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Following experimental design was done and some data as shown in table below was obtained:

pretest> intervention1> intervention2> posttest > perception survey

Number of students=60

Number of Students subjected to intervention=20

Independant variable is "Intervention2" which possible influences posttest .Perception survey is the students own evaluation of the intervention, lower the better.

pretest posttest intervention1 intervention2 perceptionrank

37  35  10  10  4.2
38  28  8   10  2
20  30  8   10  #N/A
34  22  10  10  1.7
28  21  10  10  3.2
23  19  8   10  3.5
14  8   10  10  2
26  33  7   8   3.2
24  35  8   8   2.2
33  21  7   8   1.8
29  25  7   8   2
36  20  7   8   2.9

The subjects were students in the same classroom, the interventions were also evaluated , the score for which is also tabulated above. What I am interested to find the causality between intervention2 and posttest scores.

How should I go about solving this problem , should I use t-test , correlation and how?

Update:

I forgot to add that the control group is the students who did not receive the interventions , so I have their pretest and post test scores.

Moreover I also have perception survey scores for the students who received the intervention.

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  • $\begingroup$ Are interventions 1 & 2 different experimental conditions to which the students were randomly assigned? Is one a control condition? $\endgroup$
    – Nick Stauner
    Commented Jan 9, 2014 at 7:57
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    $\begingroup$ I updated the answer @NickStauner , interventions 1,2 are similar interventions to randomly assigned students. $\endgroup$
    – stats101
    Commented Jan 9, 2014 at 9:44
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    $\begingroup$ Please tell us your complete design. Who got which intervention? How was that determined? Why did you measure perception rank? What is your N? What are you trying to find out? (Causality can only be established in some very limited circumstances). $\endgroup$
    – Peter Flom
    Commented Jan 9, 2014 at 11:48
  • $\begingroup$ @PeterFlom updated in the question $\endgroup$
    – stats101
    Commented Jan 9, 2014 at 11:57

2 Answers 2

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If you didn't have a control group, you can't really establish causality (as far as I know, though you might be able to rule it out to whatever extent your results show no relationship at all). If you edit your question or respond to my comment with clarification of whether you did, I'll edit my answer to suit.

Correlations estimate the (strength and direction) of relationships; $t$-tests estimate , and thus involve . I suspect you'll want to do all of the above to squeeze as much info as you can out of your data, but you don't really have to just to "find [some information about] the [potential] causality."

You should probably look a little further into the differences between effect size estimation and significance testing, and develop more specific questions about the potentially causal relationship(s) in question. Some of the tags on your question can help you browse related topics here, and the tag wikis themselves contain nice, brief summary intros to the topics in general. Again, if you can edit your question to be a little more specific, leave a comment to let me know, and I'll try to edit to follow up.

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    $\begingroup$ The control group is simply the one who did not recieve the treatment , in my case about 20 students from a class of 60 received the treatment. $\endgroup$
    – stats101
    Commented Jan 9, 2014 at 9:48
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    $\begingroup$ I have added a sample perception ranks $\endgroup$
    – stats101
    Commented Jan 9, 2014 at 10:39
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    $\begingroup$ perception rank is the users own evaluation of the positive effect of the intervention, lower the better. $\endgroup$
    – stats101
    Commented Jan 9, 2014 at 11:24
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    $\begingroup$ some effect on posttest only, and I cant understand " the test scores somehow incorporate the information from their evaluations...?" $\endgroup$
    – stats101
    Commented Jan 9, 2014 at 11:44
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    $\begingroup$ Both correlations and t-tests can be used to both test statistical hypotheses and as measures of effect size. They measure/test different things. $\endgroup$
    – Peter Flom
    Commented Jan 9, 2014 at 11:46
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As a rule of thumb, testing for correlation will be done if you have one sample with two metric outcomes. Then you consider the alternative hypothesis "the lower one outcome the higher (lower) the other". Correlation alone is never causality.

Here, you have two samples (students with intervention2 and intervention1) with one outcome (posttest score). You would consider the alternative hypothesis: "Is the score due to intervention2 different from the score after intervention1?" This should be tested by a Wilcoxon test, the t-test's brother for not metrically scaled data (scores) --assuming you don't have to compare the change from pre- to posttest score. Otherwise you have a 2x2 nonparametric split plot design. But that's another question.

If the students have been assigned properly to the intervention groups, you may in fact infer causality.

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  • $\begingroup$ thanks,can you explain what is "nonparametric split plot design" $\endgroup$
    – stats101
    Commented Jan 10, 2014 at 12:57
  • $\begingroup$ Split plot design means that you have independent groups of subjects and dependent measurements (pre/post) per subject that you compare altogether (pre1-post1>pre2-post2). Nonparametric means that your data cannot be evaluated by metric methods (e.g. t-test, ANOVA), mostly because they are only ordinal or the distribution is completely unknown. $\endgroup$
    – Horst Grünbusch
    Commented Jan 10, 2014 at 13:14
  • $\begingroup$ I only have two groups\subjects first who recieved the intervention and the other who did not $\endgroup$
    – stats101
    Commented Jan 10, 2014 at 13:22
  • $\begingroup$ Generally, if you want to compare the change of the two scores between intervention group 1 and intervention group 2, it's a 2x2 split plot. If you just want to compare the post scores between two groups, it's a usual 2-sample test (Wilcoxon, t-test). I hope this helps you and others. If you are still not sure about your particular data set, a statistical consultant might have a confidential look at it. $\endgroup$
    – Horst Grünbusch
    Commented Jan 10, 2014 at 13:54

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