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Mathematical Statistics- two sample t-test

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data
    At the 5% level, will the two means be equal ?


    2. Relevant equations



    3. The attempt at a solution

    I tested the variances and found out that it's very likely that the variances are equal, so this is an assumption we make when we do the second test.
    Now, I'm trying to test the means and see if they are equal.
    I'm getting confused because this is what I get:

    Two Sample t-test

    data: Var1 and Var2
    t = -2.1372 df=3998 p-value = 0.03264
    alternative hypothesis: true difference in means is not equal to 0
    95 percent confidence interval:
    -0.128821801 -0.005552541
    sample estimates:
    mean of x mean of y
    -0.0670763877 0.0001107835

    zero is not included in the but my p-value is greater than 2.5% (for a two sided test)...
    what do I do in such case ?
    my t value is more extreme than at the 2.5%.
    I think the p-value I'm given is wrong, is it possible ?

    Thanks,
    Roni.
     
  2. jcsd
  3. May 18, 2010 #2

    Mark44

    Staff: Mentor

    It seems very odd to me that you confidence interval doesn't include 0. Also, since df = 3998, your sample size must be about 4000. With that large a sample, you don't need to use the Student's t test, but could instead use a normal distribution.

    There is a lot you don't show, but I infer that you have a binomial distribution that you can approximate with a normal distribution.
     
  4. May 18, 2010 #3
    You are right, I have 2 different sets of data with 2000 r.v's in each of them.

    I think if it works with the normal dist test, it should also work with the t-test (the df is large)

    here is some of the data:

    1 -1.67695968151127 0.577070066857405
    2 -0.642441727404665 0.53974331907915
    3 1.03286440763869 -1.31999608111349
    4 2.02667778667435 -0.0361000625995362
    5 0.417407406923095 1.23635840784727
    6 0.338215656007098 0.295867822353842
    7 -0.831289368212899 -0.419309627121533
    8 0.90774100682453 1.09654263070674
    9 -1.22308033810869 0.474269912619356
    10 0.201242609350191 0.793994235426449
    11 -0.349209983009311 -0.620406980101651
    12 0.303036015889208 -1.01786836015372
    13 0.946824776561656 1.61722911478799
    14 1.04405267789264 0.95021579495309
    15 -0.5631803909092 -0.773840047711912
    16 -0.618697005188519 0.877219620279467
    17 -1.14813286274261 0.774274810083378
    18 0.361139276781852 1.29857639982538
    19 -1.69816166131597 -0.132765129167876
    20 -1.85142475578048 2.01112325343722
    21 -0.75348529082793 0.125903773730686
    22 0.373202072550282 0.210826733939696

    I don't know which distribution is used to generate these r.v's...
     
  5. May 18, 2010 #4

    Mark44

    Staff: Mentor

    Well, something is not working, it seems. If your null hypothesis is that the difference of the two means is zero (i.e. they are equal), your confidence interval should straddle 0.

    As you have things, 0 is outside the confidence interval, so if your test statistic came out as 0 you would reject the null hypothesis! That doesn't seem reasonable at all.
     
  6. May 18, 2010 #5
    I think the p-value that's given to me is already multiplied by two.
    I just did the z-test for two samples with excel and I got the same p-value.
    Now, that the p-value is less than 5%, we can reject the null hypothesis.

    Thanks for the help.
     
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