Question

Forty architecture students were each asked to judge 5
different building structures. The response variable of interest
is the judge's overall satisfaction (SAT), where a higher score
is better. We wish to compare the mean satisfaction rating across
the five buildings, so the factor of interest is BLDG.

USE R OR SAS TO SOLVE THE PROBLEMS. PLEASE INCLUDE YOUR CODE TO GET THE ANSWER.

I am not sure how the data is not understandable. Literally take the data plug it into r or sas to get the answers. SUBJ - this is one of the fourty testers. BLDG - this is what building model they rated. SAT - is there rating of said building.

(a) Why does it make sense to use the judge (denoted SUBJ in the
data set) as a blocking variable? Why should we treat this block
as a random effect?

(b) Analyze the data as a RBD, where SAT is the response, BLDG is
the treatment factor, and SUBJ is the block. Based on the appropriate
F-test, is there a significant difference in mean satisfaction rating
across the five buildings? NOTE: Use a 0.10 significance level.

(c) Based on the appropriate F-test, is there significant variation
among the judges? NOTE: Use a 0.10 significance level.

(d) Of particular interest to the investigators is whether the mean
satisfaction for building 1 differs significantly from the mean satisfaction
for the other four buildings. Use an ESTIMATE statement to test the
appropriate contrast here. NOTE: Use a 0.10 significance level.

data buildings;                                                                                                                           
input SUBJ BLDG SAT;                                                                 
cards;                                                                                                                                  
 1 1 2
 1 2 5
 1 3 6
 1 4 5
 1 5 7
 2 1 5
 2 2 6
 2 3 6
 2 4 7
 2 5 4
 3 1 4
 3 2 7
 3 3 3
 3 4 6
 3 5 7
 4 1 6
 4 2 4
 4 3 7
 4 4 5
 4 5 7
 5 1 2
 5 2 6
 5 3 4
 5 4 7
 5 5 5
 6 1 4
 6 2 6
 6 3 7
 6 4 5
 6 5 3
 7 1 7
 7 2 5
 7 3 5
 7 4 7
 7 5 4
 8 1 3
 8 2 7
 8 3 6
 8 4 7
 8 5 6 
 9 1 6
 9 2 7
 9 3 8
 9 4 6
 9 5 3
 10 1 5
 10 2 3
 10 3 3
 10 4 5
 10 5 6
 11 1 3
 11 2 6 
 11 3 4 
 11 4 4 
 11 5 3
 12 1 3
 12 2 6
 12 3 7
 12 4 5
 12 5 3
 13 1 4
 13 2 1
 13 3 7
 13 4 1
 13 5 6
 14 1 4
 14 2 6
 14 3 8
 14 4 5
 14 5 1
 15 1 4
 15 2 4
 15 3 4
 15 4 5
 15 5 5
 16 1 8
 16 2 5
 16 3 9
 16 4 9
 16 5 5
 17 1 5
 17 2 5
 17 3 6
 17 4 7
 17 5 5
 18 1 5
 18 2 4
 18 3 6
 18 4 6
 18 5 6
 19 1 2
 19 2 5
 19 3 6
 19 4 2
 19 5 8
 20 1 2
 20 2 8
 20 3 7
 20 4 8
 20 5 2
 21 1 8
 21 2 8
 21 3 8
 21 4 8
 21 5 3
 22 1 5
 22 2 4
 22 3 4
 22 4 3
 22 5 5
 23 1 6
 23 2 6
 23 3 6
 23 4 6
 23 5 4
 24 1 3
 24 2 5
 24 3 8
 24 4 5
 24 5 6
 25 1 6
 25 2 2
 25 3 5
 25 4 7
 25 5 6
 26 1 2
 26 2 7
 26 3 4
 26 4 7
 26 5 2
 27 1 7
 27 2 7
 27 3 7
 27 4 7
 27 5 7
 28 1 8
 28 2 5
 28 3 5
 28 4 6
 28 5 3
 29 1 2
 29 2 6
 29 3 7
 29 4 4
 29 5 5
 30 1 1
 30 2 5
 30 3 5
 30 4 6
 30 5 6
 31 1 9
 31 2 7
 31 3 8
 31 4 2
 31 5 8
 32 1 6
 32 2 9
 32 3 1
 32 4 8
 32 5 4
 33 1 2
 33 2 6
 33 3 8
 33 4 9
 33 5 8
 34 1 8
 34 2 4
 34 3 3
 34 4 3
 34 5 9
 35 1 2
 35 2 7
 35 3 2
 35 4 9
 35 5 2
 36 1 2
 36 2 9
 36 3 1
 36 4 8
 36 5 3
 37 1 7
 37 2 2
 37 3 3 
 37 4 3
 37 5 6
 38 1 3
 38 2 7
 38 3 3
 38 4 2
 38 5 2
 39 1 3
 39 2 3
 39 3 5
 39 4 3
 39 5 3
 40 1 9
 40 2 5
 40 3 8
 40 4 7
 40 5 8    
;
run; 

