ANOVA: Blocking
Jagruti Garg
2024-08-01
- The statement of the research/ study purpose H0: Motor skills (performance score) of elederly population are equal (Condition1 = Condition2 = Condition3) H1: at least one pair of scores differ from one another
2.The type of analysis conducted, i.e. D’Agostino test, Scatterplot of residuals, Bartlett test. etc.
- Descriptive statistics: basic information of the data, i.e. age and gender of the participants.
4.The ANOVA test
- Post-hoc analysis
6.Effect size
7.Conclusions
library(readxl)
data3 <- read_excel("Lab3.xlsx")
names(data3)## [1] "Age" "Performance_score" "Condition"str(data3)## tibble [89 × 3] (S3: tbl_df/tbl/data.frame)
## $ Age : num [1:89] 1 1 1 1 1 1 1 1 1 1 ...
## $ Performance_score: num [1:89] 36 39 35 36 37 35 36 37 35 35 ...
## $ Condition : num [1:89] 1 1 1 1 1 1 1 1 1 2 ...summary(data3)## Age Performance_score Condition
## Min. :1 Min. :15.00 Min. :1.000
## 1st Qu.:1 1st Qu.:24.00 1st Qu.:1.000
## Median :2 Median :27.00 Median :2.000
## Mean :2 Mean :27.52 Mean :2.034
## 3rd Qu.:3 3rd Qu.:32.00 3rd Qu.:3.000
## Max. :3 Max. :39.00 Max. :3.000Assumptions
library(moments)
plot(density(data3$Performance_score), main = "Density Plot")
qqnorm(data3$Performance_score)
agostino.test(data3$Performance_score)##
## D'Agostino skewness test
##
## data: data3$Performance_score
## skew = -0.11171, z = -0.45976, p-value = 0.6457
## alternative hypothesis: data have a skewnessshapiro.test(data3$Performance_score)##
## Shapiro-Wilk normality test
##
## data: data3$Performance_score
## W = 0.9755, p-value = 0.09018anscombe.test(data3$Performance_score)##
## Anscombe-Glynn kurtosis test
##
## data: data3$Performance_score
## kurt = 2.2365, z = -2.0554, p-value = 0.03984
## alternative hypothesis: kurtosis is not equal to 3Residual plot (lm - linear model)
performance.lm <- lm(Performance_score ~ Condition, data = data3)
performance.lm##
## Call:
## lm(formula = Performance_score ~ Condition, data = data3)
##
## Coefficients:
## (Intercept) Condition
## 36.686 -4.509performance.res <- resid(performance.lm) #OR performance.res <- residuals(performance.lm)
performance.res## 1 2 3 4 5 6 7
## 3.8225868 6.8225868 2.8225868 3.8225868 4.8225868 2.8225868 3.8225868
## 8 9 10 11 12 13 14
## 4.8225868 2.8225868 7.3311713 6.3311713 3.3311713 4.3311713 5.3311713
## 15 16 17 18 19 20 21
## 5.3311713 6.3311713 7.3311713 6.3311713 4.3311713 6.8397558 6.8397558
## 22 23 24 25 26 27 28
## 4.8397558 4.8397558 5.8397558 5.8397558 2.8397558 3.8397558 3.8397558
## 29 30 31 32 33 34 35
## 4.8397558 1.8225868 0.8225868 2.8225868 -0.1774132 4.8225868 0.8225868
## 36 37 38 39 40 41 42
## 1.8225868 -0.1774132 -1.1774132 -0.1774132 2.3311713 0.3311713 0.3311713
## 43 44 45 46 47 48 49
## -0.6688287 -0.6688287 -1.6688287 -1.6688287 0.3311713 1.3311713 -1.6688287
## 50 51 52 53 54 55 56
## 1.8397558 -0.1602442 -0.1602442 0.8397558 1.8397558 0.8397558 -0.1602442
## 57 58 59 60 61 62 63
## -2.1602442 -1.1602442 0.8397558 0.8397558 -5.1774132 -4.1774132 -5.1774132
## 64 65 66 67 68 69 70
## -6.1774132 -7.1774132 -6.1774132 -7.1774132 -5.1774132 -4.1774132 -4.6688287
## 71 72 73 74 75 76 77
## -5.6688287 -3.6688287 -5.6688287 -3.6688287 -2.6688287 -3.6688287 -4.6688287
## 78 79 80 81 82 83 84
## -6.6688287 -7.6688287 -4.1602442 -6.1602442 -6.1602442 -5.1602442 -4.1602442
## 85 86 87 88 89
## -7.1602442 -6.1602442 -5.1602442 -4.1602442 -8.1602442plot(data3$Performance_score, performance.res, ylab = "Residual", xlab = "Condition", main = "Independence of Observation")
abline(0, 0)
