A)
Given the data :
x | y |
2.5 | 1.21 |
1.7 | 0.86 |
2.5 | 1.78 |
1.9 | 1.23 |
2 | 0.5 |
3.2 | 1.76 |
The following results can be obtained (n: no of obs = 6) :
mean 1 | 2.3 |
mean 2 | 1.223333333 |
sd 1 | 0.547722558 |
sd 2 | 0.500666223 |
variance 1 | 0.3 |
variance 2 | 0.250666667 |
(n-1) * variance 1 | 1.5 |
(n-1) * variance 2 | 1.253333333 |
pooled variance | 0.275333333 |
pooled sd | 0.524722149 |
Thus t statistic is - (mean 1 - mean2 ) / [ pooled sd * sqrt(1/n +1/n)]
= (2.3 - 1.23 ) / [0.5247* sqrt(1/3)]
=3.55
Under null hypothesis the test follows t distribution with 6+6-2=10 degrees of freedom.
Critical point for2 tailed 95% for t distribution with 10 df is : 2.23
Thus observed t statistic is 3.55 > 2.23
thus based on the data it can be concluded that the null hypothesis is to be rejected, ie. the soy group and traditional group seem to have different mean effects.
The 95% confidence interval is given by - :
= sample mean difference +- upper 0.025 point for t dist with 10 df * pooled sd
=1.08 +- (2.22* 0.52)
=1.08 +- (1.15)
=(-0.07,2.23)
B)
The effect size is given by - :
Mean difference / pooled sd
= 1.08/0.52 = 2.05 which is roughly 2.
This implies that there is a large size effect roughly 2 times meaning the change in soy from traditional has had a huge effect.
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