Research Methodology - Dr. Krishan K. Pandey
Research Methodology - Dr. Krishan K. Pandey Research Methodology - Dr. Krishan K. Pandey
Testing of Hypotheses I 225 the values happen to be positive; one must simply know the degrees of freedom for using such a distribution. * TESTING THE EQUALITY OF VARIANCES OF TWO NORMAL POPULATIONS When we want to test the equality of variances of two normal populations, we make use of F-test 2 2 2 based on F-distribution. In such a situation, the null hypothesis happens to be H0: σp = σ 1 p , σ 2 p1 2 and σ p2 representing the variances of two normal populations. This hypothesis is tested on the basis 2 2 2 of sample data and the test statistic F is found, using σ and σ the sample estimates for σ and s1 s2 p1 2 σ p2 respectively, as stated below: d i F = σ σ X i X X i X 2 1 1 where σs = σ 1 s2 n 1 n 1 ∑ − = − ∑ − and − b 1 g 2 2 s1 2 s 2 2 2 b 2 g 2 d i 2 2 While calculating F, σs1 is treated > σ which means that the numerator is always the greater s2 variance. Tables for F-distribution ** have been prepared by statisticians for different values of F at different levels of significance for different degrees of freedom for the greater and the smaller variances. By comparing the observed value of F with the corresponding table value, we can infer whether the difference between the variances of samples could have arisen due to sampling fluctuations. If the calculated value of F is greater than table value of F at a certain level of significance for (n – 1) and (n – 2) degrees of freedom, we regard the F-ratio as significant. Degrees of 1 2 freedom for greater variance is represented as v and for smaller variance as v . On the other hand, 1 2 if the calculated value of F is smaller than its table value, we conclude that F-ratio is not significant. If F-ratio is considered non-significant, we accept the null hypothesis, but if F-ratio is considered significant, we then reject H (i.e., we accept H ). 0 a When we use the F-test, we presume that (i) the populations are normal; (ii) samples have been drawn randomly; (iii) observations are independent; and (iv) there is no measurement error. The object of F-test is to test the hypothesis whether the two samples are from the same normal population with equal variance or from two normal populations with equal variances. F-test was initially used to verify the hypothesis of equality between two variances, but is now mostly used in the * See Chapter 10 entitled Chi-square test for details. ** F-distribution tables [Table 4(a) and Table 4(b)] have been given in appendix at the end of the book. 2
226 Research Methodology context of analysis of variance. The following examples illustrate the use of F-test for testing the equality of variances of two normal populations. Illustration 19 Two random samples drawn from two normal populations are: Sample 1 20 16 26 27 23 22 18 24 25 19 Sample 2 27 33 42 35 32 34 38 28 41 43 30 37 Test using variance ratio at 5 per cent and 1 per cent level of significance whether the two populations have the same variances. Solution: We take the null hypothesis that the two populations from where the samples have been drawn have the same variances i.e., 2 2 2 H0: σ p = σ . From the sample data we work out σ 1 p s1 and 2 2 σ s2 as under: X 1i Table 9.10 Sample 1 Sample 2 dX1i − X1i X1i − X1 2 dX2i − X2i X i − X d i X 2i 2 d 2 2i 20 –2 4 27 –8 64 16 –6 36 33 –2 4 26 4 16 42 7 49 27 5 25 35 0 0 23 1 1 32 –3 9 22 0 0 34 –1 1 18 –4 16 38 3 9 24 2 4 28 –7 49 25 3 9 41 6 36 19 –3 9 43 8 64 30 –5 25 37 2 4 ∑ X1i = 220 ∑ X − X = 2 d 1i 1i 120 ∑ X i = n 1 = 10 n 2 = 12 X 1 X = n ∑ ∴ σ s 1 220 X = = 22 X 2 = 10 n ∑ ; 1i 2i d i . 2 X i X 2 1 1 = 1 n1 1 ∑ − − 2 d 2i 2i 2 420 ∑ X − X = 314 120 = 1333 10 − 1 = 2 420 = = 35 12
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226 <strong>Research</strong> <strong>Methodology</strong><br />
context of analysis of variance. The following examples illustrate the use of F-test for testing the<br />
equality of variances of two normal populations.<br />
Illustration 19<br />
Two random samples drawn from two normal populations are:<br />
Sample 1 20 16 26 27 23 22 18 24 25 19<br />
Sample 2 27 33 42 35 32 34 38 28 41 43 30 37<br />
Test using variance ratio at 5 per cent and 1 per cent level of significance whether the two<br />
populations have the same variances.<br />
Solution: We take the null hypothesis that the two populations from where the samples have been<br />
drawn have the same variances i.e., 2 2<br />
2<br />
H0: σ p = σ . From the sample data we work out σ<br />
1 p<br />
s1 and<br />
2<br />
2<br />
σ s2 as under:<br />
X 1i<br />
Table 9.10<br />
Sample 1 Sample 2<br />
dX1i − X1i<br />
X1i − X1<br />
2<br />
dX2i − X2i<br />
X i − X<br />
d i X 2i<br />
2<br />
d 2 2i<br />
20 –2 4 27 –8 64<br />
16 –6 36 33 –2 4<br />
26 4 16 42 7 49<br />
27 5 25 35 0 0<br />
23 1 1 32 –3 9<br />
22 0 0 34 –1 1<br />
18 –4 16 38 3 9<br />
24 2 4 28 –7 49<br />
25 3 9 41 6 36<br />
19 –3 9 43 8 64<br />
30 –5 25<br />
37 2 4<br />
∑ X1i = 220 ∑ X − X =<br />
2<br />
d 1i 1i<br />
120 ∑ X i =<br />
n 1 = 10 n 2 = 12<br />
X<br />
1<br />
X<br />
=<br />
n<br />
∑<br />
∴ σ s<br />
1<br />
220<br />
X<br />
= = 22 X 2 =<br />
10<br />
n<br />
∑<br />
;<br />
1i<br />
2i<br />
d i .<br />
2<br />
X i X<br />
2 1 1<br />
=<br />
1 n1<br />
1<br />
∑ −<br />
−<br />
2<br />
d 2i 2i<br />
2 420 ∑ X − X = 314<br />
120<br />
= 1333<br />
10 − 1<br />
=<br />
2<br />
420<br />
= = 35<br />
12