bartlett.test(data3$Performance_score, data3$Condition)##
## Bartlett test of homogeneity of variances
##
## data: data3$Performance_score and data3$Condition
## Bartlett's K-squared = 0.14381, df = 2, p-value = 0.9306Variance are equal. p-value = 0.9306 failed to reject the null hypothesis
Checking variance among groups
tapply(data3$Performance_score, data3$Condition, var)## 1 2 3
## 18.36508 20.94713 20.72903Ratio of largest variance to smallest variance 20.95/18.37 = 1.1 - is way less than 3 (signifies that there is no issue with failing to reject null hypothesis)
After checking all the assumptions fulfill our criteria, now I will perform ANOVA with Blocking Design
IV - 3 conditions (predictor) categorical DV - performance score (outcome) continuous
Testing Additivity of Interactions - To check there is no interaction between predictor/IV - conditions and block - age (Interaction variable - factor(Condition)*factor(Age))
Performing the linear model:
model1 <- aov(Performance_score ~ factor(Condition)*factor(Age), data = data3)
model1## Call:
## aov(formula = Performance_score ~ factor(Condition) * factor(Age),
## data = data3)
##
## Terms:
## factor(Condition) factor(Age) factor(Condition):factor(Age)
## Sum of Squares 1199.0299 1549.6499 22.6398
## Deg. of Freedom 2 2 4
## Residuals
## Sum of Squares 152.9051
## Deg. of Freedom 80
##
## Residual standard error: 1.382502
## Estimated effects may be unbalancedsummary(model1)## Df Sum Sq Mean Sq F value Pr(>F)
## factor(Condition) 2 1199.0 599.5 313.667 <2e-16 ***
## factor(Age) 2 1549.6 774.8 405.389 <2e-16 ***
## factor(Condition):factor(Age) 4 22.6 5.7 2.961 0.0246 *
## Residuals 80 152.9 1.9
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Interaction effect
factor(Condition):factor(Age) is not sifnificant (p-value < 0.05)
model2 <- aov(Performance_score ~ factor(Condition), data = data3)
model2## Call:
## aov(formula = Performance_score ~ factor(Condition), data = data3)
##
## Terms:
## factor(Condition) Residuals
## Sum of Squares 1199.030 1725.195
## Deg. of Freedom 2 86
##
## Residual standard error: 4.478884
## Estimated effects may be unbalancedsummary(model2)## Df Sum Sq Mean Sq F value Pr(>F)
## factor(Condition) 2 1199 599.5 29.89 1.4e-10 ***
## Residuals 86 1725 20.1
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Effect of condition is significant
Other Models for comparison
model3 <- aov(Performance_score ~ factor(Age), data = data3)
model3## Call:
## aov(formula = Performance_score ~ factor(Age), data = data3)
##
## Terms:
## factor(Age) Residuals
## Sum of Squares 1549.733 1374.492
## Deg. of Freedom 2 86
##
## Residual standard error: 3.997807
## Estimated effects may be unbalancedsummary(model3)## Df Sum Sq Mean Sq F value Pr(>F)
## factor(Age) 2 1550 774.9 48.48 7.97e-15 ***
## Residuals 86 1374 16.0
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1model4 <- aov(Performance_score ~ factor(Condition) + factor(Age), data = data3)
model4## Call:
## aov(formula = Performance_score ~ factor(Condition) + factor(Age),
## data = data3)
##
## Terms:
## factor(Condition) factor(Age) Residuals
## Sum of Squares 1199.0299 1549.6499 175.5449
## Deg. of Freedom 2 2 84
##
## Residual standard error: 1.445621
## Estimated effects may be unbalancedsummary(model4)## Df Sum Sq Mean Sq F value Pr(>F)
## factor(Condition) 2 1199.0 599.5 286.9 <2e-16 ***
## factor(Age) 2 1549.6 774.8 370.8 <2e-16 ***
## Residuals 84 175.5 2.1
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Age can not be considered a block as there is significant Interaction effect.
