test statistics/critical regions

madgab · #1 $Old$ June 24th, 2009, 07:51 PM

Please help me! I am stuck on this chapter and this question seemed to have it all, so its probably the best one to ask. Thank you!

Let X and Y denote the weights in grams of male and female gallinules. Assume that X is $N(\mu_{x},\sigma^{2}_{x})$ and $Y is N(\mu_{y},\sigma^{2}_{y})$

a) Given n=16 observations of X and m=13 observations of Y, define a test statistic and critical region for testing the null hypothesis $H_{0}: \mu_{x}=\mu_{y}$ against the one-sided alternative hypothesis $H_{1}: \mu_{x} > \mu_{y}$ . Let $\alpha=0.01$ and assume variances are equal.

b) Given that $\bar{x}=415.16, s^{2}_{x}=1356.75, \bar{y}=347.40 and s^{2}_{y}=692.21$ , calculate the value of the test statistic and make conclusion

c) Test whether the assumption of equal variances is valid. Let $\alpha=0.05$ .

d) Despite the fact that $\sigma^{2}_{x}=\sigma^{2}_{y}$ is accepted in part (c), let us say we suspect the equality is not valid. Thus use the test proposed by Welch.

Thank you for even looking! I am ready to tear my hair out trying to understand this entire chapter. I hope someone out there can help me!

Random Variable · #2 $Old$ June 24th, 2009, 09:06 PM

a) The test statistic is $T = \frac {\bar{X} - \bar{Y}}{S_{p} \sqrt{1/n + 1/m}}$

where $S_{p} = \sqrt{\frac{(n-1)S^{2}_{X} + (m-1)S^{2}_{Y}}{n+m-2}}$

$T$ follows a t distribution with n+m-2 degrees of freedom

For this problem the critical region is $t \ge t_{(0.01, 27)}$

c) The alternative hypothesis is $H_{1}: \sigma^{2}_{X} \neq \sigma^{2}_{Y}$

The test statistic is $F = \frac {S^{2}_{X}}{S^{2}_{Y}}$

F follows an F distribution with $r_{1} = n-1$ and $r_{2}=m-1$ degrees of freedom respectively

The critical region is $f \ge F_{(0.025, 15,12)}$ or $\frac {1}{f} \ge F_{(0.025,12,15)}$

I have to review how to do part (d).

matheagle · #3 $Old$ June 25th, 2009, 12:03 AM

a-c look fine, d is just Satherwaite's approximation.

You use as your test stat

$T = \frac {\bar{X} - \bar{Y}}{\sqrt{S_X^2/n + S_Y^2/m}}$

The NASTY degrees of freedom (which is not an integer) can be found under...
Unequal sample sizes, unequal variance at http://en.wikipedia.org/wiki/Student's_t-test

madgab · #4 $Old$ June 27th, 2009, 05:25 PM

I calculated the test statistic using the formula you provided and got 5.767 and using the degrees of freedom formula, I got r=26.628. I did some searching through the link you sent and it seems I have to use these two values for yet another test, but I can't seem to find an equation. Do you have any idea what equation that might be? Thank you very much.

madgab · #5 $Old$ June 27th, 2009, 05:30 PM

I have a couple more questions. Can the critical region in part a be found on the t test table? If so it looks like 2.473 if I'm not mistaken.

Also, how would I look up the f critical regions?

Thank you very VERY much

Random Variable · #6 $Old$ June 27th, 2009, 06:17 PM

Quote:

If so it looks like 2.473 if I'm not mistaken.

Yes, that is correct.

Quote:

Also, how would I look up the f critical regions?

F-Distribution Tables

$F_{(0.025, 15,12)} = 3.1772$
$F_{(0.025,12,15)} = 2.9633$

madgab · #7 $Old$ June 27th, 2009, 08:27 PM

Thank you. I'm sorry to bother you again. I calculated the value of the test statistic as 5.5071 but I don't know what conclusion this gives. I got the critical region of 2.473. so what does 2.473<5.5071 mean? or do I have even more calculations for this problem. i keep confusing myself with all the different values.

Random Variable · #8 $Old$ June 27th, 2009, 08:51 PM

Quote:

Originally Posted by madgab $View Post$

Thank you. I'm sorry to bother you again. I calculated the value of the test statistic as 5.5071 but I don't know what conclusion this gives. I got the critical region of 2.473. so what does 2.473<5.5071 mean? or do I have even more calculations for this problem. i keep confusing myself with all the different values.

It makes more sense to call the critical region the "rejecting region."

The value you calculated for t (5.5071) falls in that region (t > 2.473). That means that there is enough evidence at the $\alpha = 0.01$ significance level to reject the null hypothesis in favor of the alternative hypothesis.

If your calculated value had not fallen in the rejecting region, then there would not have been enough evidence at that $\alpha$ level to reject the null hypothesis.

madgab · #9 $Old$ June 28th, 2009, 11:08 AM

Thank you! That makes sense. And part c ended up showing that the assumption of equal variances IS valid. I think I have all the conclusions required. Thanks again