Question

1. You and your three suitemates are trying to decide if your four computers are equally awesome or not. After much debate, you decide the best test is to have each person's computer load 3 different applications and time how long the loading takes (in seconds). Below are the results. Roomate A B C D 13 10 20 22 17 18 MS Word 12 Fortnite Chrome a. Explain why blocking is appropriate, define parameters, and write hypotheses for the issue you want to resolve. b. Calculate all parts of the RBD printout (except P-values) using only a simple calculator. Then, write some R code to generate the RBD printout and check your answer. Finally, make a decision about the hypotheses.

Answer 1

a. Here we consider " Application" as a blocking factor. Since loading time depends on the Application so we need to incorporate blocking factor along with treatment (i.e. Roommate) Let Ti-effect due to computer used by ith roommate, i-A, B,C,D 8,-effect due to ith applicationi-MS Word, Fortnite, Chrome Null hypothesis. Hol : TA= TB = TC = TD vs. Alternative hypothesis, H1 At least one T, is dif ferent from other s. Null hypothesis, Ho2 BMS Word- Fortnite BChrome vs. Alternative hypothesis, Hn: At least one B is different from others.

2. yio-(12 + 13 + 10 + 8)/4-10.75. У20-(20 + 22 + 17-18)/4-1925, у01 (12 + 20 + 4) /3-12. У02-(13 + 22 + 3)/3-1 2.6667, У03-(10 + 17 + 5)/3-10.6667. У04-(8 + 18 + 1)/3-9. у-(12 +134 10 +8 + 20 +22 + 17 + 18+4+3+5+1)/12: 11.0833 SStotal (122+132+102+82+202222172 +182 +42+32+5212) 12 (11.0833)2550.9255 with df 12 111 SSComputer (12212.6667210.66672 +92) 12 11.0833)2 -23.59687 with df 4-1 3 SSApplication 4 (10.752+19.252+3.252) 12 (11.0833)2 512.6755 with df 3-12 SSerrorSStotal -SSComputer SSApplication 550.9255- 23.59687-512.6755 14.65313 with df (3-)(4 1) 6

MSComputer SSComputer/3 - 23.59687/3 -7.8656 MS,Application-512.6755/2 = 256.3378 M Serror SSerror/614.65313/6-2.4422 F - ratio for Computer MSComputer/MSer 7.8656/2.4422- 3.2207 F-ratio for Application = MS Application/ M Serror = 2563378/2.4422 104.9618

R code:

M=c(12,13,10,8,20,22,17,18,4,3,5,1)
f = c("A", "B", "C", "D") # treatment levels
k = 4 # number of treatment levels
n = 3 # number of control blocks
treatment = gl(k, 1, n*k, factor(f)) # matching treatment
block = gl(n, k, k*n)# blocking factor
av = aov(M ~ treatment + block)
summary(av)

Output:

Df Sum Sq Mean Sq F value Pr(>F)
treatment 3 23.6 7.86 3.216 0.104
block 2 512.7 256.33 104.864 2.15e-05 ***
Residuals 6 14.7 2.44   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Hence we fail to reject H₀₁ whereas we reject H₀₂ i.e. 4 computers are equally awesome but effect of at least one application is significantly different.