Analysis of Variance Table
anova(model1, model2)## Analysis of Variance Table
##
## Model 1: Performance_score ~ factor(Condition) * factor(Age)
## Model 2: Performance_score ~ factor(Condition)
## Res.Df RSS Df Sum of Sq F Pr(>F)
## 1 80 152.91
## 2 86 1725.19 -6 -1572.3 137.1 < 2.2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1To find out what group differs from the others
library(pgirmess)## Warning: package 'pgirmess' was built under R version 4.3.2Pairwise comparisons using t tests - Bonferroni Method
pairwise.t.test(data3$Performance_score, data3$Condition, paired = FALSE, p.adjust.method = "bonferroni")##
## Pairwise comparisons using t tests with pooled SD
##
## data: data3$Performance_score and data3$Condition
##
## 1 2
## 2 0.0017 -
## 3 6e-11 0.0002
##
## P value adjustment method: bonferroniSignificant difference between condition 1, 2, and 3 noted.
Kurskal-Wallis:
kruskalmc(Performance_score ~ factor(Condition), data = data3)## Multiple comparison test after Kruskal-Wallis
## alpha: 0.05
## Comparisons
## obs.dif critical.dif stat.signif
## 1-2 19.60357 16.25251 TRUE
## 1-3 39.31970 16.12547 TRUE
## 2-3 19.71613 15.84052 TRUETukey’s Test:
TukeyHSD(model1)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = Performance_score ~ factor(Condition) * factor(Age), data = data3)
##
## $`factor(Condition)`
## diff lwr upr p adj
## 2-1 -4.204762 -5.072310 -3.337214 0
## 3-1 -9.006912 -9.867679 -8.146146 0
## 3-2 -4.802151 -5.647707 -3.956594 0
##
## $`factor(Age)`
## diff lwr upr p adj
## 2-1 -4.516166 -5.369099 -3.663233 0
## 3-1 -10.310345 -11.177377 -9.443313 0
## 3-2 -5.794179 -6.647112 -4.941246 0
##
## $`factor(Condition):factor(Age)`
## diff lwr upr p adj
## 2:1-1:1 -2.9222222 -4.947474 -0.8969700 0.0005097
## 3:1-1:1 -8.0222222 -10.047474 -5.9969700 0.0000000
## 1:2-1:1 -2.9222222 -4.947474 -0.8969700 0.0005097
## 2:2-1:1 -8.7222222 -10.747474 -6.6969700 0.0000000
## 3:2-1:1 -12.7676768 -14.748843 -10.7865103 0.0000000
## 1:3-1:1 -9.6666667 -11.744532 -7.5888017 0.0000000
## 2:3-1:1 -13.4222222 -15.447474 -11.3969700 0.0000000
## 3:3-1:1 -18.7222222 -20.747474 -16.6969700 0.0000000
## 3:1-2:1 -5.1000000 -7.071236 -3.1287642 0.0000000
## 1:2-2:1 0.0000000 -1.971236 1.9712358 1.0000000
## 2:2-2:1 -5.8000000 -7.771236 -3.8287642 0.0000000
## 3:2-2:1 -9.8454545 -11.771369 -7.9195406 0.0000000
## 1:3-2:1 -6.7444444 -8.769697 -4.7191922 0.0000000
## 2:3-2:1 -10.5000000 -12.471236 -8.5287642 0.0000000
## 3:3-2:1 -15.8000000 -17.771236 -13.8287642 0.0000000
## 1:2-3:1 5.1000000 3.128764 7.0712358 0.0000000
## 2:2-3:1 -0.7000000 -2.671236 1.2712358 0.9674422
## 3:2-3:1 -4.7454545 -6.671369 -2.8195406 0.0000000
## 1:3-3:1 -1.6444444 -3.669697 0.3808078 0.2078478
## 2:3-3:1 -5.4000000 -7.371236 -3.4287642 0.0000000
## 3:3-3:1 -10.7000000 -12.671236 -8.7287642 0.0000000
## 2:2-1:2 -5.8000000 -7.771236 -3.8287642 0.0000000
## 3:2-1:2 -9.8454545 -11.771369 -7.9195406 0.0000000
## 1:3-1:2 -6.7444444 -8.769697 -4.7191922 0.0000000
## 2:3-1:2 -10.5000000 -12.471236 -8.5287642 0.0000000
## 3:3-1:2 -15.8000000 -17.771236 -13.8287642 0.0000000
## 3:2-2:2 -4.0454545 -5.971369 -2.1195406 0.0000001
## 1:3-2:2 -0.9444444 -2.969697 1.0808078 0.8583631
## 2:3-2:2 -4.7000000 -6.671236 -2.7287642 0.0000000
## 3:3-2:2 -10.0000000 -11.971236 -8.0287642 0.0000000
## 1:3-3:2 3.1010101 1.119844 5.0821766 0.0001161
## 2:3-3:2 -0.6545455 -2.580459 1.2713685 0.9750171
## 3:3-3:2 -5.9545455 -7.880459 -4.0286315 0.0000000
## 2:3-1:3 -3.7555556 -5.780808 -1.7303033 0.0000028
## 3:3-1:3 -9.0555556 -11.080808 -7.0303033 0.0000000
## 3:3-2:3 -5.3000000 -7.271236 -3.3287642 0.0000000Relevant stats for each factor level (only relevant output pasted below):
library(pastecs)## Warning: package 'pastecs' was built under R version 4.3.2library(compute.es)
