Module 8 Assignment

 Daniel Tafmizi

Lis 4273

Dr. Ajani 

March 3, 2024

Module 8 Assignment

Question 1)

We are comparing three groups by stress level, I need a qualitative(group name)and quantitative(level data) set

Null Hypothesis: The means of each group are the same

Alternative Hypothesis: at least one mean is different from the rest

groups <- rep(c("High", "Mod", "Low"), each = 6) #Creates vector where each variable holds 6 data

meansStress <- c(c(10,9,8,9,10,8),c(8,10,6,7,8,8),c(4,6,6,4,2,2)) #combines stress data

datafr <- data.frame(groups, meansStress) 

ggplot(datafr) + aes(x=groups, y=meansStress) + geom_jitter() #cool visual, checks that R is processing data correctly

fit <- aov(meansStress ~ groups, data = datafr) #runs anova analysis

summary(fit) 

             Df Sum Sq Mean Sq F value   Pr(>F)    

groups       2  82.11   41.06   21.36 4.08e-05 ***

Residuals   15  28.83    1.92   

Interpretation of results

Df: is 2, groups-1. DF is the number of values that can vary in a data set

Sum Sq: is 82.11, n(groupmean-totalmean)^2 + repeat on next group. Measures dispersion of all data and how it fits the model.

Mean Sq: is 41.06, Sum sq/DF. Measures average variability between group means.

F value: is 21.36, Mean sq/DF. Helps determine if there are significant differences of groups

Pr(>F): is .0000408, Uses the f ratio to quantify the difference amongst the groups.

Since the P value is tiny (less than .05), there is sufficient evidence that at least one group's mean is different from the others. Thus we reject the null hypothesis

This is to be expected given the different values of stress in high, moderate, and low levels.

Question 2)

print(zelazo)

active <-c(9.00,9.50,9.75,10.00,13.00) 

passive <-c(11.00,10.00,10.00,11.75,10.50)

none<-c(11.50,12.00,9.00,11.50,13.25)

ctr.8w <-c(13.25,11.50,12.00,13.50,11.50) Fitted data by making all vectors length = 5

df <- data.frame(active, passive, none, ctr.8w)

null hypothesis: No variation between groups

alternative hypothesis: Variation between groups

t.test(active,none)

The P-value is .2595, thus we do not reject the null hypothesis. We conclude that there is no significant difference between trained babies or un-trained babies.

t.test(active, ctr.8w)

The P-value is .04075, thus we reject the null hypothesis. We conclude that there is a significant difference between the trained babies and those tested only at 8 weeks. Stage fright?

t.test(active, passive)

The P-value is .6278, thus we do not reject the null hypothesis. We conclude that there is no significant difference between trained babies and passively trained babies. All you need in life is a can-do attitude.

For ANOVA testing We replicate what we did above for the stress analysis.

groups2<- rep(c("active", "passive", "none", "ctr.8w"), each = 5)

babydata<-c(c(9.00,9.50,9.75,10.00,13.00),c(11.00,10.00,10.00,11.75,10.50),c(11.50,12.00,9.00,11.50,13.25),c(13.25,11.50,12.00,13.50,11.50))

df2<- data.frame(groups2,babydata)

fit2 <- aov(babydata ~ groups2, data=df2)

summary(fit2)

p-value is .0796, since it is greater than .05 there is no sufficient evidence that the group means differ from each other.

For further analysis, I will do a post hoc test to see if any of the group comparisons differ.

TukeyHSD(fit2)

                            diff        lwr       upr     p adj

ctr.8w-active   2.1 -0.1821005 4.3821005 0.0769086

none-active     1.2 -1.0821005 3.4821005 0.4578266

passive-active  0.4 -1.8821005 2.6821005 0.9575505

none-ctr.8w    -0.9 -3.1821005 1.3821005 0.6780327

passive-ctr.8w -1.7 -3.9821005 0.5821005 0.1854081

passive-none   -0.8 -3.0821005 1.4821005 0.7499347

All adj P-values are greater than .05, thus we accept the null hypothesis for all comparisons. No significant variation between groups.