This document discusses hypothesis testing for single variance. It provides three examples of testing hypotheses about population variance using a chi-square distribution. The first two examples test if a sample variance is equal to or less than a hypothesized value. The third example tests if a sample variance is equal to a hypothesized value. Each example states the hypotheses, computes the test statistic, determines the critical region, and makes a conclusion about accepting or rejecting the null hypothesis.

1. HYPOTHESIS TESTING
PART-VI
SINGLE VARIANCE
NADEEM UDDIN
ASSOCIATE PROFESSOR
OF STATISTICS

2. Test Concerning Variance
Example-1
A random sample of size 20 from a normal population
has a standard deviation S=4.51. Test the hypothesis
that 𝜎2 = 36 against the alternative that 𝜎2 < 36. Use a
0.05 level of significance.

3. Solution:
1.Hypothesis
H0: 𝜎2 = 36
H1: 𝜎2 < 36
2.Level of significance α = 0.05
3.Test statistic
4.Critical Region
In case of lower tail test i.e. H1 𝑖𝑠 ˂.
Reject H0, if 𝜒2
𝑐𝑎𝑙 ≤ 𝜒2
𝑡𝑎𝑏
Where
𝜒2
𝑡𝑎𝑏 = 𝜒2
1−𝛼,(𝑛−1)
= 𝜒2
1−0.05,(20−1) = 𝜒2
0.95,(19) = 10.117
𝜒2
𝑐𝑎𝑙 ≤ 10.117 (Using chi square table)
10.117
  212
2
n S





4. 5.Computation
  
2
2 20 1 4.51
36
 

2 10.735cal 
6.Conclusion: Accept H0.
  212
2
n S





5. Example-2
A machine engine part produced by a company is claimed to
have diameter variance no longer than 0.0002 inches. A random
sample of 10 parts gave a sample variance of 0.0003. Test at 5%
level.
Solution:
1.Hypothesis
H0: 𝜎2
= 0.0002
H1: 𝜎2 > 0.0002
2.Level of significance α = 0.05
3.Test statistic
  212
2
n S





6. 4.Critical Region
In case of upper tail test i.e. H1 𝑖𝑠 ˃.
Reject H0, if 𝜒2
𝑐𝑎𝑙 ≥ 𝜒2
𝑡𝑎𝑏
Where
𝜒2
𝑡𝑎𝑏 = 𝜒2
𝛼,(𝑛−1)
= 𝜒2
0.05,(10−1) = 𝜒2
0.05,(9) = 16.919
𝜒2
𝑐𝑎𝑙 ≥ 16.919 (Using chi square table)
5.Computation
  2 10 1 0.0003
0.0002
 

2 13.5cal 
6.Conclusion: Accept H0.
  212
2
n S





7. Example-3
Ten high school seniors taking the basic English test received the
following scores:28,26,30,24,25,29,31,26,23 and 27. Past scores
at their high school have shown the score to be normally
distributed with 𝜎2 = 16 against 𝜎2 ≠ 16 at α = 0.05
Solution: X X2
28 784
26 776
30 900
24 576
25 625
29 841
31 961
26 676
23 529
27 729
∑x=269 ∑x2=7297

8. 1.Hypothesis
H0: 𝜎2
= 16
H1: 𝜎2 ≠ 16
2.Level of significance α = 0.05
3.Test statistic   212
2
n S




4.Critical Region
In case of two tail test i.e. H1 𝑖𝑠 ≠.
Reject H0, if 𝜒2
𝑐𝑎𝑙 ≤ 𝜒2
𝑡𝑎𝑏 or 𝜒2
𝑐𝑎𝑙 ≥ 𝜒2
𝑡𝑎𝑏.
Where 𝜒2
𝑡𝑎𝑏 = 𝜒2 𝛼
2
,(𝑛−1) = 𝜒2
0.05
2
,(10−1)
= 𝜒2
0.025, 9 = 19.023
𝜒2
𝑡𝑎𝑏 = 𝜒2
1−
𝛼
2,(𝑛−1)
= 𝜒2
1−
0.05
2 ,(10−1)
= 𝜒2
0.975, 9 = 2.700
𝜒2
𝑐𝑎𝑙 ≤ 2.700 or 𝜒2
𝑐𝑎𝑙 ≥ 19.023. (Using chi square table)

9. 5.Computation
 
 
   
 
2 22 10 7297 2692 6.77
1 10 10 1
n x x
S
n n
  
  
 
 10 12
16
6.77



2
3.808cal 
6.Conclusion: Accept H0.
  212
2
n S



