Elementary Statistics and Computer Application

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 3
The 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

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.=2.131
t2=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