Dependent Groups z and t Tests

There is the one final type of z and t tests we need to consider. This type of test is used when we have two scores on each of a number of individuals, such as a "before and after" test.

    This type of test is referred to under a variety of names:

*Paired Samples      *Dependent Groups

*Correlated Groups   *Repeated Measures

*Within Subjects *Participants as their own controls

    Let's use our same old scores as before, but consider that groups N and M are the first and second administration of a test to the same 5 people, such as an acheivement test given at the beginning and end of a semester of instruction.

    The good thing about this test is that you already know how to do it. It reduces to the one-sample test of a sample against a parameter, and the parameter is zero. We compute the difference (retaining signs) for each person, either  subtracting the first score from the second, or the second from the first (the only difference will be the sign of the z, and you have to keep track of the direction you choose to interpret the result). If there is no overall change between scores on the first and second administrations, we would expect that the mean difference between scores would be zero. That is the null hypothesis that is tested using the one-sample method. The reason it is one-sample is because, while we started with two scores on each person, we converted each person's two scores to a single difference score (called "D"):

.          1st   2nd

.          Test  Test     DIFFERENCE (D)

Person 1:    2   4              -2

Person 2:    7   6               1

Person 3:    9   8               1

Person 4:    8   13             -5

Person 5:    12  9               3

.

Sum of D =                      -2     .

Mean of D =           -2/5 =   -.40    .

Standard Deviation of D =      3.13    .

    Using the same method shown in lesson 5 for a one-sample test against a parameter, we compute a standard error of the mean for D, and then compute a z for the mean.

Standard Error of Mean = (3.13)/(sqrt(5)) = 1.400

z = [observed difference] / [standard error estimate]

  = [(0-(-.4)]/ [1.4000] =  -0.286

    Since our sample size is less than 30 (in this case it would be 30 "pairs"), we use the t test. The degrees of freedom is the number of pairs minus 1 (5-1 =4 = df).    Looking in the t table, we find the two-tail critical t for df=4 is 2.776 (the NIST table gives the one-tail probability in the column headings -- one-tail .025 is two-tail .05). Our obtained z (which is the same as a t in this case), is -.286. We obviously fail to reject the null hypothesis. Here is the output from SPSS. Note that if you know the terms you are learning in this class, you can decipher the output quite easily. The "sig." value is the p value, and it is the area under the t curve which is the sum of what falls above +.286 and below -.286:

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     Please be sure you understand this technique. Do ask questions for clarification. You should work the problem to be sure you can get the same numbers. There is no assignment associated with this. It is appropriate, since we are looking at "participants as their own controls", that you should serve as your own instructor! Don't fail yourself!