Trivial Quetsion - One-Sided Hypotesis Tests - P1.T2.210.2

[MissJaguar]

Hello there!

I wonder if there is a difference in result of this question if, ceteris paribus, instead of

(1)
Ho < or = 27
Ha > 27 as in the old question P1.T2.210.2, hypothesis was

(2)
H0 > or = 27
Ha < 27.

I thought that for the first case t-statistics was negative (- 1.71, since (mu hypotests - mu sample) is -2), so that at 95% cannot be rejceted.

In the second case I though that we take (mu sample - mu hypothesis), which gives t stat 1.71, so that at 95% confidence null can be rejected.

What is wrong here? I trided to find the analogy with results of question P1.T2.209.2 (of cause having in mind difference in the sample size in the two questions - large and small).

Please, please reply to my dilemma.

Thanks.

 

[David Harper CFA FRM]

Hi @MissJaguar Very interesting observation, I think one-sided tests are tricky

Start with the computed t (test) statistic, as opposed to the "lookup" value. Although it is often wrapped within an absolute value, it has a very intuitive definition; the compute t (test) statistic is the number of (standardized) standard deviations from the observed sample mean to the hypothesized (population) mean:

• compute t (test) statistic = (observed sample mean - hypothesized population mean) / standard error; so it's either negative (we observe a sample mean less than hypothesized) or positive.
• In 210.2, the test t statistic = (observed sample 29 - 27 hypothesizes)/1.16667 = + 1.714;
• i.e., the sample average of 29 is +1.714 standard deviations greater than the hypothesized mean of 27
• If, instead for example, we observed sample mean of 25, then this t statistic = (observed sample 25 - 27 hypothesizes)/1.16667 = -1.714

Consequently, hypothesis-wise, we are in the right-tail regardless of one versus two-tailed test.

• However, a left-tailed (one tailed) test is not very useful when our observation falls in the right-tail (I don't see why we'd do it, maybe I am not fully aware ... ). A left-tailed (one-tailed) test assumes a null of: population mean >= 27, but 29 is then obviously the acceptance region, by definition. We obviously can't reject this null, but we don't need any computations either!
• As we observe 29 in the right-tail (to the right of hypothesized 27), if we are not going to conduct a typical two-tailed test, the appropriate null follows from a right-tailed test: null is that true (population) mean is less than or equal to 27. And becomes +1.714 is greater than +1.65, we are in the right-tail rejection region. We can reject, which is what we were hoping to do, in favor of accepting the idea that 29 supports our suspicion that the true mean is higher than 27 (compared to the useless left tail test which tries to find a way for a 29 to support the idea that the true mean is less than 27. It obviously does not!)

To further illustrate, the left-tailed test is the appropriate test if, for example, we observed a sample mean of 25 (compared to hypothesized true mean of 27). In this case,

• computed (test) statistic = -1.714; i.e., we observe sample mean "to left of" (less than)
• The lookup t value is still 1.65, due to symmetry, but the left-tail rejection region is actually any value less than -1.65 (left tail of the distribution)
• The null is rejected because (when we are in the reject region) -1.714 < -1.65; but as the student's t is symmetrical, we never express is that way, we use the same absolute value comparison
• But here our test is saying "25 sufficiently supports the suspicion that the true mean is less than 27" (left-tailed one-tailed), compared to the useless left-tailed test that can't use 25 to support the suspicion that the true mean is greater than 27. I hope that helps!

 

[MissJaguar]

Hey David,

Many thanks for your reply and apologizes for late thank you note - was quite busy and had no chance to sit for FRM.

Thanks for your valuable input as always