TESTING OF SINGLE VARIANCE

      

               

TESTING OF SINGLE VARIANCE
            If we want to test variance of the population with that of sample variance then this is useful. If we denote that  is variance of population and S2 is for the sample then the null and alternative hypotheses are written as below.
            H0: S2 =  
                Ha: S2 ≠  
For one tailed test the alternative hypothesis will be
            Ha: S² >     or
Ha: S² <     
            The logic behind this test is that here we want to control not only mean but variance also. The testing of population mean and sample mean is carried out using t-test. But here we want to test variance of population whose size is unknown with the variance of sample (here ‘n’ is known).
Example:
1.      The manufacturer of golf ball wants to control the variance of weight of balls.
2.      The variation in the thickness of silicon disc is controlled by the firm.
3.      The gap between wheel teeth gear should be within specification limit. Higher and lower gap can cause problem in performance of gear.
Formula:
            This formula is solved by finding out X2- statistics from the variances. Here (n-1) is degree of freedom for the sample.

The testing Criteria:
1.      If alternative hypothesis is S2 ≠   ten
X²Cal> X²α/2 then reject null hypothesis. However X²Cal< X²(1- α/2) then also reject H0.
2.      For S² <  type alternative hypothesis it is necessary to find out X Critical value using X²(1-α). If X²Cal< X²(1- α) reject null hypothesis.
3.      For S² >  If type alternative hypothesis find out X²α value which is critical value.
If X²Cal> X²α th3en rejects null hypothesis.
Graph ↓:

In some cases it is expected that the variance should not be more than a particular value. In that case we have to formulate the null hypothesis as H0 : S2 £  against Ha: S² >. In the case of quality management, the precision as measured by variance of an instrument is not more than 0.16. Here if we have taken 13 measurements of the instrument on the same subject as 2.5, 2.3, 2.4, 2.3, 2.5, 2.7, 2.5, 2.7, 2.5, 2.6, 2.6, 2.7, 2.5. Let us carry out the hypothesis testing at 1 % level of significance to test whether the variance observed is not more than 0.16. Here the expectation 0.16 is the variance of population that should be mentioned in the null hypothesis at .
H0 : S2 £ 0.16 against Ha: S² > 0.16
Let us find out variance of the given sample by using formula:
This formula is used here because of given numbers are small and there is no possibility that the mean of the given numbers is integer.
X
X2
2.5
6.25
2.3
5.29
2.4
5.76
2.3
5.29
2.5
6.25
2.7
7.29
2.5
6.25
2.7
7.29
2.5
6.25
2.6
6.76
2.6
6.76
2.7
7.29
2.5
6.25
32.8
82.98

χ2cal= =
The critical value of the χ2 at 1% of level of significance and 13-1=12 degree of freedom is 21.0261. As the calculated value is lesser than critical value we accept null hypothesis. Thus the variance of the instrument is not more than 0.16.
Question 1: A manufacturer of golf balls wants to control the weight of the ball in such a way that the variance of weight should not be more than 1.8 mg². For this purpose he has taken 15 samples of the balls which resulted into variance of 2.8 mg². Is this sufficient to reject the claim of manufacturer at 5% level of significance?
The null and alternative hypotheses are given as H0 : S2 £ 1.8 against Ha: S² > 1.8. n= 15. Here S²= 2.8mg² and  = 1.8mg². Putting into following formula:
=
χ2cal=21.78
Critical value of the   at 15-1=14 degree of freedom and level of significance 5% is 23.6848. As calculated value is lesser than critical value we accept null hypothesis and hereby assume that the variance of the sample i.e. 2.8 is approximately equal to 1.8mg². In other words the difference between sample variance and expected variance is insignificant at 5 % level of significance.

Question 2: The diameter of the shaft should have precision at each cross section measured longitudinally. The variance allowed is 0.012mm². The sample of 20 reveals that standard deviation of the diameter is 0.12mm. Give your opinion whether the variation is within limit?
Given:  = 0.012mm², S=0.12
S²= 0.12²=0.0144mm², n=20
H0 : S2 = = 0.012    ,  Ha: S² <
=
χ2cal=22.80
χ2table at 5% level of significance means we expect this at 0.995 probability to be within limit. Thus for 0.995 area and degree of freedom = (20-1) =19 the value of   
χ2critical = 10.117.
            As the calculated value of χ2 is greater than critical value (10.117) we reject the null hypothesis. Here we expect that the variance of the shaft will not in the limit.

Question 3: The client purchaser suspects that the gap between fan hole and shaft is more than specific limit as measured at 8 directions. The variance allowed is 0.036mm². Following measurements are taken by the representative.
22.6, 23.1, 22.6, 22.8, 22.7, 22.8, 22.6, 22.9, 23.2, 22.5, 22.5, 22.6, 22.4, 22.7, 22.5, 22.6, 22.4, 22.9, 22.7, 22.6.  Test the Hypothesis that the variance is more than limit. Use 5 % level of significance.
H0 : S2 £ =0.036  against Ha: S² >
Let us find out S2  using formula of deviation from assumed mean. Assume a = 20.
X
d=(X-a)=(X-20)
d^2
22.6
0.6
0.36
23.1
1.1
1.21
22.6
0.6
0.36
22.8
0.8
0.64
22.7
0.7
0.49
22.8
0.8
0.64
22.6
0.6
0.36
22.9
0.9
0.81
23.2
1.2
1.44
22.5
0.5
0.25
22.5
0.5
0.25
22.6
0.6
0.36
22.4
0.4
0.16
22.7
0.7
0.49
22.5
0.5
0.25
22.6
0.6
0.36
22.4
0.4
0.16
22.9
0.9
0.81
22.7
0.7
0.49
22.6
0.6
0.36
Total
13.7
10.25
d bar =     =    = 0.685
       = 0.0455.  Statement of hypothesis:
=
χ2critical at α =0.05 and degree of freedom i.e. 20-1=19. As the values in the table of χ2 gives right side probabilities hence we should see this α =0.05. χ2critical = 30.144.
As the calculated value is lesser than critical value, we accept null hypothesis. The variation 0.0455 can be assumed to be appropriate and equal to 0.036.
Question 1:
A firm manufactures shaft expects that the length should be 150 mm. It is observed that actual length varies more and less than the specifications. A manager took random samples of 17 shafts and found that the variance is 15.00 Sq. mm. He now wants to establish variance and standard deviation limits of the population with 95% confidence. Also wants to know the maximum error in the shaft length that the manufacturing process can produce. Help him in finding out.
 Question 2:
A large candy manufacturer produces packs of candy targeted to weight 52 grams. A quality control manager was concerned that the variation in the actual weights was larger than acceptable. That is, he was concerned that some packs weighed significantly less than 52 grams and some weighed significantly more than 52 grams. In order to estimate σ the standard deviation of the weights of the nominal 52 grams packs, he took a random sample of n = 10 packs off of the factory line. The random sample yielded a sample variance of 4.2 grams.
i.                    Use the random sample to derive a 95 % confidence interval for σ. (Note: the interval is for σ not σ²).
ii.                  What assumptions did you make about the sample in order to make your estimate?





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