Problem:

Assume a sample of 25 workstations was selected from a production
batch and that they achieve an average time of 28 seconds on a
benchmark with a standard deviation of 15 seconds.

Find a 95% confidence interval for the average benchmark time for 
the entire production batch.

Solution:

Since the sample size is small the t table is required. Also SD(ybar)
is 15/sqrt(25)=15/5=3. Since the required t value is 2.064 then the
95% CI is 28 plus/minus 2.064*3 or 28 plus/minus 6.182.


Problem: 

Assume a sample of 9 COBOL programs. Each program is rigorously tested and
the average fault density found to be 0.5 faults per line of code with a 
standard deviation of 0.15.

a) 	Find a 95% confidence interval for the average fault density for 
	the entire system.
b)	What if the sample size was 100. What is the 95% confidence interval
	in this case.
c)	Estimate the number of faults that may be expected for the entire
	system if the system has 10,000 lines of code.

Solution:

a)	SD(ybar)=0.15/sqrt(9)=0.15/3=0.05. The required t value is 2.306.
	The 95% CI is 0.5 plus/minus 2.306*0.05 or 0.5 plus/minus 0.1153.
b)	In this case the sample is large and the 95% CI is 0.5 plus/minus
	1.95*SD(ybar). Since SD(ybar)=0.15/sqrt(100)=0.015 then the 95%
	CI is 0.5 plus/minus 0.03.
c)	5,000 faults should be expected.