Here is a simple example of repeated measures where we will use Graphpad and onl
ID: 3217577 • 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. n a crossover des Here is the data Use graphpad to analyze a paired t test? observed value of the test statistic? 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
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d = 0
Alternative hypothesis: d 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
s = sqrt [ ((di - d)2 / (n - 1) ]
s = 7.463
SE = s / sqrt(n)
S.E = 3.34
DF = n - 1 = 5 -1
D.F = 4
t = [ (x1 - x2) - D ] / SE
t = - 0.06
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 4 degrees of freedom is more extreme than 0.06; that is, less than - 0.06 or greater than 0.06
Thus, the P-value = 0.95
Interpret results. Since the P-value (0.95) is greater than the significance level (0.05), we cannot reject the null hypothesis.
From the above test we have sufficient evidence in the favor of the claim that there is no differnce between two groups.