Page 445 - DMTH404_STATISTICS

Unit 32: Hypothesis Concerning Standard Deviation

Solution. Notes

We have to test H : s = s against s > s .
0 1 2 1 2
2
2
It is given that S = 15.2, S = 12.8, n = 12 and n = 10.
1 2 1 2
The unbiased estimates of respective population variances are
12 2 10
2
s = ´ 15.2 16.58 and s = ´ 12.8 14.22.
=
=
1
2
11 9
16.58
Thus, F = = 1.166.
cal
14.22
The value of F from tables at 5% level of significance with 11 and 9 d.f. is 3.10. Since this value is
greater than F , there is no evidence against H .
cal 0
Example 6: The increase in weight (in 100 gms) due to food A and food B given to two
independent samples of children was recorded as follows. Test whether (i) mean weights and (ii)
standard deviations of the two samples are equal.
Sample I : 6, 12, 10, 14, 12, 12, 10, 7, 5, 7.

Sample II : 9, 11, 8, 5, 6, 12, 7, 13, 10.
Solution.
We shall first test H : s = s against s > s .
0 1 2 1 2
95 81
The means of the samples are X = = 9.5 and X = = 9.0 , respectively.
2
1
10 9
2
æ
n k å X ki 2 ö å X ki 2 n k 2
2
s = ç - X k ÷ = - X
We can write k k (k = 1, 2)
n - 1 n n - 1 n - 1
k è k ø k k
987 10 2 2 789 9 2
2
Thus, we have s = - ´ 9.5 = 9.39 and s = - ´ 9 = 7.50.
1 2
9 9 8 8
9.39
Further, the test statistic is F = = 1.25.
7.50
The critical value of F at 5% level of significance and (9,8) d.f. is 3.39, therefore, there is no
evidence against H . Hence, s and s may be treated as equal.
0 1 2
To test H :  =  against Ha ;    , we note that samples are small, t-test is to be used. Since
0 1 2 1 2
s = s = s (say), its unbiased estimate is
1 2
2
n - 1 s + n - 1 s 2 9 9.39 8 7.50
+
´
´
( 1 ) 1 ( 2 ) 2
s = = = 2.92.
-
+
n + n - 2 10 9 2
1
2
X - X 2 n n 9.5 9.0 10 9
´
-
1
The test statistic is t = 1 2 = = 0.37.
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s n + n 2 2.92 10 9
+
1
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