Answer 1

I have included all the answers in the R script below:

 # After the data is loaded into R head(buildings) # Extracting subject and building as factors SAT <- buildings[, "SAT"] SUBJ <- factor(buildings[, "SUBJ"]) BLDG <- factor(buildings[, "BLDG"]) # (a) Why does it make sense to use the judge (denoted SUBJ in the # data set) as a blocking variable? Why should we treat this block # as a random effect? # Let us ignore the blocks and just do a one-way CRD ANOVA av.CRD <- aov(SAT~BLDG) summary(av.CRD) 

 Df Sum Sq Mean Sq F value Pr(>F) BLDG 4 33.6 8.392 1.956 0.103 Residuals 195 836.7 4.291 # The p-value of the obtained F-statistic for BLDG # for the buildings is 0.103. From here, we can be tempted # conclude that there is no significant difference in the SAT # ratings of the 5 buildings. BUT THIS IS WRONG. # Why is this wrong? # Because we forgot to account for the variation in the ratings # of these 40 Judges. The inherent biases in the ratings they give # can influence the ratings of the Buildings. These are said to have # random effects. We can account for these variations by treating SUBJ as # a blocking variable. # Blocking is used to remove the effects of a few of the most important # nuisance variables. Randomization is then used to reduce the contaminating # effects of the remaining nuisance variables. For important nuisance variables, # blocking will yield higher significance in the variables of interest # than randomizing 

# b) Analyze the data as a RBD, where SAT is the response, BLDG is # the treatment factor, and SUBJ is the block. Based on the appropriate # F-test, is there a significant difference in mean satisfaction rating # across the five buildings? # Now perform the Analysis of variance # SAT = response # BLDG = treatment factor # SUBJ is the blocking factor av = aov(SAT~BLDG+SUBJ) summary(av)

 Df Sum Sq Mean Sq F value Pr(>F) BLDG 4 33.6 8.393 2.032 0.0926 . SUBJ 39 192.3 4.931 1.194 0.2236 Residuals 156 644.4 4.131 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 # Ans (b) # The p-value of the obtained F-statistic for BLDG # for the buildings is 0.0926. Since it is less # than the significance level alpha=0.1, we can # conclude that there is significant difference # in mean satisfaction rating across the five buildings # Ans (c) # The p-value of the obtained F-statistic for SUBJ # for the buildings is 0.2236. Since it is more # than the significance level alpha=0.1, we can # conclude that there is NO significant variation # among the judges

# (d) Testing whether the mean satisfaction for building 1 # differs significantly from the mean satisfaction # for the other four buildings # Choosing the appropriate contrasts (they should sum up to 0) contrasts(BLDG) <- matrix(c(1, -1/4, -1/4, -1/4, -1/4), nrow=5, ncol=1) # This will give us the the two-sided P-value of t-test # about contrast. # Here we fit the Multiple Linear Regression model # and explore the summary output summary(lm(SAT ~ BLDG + SUBJ))$coef["BLDG1", ]

 Estimate Std. Error t value Pr(>|t|) -0.64500000 0.28743561 -2.24398082 0.02624094 # Ans (d) # Since the p-value = 0.0262 is very less than the # the significance level alpha = 0.1, we can conclude # YES, the mean satisfaction for building 1 # differs significantly from the mean satisfaction # for the other four buildings

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