Question

From this Sample
No A.   B.  
   1. 7.   4.  
   2. 8.   5.  
   3. 9.   3.  
   4. 5.   6.  
   5. 7.   2.  
   6. 6.   4.  
Test of hypothesis for difference of two means by using
1. t Test
2. ANOVA
Take 5% type 1 error, then compare the results.
Show that t² is equal to f.

Answer 1

(1)

H0:

HA:

From the given data,the following statistics are calculated:

1 = 7

2 = 4

s1 = 1.4142

s2 = 1.4142

n1 = 6

n2 = 6

= 0.05

Pooled standard deviation is given by:

Test Statistic is given by:

df = 6 + 6 - 2 = 10

From Table, critical values of t = 2.228

Since calculated value of t = 3.674 is greater than critical value of t = 2.228, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that there is significant difference betwen the two means.

(2)

H0:     :

HA:    

From the given data,the following Table is calculated:

	Sample A	Sample B	Total
N	6	6	12
	42	24	66
Mean	42/6 = 7	24/6 = 4	66/12 = 5.5
	304	106	410
Std. Dev.	1.4142	1.4142	2.0671

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F stat
Between treatments	27	1	27/1=27	27/2 = 13.5
Withintreatments	20	10	20/10 = 2
Total	47	11

F Stat = 27/2 = 13.5

Degrees of Freedom = (1,10)

By Technology, p - value = 0.0043

Since p - value = 0.0043 is less than = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that there is significant difference betwen the two means

We note:

t = 3.674

F = 13.5

Thus, we get:

t2 = 3.6742 = 13.5 = F

Thus, we show t2 is equal to F.