Question

Certified Behavior Analysts claim that their procedures are more effective than any other. Their procedures are clear-cut analytics that anyone can learn and apply. It’s not an art they proclaim, but a science. You decide to put it to test. You interview and survey the parents of autistic kids, loved ones of the depressed, and former sufferers of phobias, who had been treated with Behavior Therapy, Drug Therapy, and Play Therapy. Your survey generates a score from 1-100 with higher values indicating greater effectiveness of the therapy and 85 and above indicates complete resolution of the problem. Behavior Therapy: 58, 65, 71, 59, 81, 74, 83, 63 Drug Therapy: 59, 68, 32, 44, 38, 41, 30, 51 Play Therapy: 58, 61, 50, 60, 64, 62, 85, 57

1. Using the above data calculate the one-way analysis of variance in JASP. Report the F-ratio in APA style, including the p-value and eta-squared (treatment magnitude).

2. If the one-way ANOVA shows a significant difference among the groups, then calculate post hoc test to determine which groups are different – be sure you compare each group to each other group. Report the significant differences among the groups. If the one-way ANOVA does not show a significant difference, then state no difference.

Answer 1

we can analyse this using the open source statisitcal package R , the complete R snippet is as follows

BehaviorTherapy<-c( 58, 65, 71, 59, 81, 74, 83, 63 )
DrugTherapy <- c( 59, 68, 32, 44, 38, 41, 30, 51 )
PlayTherapy<-c( 58, 61, 50, 60, 64, 62, 85, 57)

library(reshape2)

df <- data.frame(BehaviorTherapy,DrugTherapy,PlayTherapy)

df <- melt(df)

## anova results
a <- aov(lm(value~variable,df))
summary(a)

# tukey test
TukeyHSD(a)

The results are

summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
variable 2 2404 1201.8 9.766 0.001 ** # as the p value is less than 0.05 , hence the results are signficant and we shall conduct a post hoc analysis
Residuals 21 2584 123.1   
---
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, df))

$variable
diff lwr upr p adj
DrugTherapy-BehaviorTherapy -23.875 -37.85562 -9.89438 0.0008791 # as the p value is less than 0.05 , hence the results are signficant
PlayTherapy-BehaviorTherapy -7.125 -21.10562 6.85562 0.4191170 # as the p value is not less than 0.05 , hence the results are not signficant
PlayTherapy-DrugTherapy 16.750 2.76938 30.73062 0.0171808 # as the p value is less than 0.05 , hence the results are signficant