I am trying to undertake some Z-Tests to see the level of effect in Standard Deviations of 8 separate treatments compared to a control mean.

However there is one thing that has constantly confused me in all my attempts. How do I set the mean rating of my control as the mean to compare all the other too?

My Data

             Control  T1      T2      T3     T4     T5     T6      T7     T8
Mean Ratings 4.74     -1.77   -1.88   7.77   4.08   6.90   13.66   17.13  19.44

The sample population for the control and each treatment is circa 105, is normally or approximately normally distributed and each participant contributed one score to each mean rating.

Calculating Z-Scores - Method 1

Using SPSS if I run a simple two tailed Z Test on the above using Analyse > Descriptive Statistics > Descriptives I get:

            N     Minimum   Maximum   Mean      Std. Deviation
VAR00023    9     -1.88     19.44     7.7856    7.63361
Valid N (listwise)  9                  

Control: -.39897
T1:      -1.25177
T2:      -1.26618
T3:      -.00204
T4:      -.48543
T5:      -.11601
T6:      .76955
T7:      1.22412
T8:      1.52673

The above tells me that none of my treatments were significant at α=0.05

However this does not seem right as its including my control mean (4.74) in the equation.

Calculating Z-Scores - Method 2

If I run the same test without the control mean I get:

Descriptive Statistics                  
              N    Minimum   Maximum    Mean    Std. Deviation
VAR00002      8    -1.88     19.44      8.1663  8.06883
Valid N (listwise)  8               

T1:      -1.23144
T2:      -1.24507
T3:      -.04911
T4:      -.50642
T5:      -.15693
T6:      .68086
T7:      1.11091
T8:      1.39720

The above tells me much the same as the previous method, that none of my treatments were significant at α=0.05, however there are variations in each score

This does not seem right either as I have not specified the control mean (4.74) that I want all of the Treatments to be compared too?

My Question

Can anyone tell me which is the right track, or the closest to it, and where I might be going wrong?

Neither method I have tried appears to be right for the aforementioned reasons.


Thanks all for taking the time to look at this. I have struggled with this for a good few days now and would really appreciate if anyone could explain the correct way to undertake a Z-Test using the above data so that I can see where my reasoning has been going wrong. Thanks

NOTE: I understand from the answer to previous question I asked that I will have to correct for Family Wise Error Rate, that will be my next challenge to figure out, I just want to see if I can get this part figured out first.

  • $\begingroup$ I don't see any place where you have informed SPSS that each of the nine values is the mean of 105 others, nor do I see where you have specified the nine associated standard deviations (or variances), whose values are essential for doing the tests you propose. I also don't see any indication of "significance" in any of the output you have shared with us. Consequently I suspect that you really want to do something completely different. What exactly is your research question? What exactly do the original (non-summarized) data look like? $\endgroup$ – whuber Aug 17 '16 at 15:36
  • $\begingroup$ Hi whuber, I believe this question stats.stackexchange.com/questions/229698/… and especially this question stats.stackexchange.com/questions/229794/… which I asked previously will show you what I am trying to achieve. --- Regarding the first two points you raised I was not even aware that this was necessary or possible in SPSS and I have looked at and followed every tutorial I could find. I will start looking again now. 1/2 $\endgroup$ – Deepend Aug 17 '16 at 16:42
  • $\begingroup$ Regarding "I also don't see any indication of "significance" in any of the output you have shared with us." This is just the first of a series of similar tests I will have to do and while there might be no obvious effect of any of the treatments here I would like to know how to undertake such a test for the future as well. Thanks for all your help. 2/2 $\endgroup$ – Deepend Aug 17 '16 at 16:44

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