Examples
Example 2
- In a fertilizer trial the grain yield of paddy ( Kg/plot) was observed as follows
- Under ammonium chloride 42,39,38,60 &41 kgs
- Under urea 38,42,56,64,68,69,&62 kgs.
- Find whether there is any difference between the sources of nitrogen?
Solution
- Ho: µ1=µ2 (i.e) there is no significant difference in effect between the sources of nitrogen.
- H1: µ1≠µ2 (i.e) there is a significant difference between the two sources
- Level of significance = 5%
- Before we go to test the means first we have to test their variances by using F-test.
- F-test
- Ho:., σ1212=σ2222
- H1:., σ1212≠σ2222
Ftab(6,4) d.f. = 6.16
We accept the null hypothesis H0. (i.e) the variances are equal. Use the test statistic
where
- The degrees of freedom is 5+7-2= 10. For 5 % level of significance, table value of t is 2.228
- Inference:
- t <ttab
We accept the null hypothesis H0
- We conclude that the two sources of nitrogen do not differ significantly with regard to the grain yield of paddy.
Example 3The summary of the results of an yield trial on onion with two methods of propagation is given below. Determine whether the methods differ with regard to onion yield. The onion yield is given in Kg/plot.
Method I
|
Method II
|
n1=12
|
n2=12
|
|
|
SS1=186.25
|
SS2=737.6667
|
|
|
Solution
- Ho:., µ1=µ2 (i.e) the two propagation methods do not differ with regard to onion yield.
- H1 µ1≠µ2 (i.e) the two propagation methods differ with regard to onion yield.
- Level of significance = 5%
- Before we go to test the means first we have to test their variability using F-test.
- F-test
- Ho:., σ1212=σ2222
- H1:., σ1212≠σ2222
Ftab(11,11) d.f. = 2.82
- We reject the null hypothesis H0.we conclude that the variances are unequal.
- Here the variances are unequal with equal sample size then the test statistic is
where t =1.353 The table value for =11 d.f. at 5% level of significance is 2.201 Inference:
- t<ttab
- We accept the null hypothesis H0
- We conclude that the two propagation methods do not differ with regard to onion yield.
Example 4: The following data relate the rubber yield of two types of rubber plants, where the sample have been drawn independently. Test whether the two types of rubber plants differ in their yield.
Type I
|
6.21
|
5.70
|
6.04
|
4.47
|
5.22
|
4.45
|
4.84
|
5.84
|
5.88
|
5.82
|
6.09
|
5.59
|
6.06
|
5.59
|
6.74
|
5.55
|
|
|
|
|
|
|
|
|
Type II
|
4.28
|
7.71
|
6.48
|
7.71
|
7.37
|
7.20
|
7.06
|
6.40
|
8.93
|
5.91
|
5.51
|
6.36
|
Solution
- Ho, µ1=µ2 (i.e) there is no significant difference between the two rubber plants.
- H1 µ1≠µ2 (i.e) there is a significant difference between the two rubber plants.
- Level of significance = 5%
Here
n1=16 n2=12 Before we go to test the means first we have to test their variability using F-test.
F-test
Ho:., σ12=σ22
H1:., σ12≠σ22
- Ftab(11,15) d.f.=2.51
- We reject the null hypothesis H0. Hence, the variances are unequal.
- Here the variances are unequal with unequal sample size then the test statistic is
t1=t(16-1) d.f.=2.131t2=t(12-1) d.f .=2.201 Inference t>tw
- We reject the null hypothesis H0. We conclude that the second type of rubber plant yields more rubber than that of first type
|
Last modified: Tuesday, 31 January 2012, 10:25 PM