# t-test degrees of freedom

I'm having difficulty in understanding this fairly common question:

"A teacher is trying to determine whether or not a new teaching method is effective in helping students understand a challenging concept. The teacher evenly divides 30 students into two randomly selected groups. the first group will be taught using a traditional method, while the second group will be taught using the new method. At the end of the unit, all of the students will take the same exam. Assuming a 95 percent level of confidence, which of the following decisions should be made regarding hypothesis below?"

$$H_0:μ1=μ2$$ $$H_a:μ1<μ2$$

(A) Reject the mull hypothesis if the test statistic is greater then -1.761

(B) Reject the mull hypothesis if the test statistic is less then -1.761

(C) Reject the mull hypothesis if the p-value >0.05

(D) Reject the mull hypothesis if $$\bar{x_1}-\bar{x_2}=-1.5$$

(E) Reject the mull hypothesis if $$\bar{x_1}-\bar{x_2}=-2.5$$

the correct answer is B: in which $$df=15-1=14$$ was used to find the above critical value.

But in my opion, the answer used a wrong test (t-test for population mean claim) here.

this is a question for testing difference of means in two populations. the hypothesis should be:

$$H_0:μ1−μ2=0$$ $$H_a:μ1−μ2<0$$

and $$t=\frac{\bar{x_1}-\bar{x_2} -0}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$$ which should have a $$df=n1+n2−2=15+15−2=28$$

Can anyone help verify my belief?

the original question is here:

Questoin 10

• Which version of the Student t-test is being applied? Until that is specified, this question has no definite answer.
– whuber
Oct 6, 2022 at 14:34
• I think both t-test for population mean claim and t-test for difference in populations means can be used. but the methods to calculate the degree of freedom are different. I wonder if both test will produce same result, if not which test is the correct test. Oct 6, 2022 at 15:00
• I believe t-test for difference in populations means is more appropriate. I believe the degree of freedom should be calculated as $df=n_1+n_2-2=28$. but the answer was $df=15-1=14$ Oct 6, 2022 at 15:10
• For what it's worth, I agree with the original poster, @techie11, and think none of the answers would be correct under the usual assumptions of a question like this. I would assume the degrees of freedom should be 28, and the cutoff for the t statistic would be 1.701 or -1.701, with a one-sided test. Oct 6, 2022 at 15:45
• The link goes to a page that is not available. This is a test review book, so enough said: don't assume the questions are good or the answers are reliable.
– whuber
Oct 6, 2022 at 17:15

## 1 Answer

Starting with the null and alt hypotheses

𝐻0: μ1 − μ2=0
𝐻𝑎: μ1 − μ2<0.

and

𝐻0: μ1 = μ2
𝐻𝑎: μ1 < μ2

Are the same thing because if μ1 = μ2 then μ1 − μ2 = 0 and likewise if μ1 < μ2 then μ1 − μ2 < 0, they are different ways of stating the same thing.

AS for the test, it should be the two sample difference of means t-test and you are correct the df should be (n1+n2)-2=28