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Here is a simple example of repeated measures where we will use Graphpad and onl

ID: 3218555 • Letter: H

Question

Here is a simple example of repeated measures where we will use Graphpad and online ANOVA. Two treatments and each subj. received both the treatments (e.g in a crossover design). Here is the data Use graphpad to analyze a paired t test? 1 1.) observed value of the test statistic? 2.) Whats the conclusion about the mean response of 2 treatments? why? Now same study with three groups. Run a repeated measures ANOVA a) global p-value for testing differences in means? Whats the conclusion? Why? b) In the above| design, reanalyze it with 15 different subj with no repeated measures? How will the results be different?

Explanation / Answer

1) Null hypothesis H0: mewd = 0

Alternative hypothesis H1:mewd not = 0

Paired Samples Statistics
Mean   N   Std. Deviation   Std. Error Mean
Pair 1   GroupA   34.4000   5   7.82943 3.50143
GroupB   34.6000   5   6.91375 3.09192

Paired Samples Test
Paired Differences
95% Confidence Interval of the Difference              
Mean   Std. Deviation   Std. Error Mean   Lower   Upper t   df   Sig.(2-tailed)
Pair 1   GroupA - GroupB   -.20000   7.46324 3.33766 -9.46684   9.06684   -.060   4   .955

therefore the sginifant value 0.955>0.05

therefore H0 is accepted i.e we may conclude that the mean difference between group A and group B is zero

2) Nullhypothesis H0:mewA = mewB=mewc i.e there is no significance difference between the means of theree groups

Alternative hypothesis H1:mewA not = mewB not =mewc(two tailed test)

               Descriptives
Group
95% Confidence Interval for Mean
N   Mean   Std. Deviation   Std. Error   Lower Bound   Upper Bound
GroupA   5   34.4000   7.82943 3.50143 24.6785   44.1215   
GroupB   5   34.6000   6.91375 3.09192 26.0154   43.1846
GroupC   5   52.6000   12.13672 5.42771 37.5303   67.6697
Total   15   40.5333   12.29905 3.17560 33.7223   47.3443   

       ANOVA One-way classification table
Group
Sum of Squares   df   Mean Square   F Sig.
Between Groups   1092.133 2 546.067 6.389   .013
Within Groups 1025.600 12 85.467      
Total 2117.733 14          

therefore the signficant value 0.013< 0.05

therefore the result is significant at p value i.e we may conclude that null hypothesis is rejected i.e there is a significnat diffrence between the three groups