One-factor Analysis of Variance
Joshua Freimark
03/01/2020
Problem 1
- Analyze the data using \(\alpha=.05\). Perform multiple comparisons if necessary. Also, assess the reasonableness of the normality assumption using a normal probability plot of the residuals. Assess the reasonableness of the equal variance assumption using Levene’s test.
library(readxl)
HW5_prob1 <- read_excel("C:/Users/Joshu/OneDrive/Desktop/WSU courses/STAT511/test 2 material/HW5/HW5_prob1.xlsx")
attach(HW5_prob1)- The interaction term is not significant, therefore we do not reject \(H_{0}:\) and conclude that fetalizer and soil do not interact.
mod1 = aov(y ~ soil * fert, data = HW5_prob1)
summary(mod1)## Df Sum Sq Mean Sq F value Pr(>F)
## soil 2 141.75 70.87 4.573 0.0248 *
## fert 1 121.50 121.50 7.839 0.0118 *
## soil:fert 2 14.25 7.12 0.460 0.6387
## Residuals 18 279.00 15.50
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1- Verify that the 2 factors do not interact with an interaction plot.
interaction.plot(soil,fert,y)
Tukey’s Procedure
- The simultaneous pairwise comparisons indicate that the differences S2−S1 and S3−S2 are not significantly different from 0 (their confidence intervals include 0), and all the other pairs are significantly different (S3-S1).
soil_hsd = TukeyHSD(mod1, "soil", conf.level = .95)
soil_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = y ~ soil * fert, data = HW5_prob1)
##
## $soil
## diff lwr upr p adj
## S2-S1 4.500 -0.5239382 9.523938 0.0837066
## S3-S1 5.625 0.6010618 10.648938 0.0269159
## S3-S2 1.125 -3.8989382 6.148938 0.8368074fert_hsd = TukeyHSD(mod1, "fert", conf.level = .95)
fert_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = y ~ soil * fert, data = HW5_prob1)
##
## $fert
## diff lwr upr p adj
## FB-FA 4.5 1.12324 7.87676 0.0118426plot(soil_hsd)
The residuals plot displays normality in the residuals, so we can say the normality assumption is reasonable.
The equal variance assumption is accepted based on the random distributions of the redisual vs. fitted plot.
qqnorm(residuals(mod1))
qqline(residuals(mod1), col = "red")
plot(fitted(mod1), residuals(mod1))
detach(HW5_prob1)Problem 2
HW5_prob2 <- read_excel("C:/Users/Joshu/OneDrive/Desktop/WSU courses/STAT511/test 2 material/HW5/HW5_prob2.xlsx")
attach(HW5_prob2)
mod2 = aov(y ~ company * region, data = HW5_prob2)
summary(mod2)## Df Sum Sq Mean Sq F value Pr(>F)
## company 2 26.6 13.32 0.461 0.63445
## region 3 446.0 148.67 5.142 0.00462 **
## company:region 6 108.2 18.03 0.624 0.71013
## Residuals 36 1040.8 28.91
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1interaction.plot(company,region,y, col = 1:5)
Tukey’s Procedure
- Fail to reject \(H_{0}:\) for each pairwise comparison for the companies. There is a zero contained in each confidence interval therefore, we fail to reject \(H_{0}\).
company_hsd = TukeyHSD(mod2, "company", conf.level = .95)
company_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = y ~ company * region, data = HW5_prob2)
##
## $company
## diff lwr upr p adj
## B-A 1.82500 -2.821738 6.471738 0.6064798
## C-A 0.90625 -3.740488 5.552988 0.8826786
## C-B -0.91875 -5.565488 3.727988 0.8796366plot(company_hsd)
- WW-NE, SE-NE are significantly different from 0 and we reject \(H_{0}:\). All otherpairwise compairisons are not significantly different from 0.
region_hsd = TukeyHSD(mod2, "region", conf.level = .95)
region_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = y ~ company * region, data = HW5_prob2)
##
## $region
## diff lwr upr p adj
## NE-MW 4.125000 -1.787029 10.0370287 0.2548209
## SE-MW -4.183333 -10.095362 1.7286954 0.2436859
## WW-MW -1.900000 -7.812029 4.0120287 0.8224546
## SE-NE -8.308333 -14.220362 -2.3963046 0.0030220
## WW-NE -6.025000 -11.937029 -0.1129713 0.0444000
## WW-SE 2.283333 -3.628695 8.1953620 0.7272263plot(region_hsd)
The normal probability plot (qqplot) of the residuals displays very little departure from normaility so we will say the normality assumption is satisified
The residual vs. fitted plot displays no trend so the equal variance assumption is satisified.
