# T test.

• Apr 27th 2011, 12:07 PM
statsmuncher
T test.
Hello, I was told by my lecturer that I need to do some T-tests on my data in order to find out if it is significant or not. The data is basically student's grades and I am trying to prove that intrinsic students achieve higher grades in comparison to extrinsic students.

So I used an online calculator to calculate the T test on this data:

Data 1 (extrinsic students):

49.5
52
55
55.5
55.5
59
59.5
60
60.5
63.5
65.75
66
67
69.5
76.5
78
78

Data 2: (Intrinsic students)

55
59.5
60
63.5
64
65
65.5
68
69
72.5
76.5
76.5
77.5
77.5
80
82
83

For this data I set it as of unequal variance and a 2 tail test. I am not too sure If i used the right tail test. The following figure was returned:

Online calculator: = 0.01967
Excel: 0.009833876

I have no idea what either of these numbers mean, can anyone help?

Thanks,
Sean
• Apr 27th 2011, 02:16 PM
pickslides
These numbers will mean nothing to you unless you understand how a T-test works.

Because you have 2 groups of data you will need a 'two sampled T-test' and because you are trying to find if one group is better than the other (not just different) you will need only 1-tail.

Now I suggest you research the type of test I have prescribed.
• Apr 27th 2011, 03:59 PM
Alonso
Based on your data (group 1 vs group 2), I believe that you are testing the alternative hypothesis that the mean for the group 1 is different from the mean of the group 2, or vice-versa - assuming that this difference can be in any direction. In other words, your null hypothesis is that the difference between those means is zero. Running your data in R, I got a two-sided P-value of 0.0196677511603729 (t=-2.4559, with 32 degrees-of-freedom). The value of 0.0098 you mention refers to the one-sided test, in which your alternative hypothesis is that the mean for the group 1 (extrinsic) is smaller (one direction only) than group 2, that is, your alternative is that the difference (mean 1)-(mean 2) < 0. I got these results assuming equal variances. Use of Welch's approximation (unequal variances) gives slightly different results: (P = 0.0193 and P = 0.0097 , t= -2.4559 and 33.9832 degrees-of-freedom)
• Apr 28th 2011, 02:50 AM
statsmuncher
Thankyou for the help
Thanks for the help so far, I have done further research as suggested. I going to summarize what I have learned and hopefully it is not horribly wrong (Angry)

My Null Hypothesis is: There is no statistical difference between the intrinsic and extrinsic student grades.

• Two sample test: As there are two sets of data

• unequal variance

• 1 tail: As I am trying to prove Intrinsic students get higher grades than extrinsic students.

Result (using excel): 0.009835258 = 00.1

Based upon this, I can reject the Null hypothesis as it is below the 0.5 figure and say that my data is significant.

Ok so that is my understanding so far, I would really appreciate if someone could advise me if my understanding is correct (or not!).

Thanks
• Apr 28th 2011, 02:12 PM
pickslides
You are on the right track,

The hypothesis formulation here is

$\displaystyle H_0:$ the mean of intrinsic students equals the mean of extrinsic students.

$\displaystyle H_a:$ the mean of intrinsic students is greater than the mean of extrinsic students.

If the one tailed test returned a p-value of $\displaystyle 0.01$ then reject $\displaystyle H_0$ at $\displaystyle \alpha = 0.05$

(be careful of your rounding, above you have written "Result (using excel): 0.009835258 = 00.1", this is not true)
• Apr 30th 2011, 04:14 PM
statsmuncher
Quote:

Originally Posted by pickslides
You are on the right track,

The hypothesis formulation here is

$\displaystyle H_0:$ the mean of intrinsic students equals the mean of extrinsic students.

$\displaystyle H_a:$ the mean of intrinsic students is greater than the mean of extrinsic students.

If the one tailed test returned a p-value of $\displaystyle 0.01$ then reject $\displaystyle H_0$ at $\displaystyle \alpha = 0.05$

(be careful of your rounding, above you have written "Result (using excel): 0.009835258 = 00.1", this is not true)

Thank you for your help, it is greatly appreciated! :)