Paired sample T-Test

We said we would look at an example where the samples are not independent.  (SPSS calls these paired samples.  Sometimes they are called matched samples.)  Let's say we wanted to compare the educational level of the respondent's father and mother.  PAEDUC is the years of school completed by the father and MAEDUC is years of school for the mother.  Clearly our samples of fathers and mothers are not independent of each other.  If the respondent's father is in one sample, then his or her mother will be in the other sample.  One sample determines the other sample.  Another example of paired samples is before and after measurements.  We might have a person's weight before they started to exercise and their weight after exercising for two months.  Since both measures are for the same person we clearly do not have independent samples.  This requires a different type of t test for paired samples.

Click on "Analyze", then point your mouse at "Compare Means", and then click on "Paired-Samples T Test".  Scroll down to MAEDUC in the list of variables on the left and click on it to move it to the Current Selections box as Variable 1.  Now click on PAEDUC to move it to the Current Selections box as Variable 2.  Click on the arrow to the left of the Paired Variables box to move this pair of variables into the box in the middle of the window.  Click on “OK".  The output table shows the mean years of school completed by mothers (11.47) and by fathers (11.33), as well as the standard deviations.  The t-value for the paired-samples t test is 1.822 and the 2-tailed significance value is 0.069.  (We may have to scroll down to see these values.)  This is the probability of getting a t-value this large or larger just by chance if the null hypothesis is true.  Since this probability is greater than .05, we won't reject the null hypothesis.  There is no statistical basis for saying that the respondents' fathers and mothers have different educational levels.  However, notice that if we were using a one-tailed test, then we would divide the two-tailed significance value of .069 by 2 which would be .0345.  For a one-tailed test, we would reject the null hypothesis since the one-tailed significance value is less than .05.