qqnorm(residuals(mod2))
qqline(residuals(mod2), col = "red")
plot(fitted(mod2), residuals(mod2))
detach(HW5_prob2)Problem 3
prob3 <- read_excel("C:/Users/Joshu/OneDrive/Desktop/WSU courses/STAT511/test 2 material/HW5/prob3.xlsx")
prob3$isolation = as.factor(prob3$isolation)
attach(prob3)\(H_{0}:\) Level of reinforcement and level of isolation do not interact.
\(H_{a}:\) Level of reinforcement and level of isolation do interact.
- Although we cannot draw any definitive conclusions from the profile plot, we can see interaction between the 2 factors.
- Level is statisitcally significant in mod3, therefore \(H_{0}:\) is rejected. Level and isolation interact with eachother.
mod3 = aov(value ~ level + isolation)
summary(mod3)## Df Sum Sq Mean Sq F value Pr(>F)
## level 1 225.0 225.00 4.845 0.0351 *
## isolation 2 181.6 90.78 1.955 0.1581
## Residuals 32 1486.0 46.44
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1interaction.plot(x.factor = isolation, trace.factor = level, response = value,col=1:2, fun = mean, xlab = "Isolation time", ylab = "mean value", legend = TRUE)
Tukey’s procedure
- No interaction is displayed between the levels.
isolation_hsd = TukeyHSD(mod3, "isolation", conf.level = .95)
isolation_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = value ~ level + isolation)
##
## $isolation
## diff lwr upr p adj
## 40-20 5.333333 -1.503101 12.169767 0.1502797
## 60-20 1.500000 -5.336434 8.336434 0.8526731
## 60-40 -3.833333 -10.669767 3.003101 0.3641799level_hsd = TukeyHSD(mod3, "level", conf.level = .95)
level_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = value ~ level + isolation)
##
## $level
## diff lwr upr p adj
## v-n 5 0.3731014 9.626899 0.0350534- The normal probability plot shows observations near the best fit line which means we can assume the normality assumption is satisified.
qqnorm(residuals(mod3))
qqline(residuals(mod3), col = "red")
- The residuals display no pattern and seem to be randomly distrubted. Therefore the the equal variance assumption is satisified.
plot(fitted(mod3), residuals(mod3))
detach(prob3)Problem 4
HW5_prob4 <- read_excel("C:/Users/Joshu/OneDrive/Desktop/WSU courses/STAT511/test 2 material/HW5/HW5_prob4.xlsx")
attach(HW5_prob4)\(H_{0}: \mu_{1} = \mu_{2} = \mu_{3}\)
\(H_{0}:\) At least one of the means are different.
F-value = 103.75
P-value = 1.32e-10 ***
Therefore we reject \(H_{0}\) and conclude that there is enough evidence to say that at least one of the treatments are different.
mod4 = aov(Score ~ Method + Block)
summary(mod4)## Df Sum Sq Mean Sq F value Pr(>F)
## Method 2 1295.0 647.5 103.754 1.32e-10 ***
## Block 9 433.4 48.2 7.716 0.000132 ***
## Residuals 18 112.3 6.2
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1interaction.plot(Method,Block,Score, col = 1:9, pch = c(1:9, 0, letters))
We reject \(H_{0}\) for each of Tukey’s comparisons
Method_hsd = TukeyHSD(mod4, "Method", conf.level = .95)
Method_hsd## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = Score ~ Method + Block)
##
## $Method
## diff lwr upr p adj
## "B"-"A" 4.0 1.148709 6.851291 0.0057634
## "C"-"A" 15.5 12.648709 18.351291 0.0000000
## "C"-"B" 11.5 8.648709 14.351291 0.0000000- Both the normality assumption and equal variance are satisified for the same reasons as problem 1, 2, 3 each sataisifed the assumptions.
qqnorm(residuals(mod4))
qqline(residuals(mod4), col = "red")
plot(fitted(mod4), residuals(mod4))