2. An animal caretaker at an aquarium wishes to determine which kind of reward f
ID: 3317833 • Letter: 2
Question
2. An animal caretaker at an aquarium wishes to determine which kind of reward for good behavior keeps her seals most motivated. She divides the seals up into groups of 5, and in each group she uses a different kind of reward: one group receives a normal fish, another receives double the normal amount, two groups receive fish of different species (identified by their color, red and blue), and one group received no fish. The seals then have their motivation measured by a professional seal psychologist with the Talent Utilization Numerical Assessment score (a higher score indicates higher motivation). She performs ANOVA to test the hypothesis that the different treatments are different and finds a p-value of 1.07x10-7 for her F test and an MSE of 25.72. The treatment means are the following:
One Fish: 51.5
Two Fish: 60.6
Red Fish: 57.7
Blue Fish: 53.6
No Fish: 31.6
(a) Create a diagram showing which treatments are not significantly different from each other using Tukey's Method with FW = 0.05. (Note that each treatment has the same sample size (there are 5 groups of 5, or 25 total test subjects), so your least significant difference is the same for every comparison.)
(b) What would the least significant difference be if we used Fisher's LSD with = 0.05?
(c) What would the least significant difference be if we used the Bonferroni method with FW = 0.1?
Explanation / Answer
there are total 25 subject , so total df=25-1=24 and
there are 5 gropus, group df=5-1=4
error df=24-4=20
at 5% level of significance i.e. alpha=0.05, critical HSD Q(0.05,5,20)=4.23
HSD statistic q=(difference of mean)/sqrt(MSE/n)
pairs 1&2, 1&3, 1&4, 2&3, 2&4, and 3&4 are not significant
(b) Fisher LSD=sqrt(2*mse/r)*t(alpha,error df)=sqrt(2*25.72/5*t(0.05,20)= sqrt(2*25.72/5*2.09=6.7
(c) bonferroni method fishter LSD=sqrt(2*mse/r)*t(alpha/10,error df)= =sqrt(2*25.72/5*t(0.05/10,20)= sqrt(2*25.72/5*3.15=10.1
here there are 5C2=10 pair of comparison, so level of signifiance will be=alpha/10
mean group 1 One Fish 51.5 group 2 Two Fish 60.6 group 3 Red Fish 57.7 group 4 Blue Fish 53.6 group 5 No Fish 31.6 1&2 1&3 1&4 1&5 2&3 2&4 2&5 3&4 3&5 4&5 difference of mean -9.1 -6.2 -2.1 19.9 2.9 7 29 4.1 26.1 22 sqrt(MSE/r)=sqrt(25.72/5)=2.268 -4.0123 -2.7337 -0.9259 8.7743 1.2787 3.0864 12.7866 1.8078 11.5079 9.7002Related Questions
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