IT233: Applied Statistics
TIHE 2005
Confidence Intervals and Tests of Hypotheses
 Confidence intervals and tests for 
 Confidence intervals and tests for   
1
2
 Confidence intervals and tests for  2
 Confidence intervals and tests for  2 /  2
1
2
Confidence Intervals and Tests Concerning σ 2
Confidence Intervals for σ2
Theorem: If S 2 is the variance of a random sample of size n from a normal
population, a 1 100% confidence interval for  2 is
 n 1 S 2   2   n 1 S 2
2
where 2 and  2 
1
2
leaving areas 
2 
1
2
2
are  2 - values with   n 1 degrees of freedom,
2

2 and 1  2 , respectively, to the right.
Proof:
The confidence interval for  2 can be derived by using the statistic
n  1 S 2

 
2
2

1
Consider the following figure:
 2
 2
1

2
1
0

2
2
2
2
To derive the confidence interval, we begin by giving a probability statement
based on the above figure that
P  12    2  2   1  

2
2
Substituting for  2 , we write


n 1 S 2

2
2 
P  1   


   1 
2

2
2



To isolate  2 in the centre of the inequality, we divide each term by
(n 1)S 2 and then inverting each term, we obtain

2
n 1 S 2


2  n  1 S 
P
  2
2
  1 





1
2
2 


The desired 1   100% confidence interval for  2 can be read as:
 n 1 S 2   2   n 1 S 2
2
2

2
1  
2
2
Example 1: A manufacturer of batteries claims that his batteries will last,
on average, 3 years with a variance of 1 year. If 5 of these batteries have
lifetimes 1.9, 2.4, 3.0, 3.5, 4.2 years construct a 95% C.I. for  2 and decide
if the claim that  2  1 is valid.
Solution:
First we find the summary data:

NOTE:
 xi2 = 48.26
 xi = 15,
n = 5,
 
n xi2   xi
2
S 
n  n 1
2

5  48.26  152
= 0.815
5 4
You could use your Calculator to find:

S 2   n 1
We have,

2
  0.902773504 2  0.815
  0.05
  n 1  4

2
2  0.025
 11.143
2
2
 
1
2
2
 0.975
 0.484
 the 95% Confidence Interval for  2 is
 n 1 S 2   2   n 1 S 2
2
2
2 
1
2

4  0.815
4  0.815
 2 
11.143
0.484

0.293   2  6.736
3
To answer the last part of the problem, that is, “Is the claim that  2  1
valid?”, we use the confidence interval method.
Since above interval reveals that the value  2  1 lies in the
internal, we can accept the claim with a 95% confidence.
Answer:
Notes: (1) Do not use confidence interval method for testing any claim, if the test
is one-tailed.
(2) You could also use the following traditional method.
Test of Hypothesis for  2
The traditional method of hypothesis testing for the above example is as
follows:
Null hypothesis:
H :  2 1
0
Alternative hypothesis: H :  2  1 (Two-tailed test)
1
Critical Region:
  0.05
  n 1  4

 2 =0.025
2
2  0.025
 11.143
 2 = 0.025
2
2
 2   0.975
 0.484
1
0
Acceptance
Region
0.484
11.1434
2
 the null hypothesis is rejected when  2  0.484 and  2  11.143.
4
Test Statistic
2 

 n 1 S 2
2
4  0.815
1
 3.26
Conclusion: Since the value of test statistic  2  3.26 falls in the acceptance
region, we accept H0 :  2  1 .
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