The following data are from a completely randomized (between-subjects) design: F

Question

The following data are from a completely randomized (between-subjects) design:

Five psychologists analyze this data set individually, each with different goals in mind. Your task is to duplicate the results obtained by each.

a. Psychologist 1 formulates three planned comparisons of interest: Group 1 versus Group 2, Group 1 versus Group 3, and Group 2 versus Group 3. Perform these planned comparisons, assuming homogeneity of variance.

b. Psychologist 2 performs an omnibus test and a comparison of the averages of Groups 1 & 2 versus Group 3.

Note: A full answer includes the null hypothesis for each test, the contrast specifications (c = ...), the test statistic with degrees of freedom, and the critical values (which depend on the method you use).

In addition, provide a very brief explanation of why you chose the specific method you did to analyze the data.

Group 1 Group 2 Group 3 48 59 68 54 46 62 47 49 53 54 63 59 62 38 67 57 58 71

Explanation / Answer

We shall analyse this using the open source R statistical tool

H0 : The mean values of group1 and group2 are not different

H1 : The mean values of group1 and group2 are statistcaily different

Likewise we can formulate 2 more hypothesis for the other 2 groups combos

For the omnibus test

H0 : there is no difference in the mean values of the groups

H1 : there is a significant difference in the mean values of atleast 2 groups

The complete R snippet is as follows

data.df<- read.csv("C:\Users\586645\Downloads\Chegg\psyco.csv",header=TRUE)
str(data.df)

# performing the t test individually one at a time
attach(data.df)
t.test(Group.1,Group.2)
t.test(Group.1,Group.3)
t.test(Group.2,Group.3)

#perform the omnibus 1 way anova analysis

library(reshape2)
data.df<- melt(data.df)

# perform anova analysis
a<- aov(lm(value~ variable,data=data.df))

#summarise the results
summary(a)

TukeyHSD(a)

The results are

> t.test(Group.1,Group.2)

   Welch Two Sample t-test

data: Group.1 and Group.2
t = 0.33425, df = 8.1327, p-value = 0.7466 , , as the p value is not less than 0.05 hence not significant
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-8.819196 11.819196
sample estimates:
mean of x mean of y
53.66667 52.16667

> t.test(Group.1,Group.3)

   Welch Two Sample t-test

data: Group.1 and Group.3
t = -2.7209, df = 9.7223, p-value = 0.02205 , , as the p value is less than 0.05 hence significant
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-17.613505 -1.719828
sample estimates:
mean of x mean of y
53.66667 63.33333

> t.test(Group.2,Group.3)

   Welch Two Sample t-test

data: Group.2 and Group.3
t = -2.3661, df = 8.9775, p-value = 0.04224 , as the p value is less than 0.05 hence significant
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-21.8466540 -0.4866793
sample estimates:
mean of x mean of y
52.16667 63.33333

for the omnibus test

> summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
variable 2 440.8 220.39 4.005 0.0404 * , as the p value is less than 0.05 hence the omnibus test is signifcant , we can now perform a tukeyhsd test to check which groups are statisically different
Residuals 15 825.5 55.03   
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

> TukeyHSD(a)
Tukey multiple comparisons of means
95% family-wise confidence level

Fit: aov(formula = lm(value ~ variable, data = data.df))

$variable
diff lwr upr p adj
Group.2-Group.1 -1.500000 -12.6250728 9.625073 0.9349151
Group.3-Group.1 9.666667 -1.4584061 20.791739 0.0936399
Group.3-Group.2 11.166667 0.0415939 22.291739 0.0490903 , this is different as the p value is less than 0.05

The degrees of freedom are also generated as a part of the analysis and are denoted by df

we are assuming an alpha =0.05 for the question

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