Math Probability and Statistics II
Chapter 9.2 Topic: The Two-Sample t Test and Confidence Interval
The test statistic used for testing difference of means is given by
_ _
t = (μ1-μ2) –(x - y) /√ S^2(1/n1+1/n2) where S^2 = n1s1^2 + n2s2^2 / n1+n2-2
~ t with n1+n2-2 degrees of freedom.
Here x(bar) is the sample mean of advanced player and y(bar) is sample mean of
intermediate
H0: μ1 = μ2 ie There is no significant difference between average force for advanced
player and that for intermediate players
Its two tailed test .
Critical Value is 2.179
95% Confidence Interval:
S^2 = 6*11.3*11.3 + 8*8.3*8.3 / 12 = 109.772.
Since its 95% confidence interval
P(|t| < 2.179) = 0.95
_ _
| (μ1-μ2) –( x – y ) /√ S^2(1/n1+1/n2) | < 2.179
| (μ1-μ2) - (40.3 - 21.4) /√ 109.772 (1/6+1/8) | < 2.179
| (μ1-μ2)-18.9 | < 12.328
-12.328 < (μ1-μ2) - 18.9 < 12.328
-12.328 + 18.9 < (μ1-μ2) < 12.328 +18.9
6.571 < (μ1-μ2) < 31.228
Since the difference in means (μ1-μ2) is always positive there is enough evidence that
the averages are different.
Yes. I would have reached the same conclusion even if the labels are interchanged. If the
labels are interchanged the test statistic would have been
_ _
t = (μ2-μ1) –(y - x) /√ S^2(1/n1+1/n2) where S^2 = n1s1^2 + n2s2^2 / n1+n2-2
~ t with n1+n2-2 degrees of freedom.
Then the confidence interval would have been
| (μ2-μ1) - (-18.9) | < 12.328
|(μ2-μ1) + 18.9) | < 12.328
-12.328 < ((μ2-μ1) + 18.9) <12.328
-31.228< (μ2-μ1) < -6.5712
In the above interval too there is never a possibility that difference is zero and the average
means are equal.
Therefore there is enough evidence that the means are different.
NOTE : Had the interval been -ve value < (μ2-μ1) < +ve value , then there is
possibility that difference can be zero.