This is Level I material and yet I got totally confused. Please someone help me.
When doing a one-tailed t-test and our calculated t value is a negative number do we compare it with the negative of the critical t value from the t-table or with the positive?
Eg. My calculated t value is: - 2.6 The critical t-value is 2.42 from the table.
If we look at the negative tail we can reject the null because -2.6 is less than -2.42
However -2.6 is less than + 2.42 so if we look at the upper tail we will fail to reject the null.
What is the rule here? I recall it has to be decided in advance, which tail (upper or lower) to use, but I never saw this defined in a question.
Upper-tail test- compare the calculated test statistic with the positive critical value. If the calculated value exceeds the positive critical value, reject Ho.
Lower-tail test- compare the calculated test statistic with the negative critical value. If the calculated value does not exceed (is less than) the negative critical value, then reject Ho.
Two-tail test- compare the calculated test statistic magnitude with the magnitude of the critical value. If the calculated value has a large magnitude, reject Ho.
Don’t get confused by the sign of the calculated test statistic. Decide on the “tailedness” of the test and let that dictate which “procedure” you use.
With the negative, because you are assessing a value at the left of the mean, so use the left tail.
Yup, correct.
Nope, wrong.
The T probability distribution is symmetric centered at 0. So, you can assess values at the left of the center, or right. If you evaluate values at the left of the center, the T-calculated will result negative, so compare it with the left tail critical value. If you evaluate values at the right of the center, the T-calculated will result positive, so compare it with the positive tail critical value. You can use this as a “rule”, however is not more than a fact.
As Tickersu said above, you may use absolute value of the T-calculated and compare it with the absolut value of the T-critic. As you may have noticed, both the negative and positive critical values are identical to their counterparts. This is true because T probability distribution is symmetrical.
I wouldn’t be so quick to tell Moosey this is incorrect. If doing an upper-tail test and the critical value is +2.42, but our calculated test statistic is -2.6, we would fail to reject the null. You would reject the null in the case of a two-tail test or a lower-tail test, given this calculated test statistic of -2.6 (since the lower critical value would be -2.42).
This was my point earlier: do not let the calculated test statistic guide your choice of upper or lower tailed test (that’s like peeking). Decide from the question (or theory/good judgement, if you’re actually doing research) whether or not the critical value should be a positive (upper tail), negative (lower tail), or both/absolute value (two tail). If you follow this, you’re less likely to mix up the “tailedness” of the test.
That implies that the other methods are harder…Also, your suggestion can lead to wrong conclusions provided you’re doing an upper (lower) tail test and have a test statistic that’s negative (positive). The correct conclusion is failing to reject Ho (say if -2.8 < +1.96 in an upper tail test), whereas your method of comparing absolute values would lead to rejecting the null (which would be incorrect).
How would you get a negative T-statistic from an upper-tail test assessment? Can you provide an example?
An example of mine. Suppose you are doing a hypothesis testing from a sample of “The worst returns of our portfolio ever”, so the sample is full of negative values like (-15%, -8%, etc). Therefore, this sample mean would also be negative.
When you apply a T-test over this sample you can do an upper-tail test or a lower-tail test. If I do an upper-tail test, will I get a negative T-calculated? I think no, because the value I’m assessing is above the mean.
That’s why OP would be wrong if he compares its upper-tail t-critic with his lower-tail t-calculated.
Perhaps my example is not even exhaustive and there is a case when the opposite happens, so you would be 100% right
Because the calculated test statistic is independent of the critical value chosen (which depends on upper, lower, or both tails). So, this happens any time I believe the parameter value exceeds some null hypothesis value but my sample statistic lies below that null hypothesis value.
Sure. So let’s say you were testing with the null hypothesis is that the mean return is positive-- Ho: mu<=0 Ha: mu>0
When you calculate the test statistic using the sample data you take the negative sample mean (as you mentioned) and subtract off the null hypothesis value of 0 (-5%-0%, for example) and you have a negative number in the numerator which makes the calculated test statistic negative (the denominator is always nonnegative). So let’s say this gives us a calculated test statistic of -2.9. If we were doing an upper-tail test using +1.645 as our cutoff (5% in one tail), then -2.90 < +1.645 and we fail to reject Ho. We can’t conclude that the true mean return is positive.
More generally, you’ll get a negative (positive) test statistic when the sample statistic (sample mean in our case) is less (greater) than the hypothesized value of the parameter (mu in our case). This is independent of whether or not you choose to do an upper or lower-tail test (or two tail).
The OP still wouldn’t be wrong, though. It really depends on whether or not the calculated test statistic comes out positive or negative and whether you independently chose a critical value based on an upper or lower tail test. This isn’t a case that is uncommon or an exception to the rule. There are plenty of times that the researcher might conduct a test where the tail “disagrees” with the sign on the calculated test statistic. I would try not to think of a calculated test statistic in terms of “tailedness.” This seems to be where the confusion is stemming from. The tail refers to the choice made a priori and relates to the critical value.
In the test, they’ll tell you in the question (or you need to read between the lines). They might say, “…manager X believes his fund return increases with increases in unemployment…” In a regression context, you’d say, “Ok, he thinks the slope between his fund return (Y) and unemployment (X) is positive. So the null is Ho: b1<=0 Ha: b1>0; we’re looking for evidence to support that the slope is positive (fund return increases on average as unemployment increases).”
In real life you often use subject matter knowledge or practical implications to guide your choice (usually before you see any data…). Suppose you’re doing a study to investigate the if portfolio managers returns are impacted by their age. As an employer, you might say, “The older PMs are getting a lot of flack around the office, so we want to check if increasing age of a PM has a negative impact on a portfolio’s returns. Therefore, I’m going to conduct a one-tail test to see if there’s evidence to say the relationship between portfolio returns and portfolio manager’s age is negative (older guys/gals get worse returns, on average).” So, we collect our data, run our regression (just an example) and do a one tail (negative test). Therefore, we use -1.645 (for example) as our critical value. If we calculate a test statistic less than this, we have sufficient evidence to say there exists a negative linear relationship between portfolio return and PM age. However, we don’t know this going in. It’s entirely possible that older PMs have more knowledge and expertise and generate higher returns on average. If this occurs in our sample, then we still chose a negative tail-test for our purpose (what mattered to our firm and independent of the data at hand) but we calculated (observed in the current data) a positive test statistic of 1.5, for example. This would indicate that we don’t have evidence to say the older PMs have worse returns on average (more or less).
