PROTEUS

              

 

 

 

 

 

 

 

 

 

# 53   Detailed solution

 

 Solution assuming data were UNPAIRED

               7.8            8.9

               5.0            7.1

               2.1            4.5

               5.5            7.7

             10.0           11.9

mean      6.08            8.02

S            2.98613       2.70037

S2           8.917          7.292

 

Analysis:

                                     | 1 - 2 |                1.94 

                          t  =    ----------------      =    -----------      =       1.53       

                        ﴾S1 - 2﴿              1.273

      

 

   Calc t :   1.53

   t(0.05)(8df) : 2.31     therefore cannot reject null hypothesis

 

    The mean difference between the two groups of values was 1.94, and using a t-test for UN-paired data, we can conclude that a difference of this magnitude could have occurred by chance alone more than 5% of the time, and is therefore not statistically significant.  The treatment did not appear to have any real effect in increasing muscle strength.

 

 

  1 - 2: 1.53,     t:   1.08      8 df,         P>0.05  

 

 

 

      

       Solution assuming data were PAIRED

              pre           post         diff (post-pre)

               7.8            8.9         1.1 

               5.0            7.1         2.1

               2.1            4.5         2.4

               5.5            7.7         2.2

             10.0           11.9         1.9

                     mean                 1.94

                       S                      0.50299

                       S2                    0.2530

 

                         đ                1.94

           t  =    ------        =     ------------  =   8.624

                        SEđ               0.225          

 

   Calc t   :   8.624

   t(0.001)(4df)  :    8.61   Therefore reject null hypothesis

 

The mean difference between the pre and post treatment values was 1.94, and using a t-test for PAIRED data, we can conclude that a difference of this magnitude would occur by chance alone less than 1 in 1,000 trials.  Therefore we can reject the null hypothesis that there is no treatment effect, and safely consider the treatment as effective in increasing muscle strength.

 

     đ: 1.94,      t:   8.624      4 df,         P<0.001    

 

 

 
NOTE:  This is a very unusual situation and has been artificially created to illustrate some of the important understandings of the t-tests and what their component parts represent.   Clearly the more complete information (that the data are paired) is correct.    You will observe that while the mean differences is the same in both cases, the UNpaired data are a long way from being able to claim statistical significance, whereas the PAIRED data show a very high significance.  the explanation of course lies in the denominator for the t-statistic.  The UNpaired data allowed ALL the variation to be included i the SEmean including the irrelevant variation between individuals at the start of the trial.  The Paired data, on the othe hand, include in the SEđ only the variation that arose from each individual personal shift due to the treatment.

If the data are PAIRED, then you must used the t-test for paired data;  if the data are NOT paired, then you must use the t-test for Unpaired data.