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Math Help - T test.

  1. #1
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    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
    Last edited by mr fantastic; April 27th 2011 at 07:40 PM. Reason: Title.
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  2. #2
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    pickslides's Avatar
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    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.
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  3. #3
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    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)
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  4. #4
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    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

    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
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  5. #5
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    You are on the right track,

    The hypothesis formulation here is

    H_0: the mean of intrinsic students equals the mean of extrinsic students.

    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 0.01 then reject H_0 at \alpha = 0.05

    (be careful of your rounding, above you have written "Result (using excel): 0.009835258 = 00.1", this is not true)
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  6. #6
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    Quote Originally Posted by pickslides View Post
    You are on the right track,

    The hypothesis formulation here is

    H_0: the mean of intrinsic students equals the mean of extrinsic students.

    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 0.01 then reject H_0 at \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!
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