* Sqrt(0.002) ]
= [ 0.122,0.278]

c)
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Samle Proportion = 0.2
ME = 0.05
n = ( 1.96 / 0.05 )^2 * 0.2*0.8
= 245.862 ~ 246