by(data3$Performance_score, data3$Condition, stat.desc)## data3$Condition: 1
## nbr.val nbr.null nbr.na min max range
## 28.0000000 0.0000000 0.0000000 25.0000000 39.0000000 14.0000000
## sum median mean SE.mean CI.mean.0.95 var
## 898.0000000 33.0000000 32.0714286 0.8098739 1.6617239 18.3650794
## std.dev coef.var
## 4.2854497 0.1336220
## ------------------------------------------------------------
## data3$Condition: 2
## nbr.val nbr.null nbr.na min max range
## 30.0000000 0.0000000 0.0000000 20.0000000 35.0000000 15.0000000
## sum median mean SE.mean CI.mean.0.95 var
## 836.0000000 27.5000000 27.8666667 0.8356061 1.7090064 20.9471264
## std.dev coef.var
## 4.5768031 0.1642393
## ------------------------------------------------------------
## data3$Condition: 3
## nbr.val nbr.null nbr.na min max range
## 31.0000000 0.0000000 0.0000000 15.0000000 30.0000000 15.0000000
## sum median mean SE.mean CI.mean.0.95 var
## 715.0000000 24.0000000 23.0645161 0.8177276 1.6700226 20.7290323
## std.dev coef.var
## 4.5529147 0.1973991n - sample size, M - mean, SD - standard deviation Condition 1 (n = 28, M = 32.07 , SD = 4.3) Condition 2 (n = 30, M = 27.87, SD = 4.6) Condition 3 (n = 31, M = 23.06, SD = 4.6)
Option 1 will compare Condition 1 and 2 ~ Performace_score vs Condition
mes(27.87, 32.07, 4.6, 4.3, 30, 28)## Mean Differences ES:
##
## d [ 95 %CI] = -0.94 [ -1.48 , -0.4 ]
## var(d) = 0.08
## p-value(d) = 0
## U3(d) = 17.31 %
## CLES(d) = 25.26 %
## Cliff's Delta = -0.49
##
## g [ 95 %CI] = -0.93 [ -1.46 , -0.39 ]
## var(g) = 0.07
## p-value(g) = 0
## U3(g) = 17.63 %
## CLES(g) = 25.55 %
##
## Correlation ES:
##
## r [ 95 %CI] = -0.43 [ -0.62 , -0.2 ]
## var(r) = 0.01
## p-value(r) = 0
##
## z [ 95 %CI] = -0.46 [ -0.73 , -0.2 ]
## var(z) = 0.02
## p-value(z) = 0
##
## Odds Ratio ES:
##
## OR [ 95 %CI] = 0.18 [ 0.07 , 0.48 ]
## p-value(OR) = 0
##
## Log OR [ 95 %CI] = -1.71 [ -2.69 , -0.72 ]
## var(lOR) = 0.25
## p-value(Log OR) = 0
##
## Other:
##
## NNT = -6.14
## Total N = 58Option 2 will compare Condition 2 and 3 ~ Performace_score vs Condition
mes(23.06, 27.87, 4.6, 4.6, 31, 30)## Mean Differences ES:
##
## d [ 95 %CI] = -1.05 [ -1.58 , -0.51 ]
## var(d) = 0.07
## p-value(d) = 0
## U3(d) = 14.79 %
## CLES(d) = 22.98 %
## Cliff's Delta = -0.54
##
## g [ 95 %CI] = -1.03 [ -1.56 , -0.5 ]
## var(g) = 0.07
## p-value(g) = 0
## U3(g) = 15.1 %
## CLES(g) = 23.27 %
##
## Correlation ES:
##
## r [ 95 %CI] = -0.47 [ -0.64 , -0.25 ]
## var(r) = 0.01
## p-value(r) = 0
##
## z [ 95 %CI] = -0.51 [ -0.77 , -0.25 ]
## var(z) = 0.02
## p-value(z) = 0
##
## Odds Ratio ES:
##
## OR [ 95 %CI] = 0.15 [ 0.06 , 0.4 ]
## p-value(OR) = 0
##
## Log OR [ 95 %CI] = -1.9 [ -2.87 , -0.93 ]
## var(lOR) = 0.25
## p-value(Log OR) = 0
##
## Other:
##
## NNT = -5.87
## Total N = 61Option 3 will compare Condition 1 and 3 ~ Performace_score vs Condition
mes(23.06, 32.07, 4.6, 4.3, 31, 28)## Mean Differences ES:
##
## d [ 95 %CI] = -2.02 [ -2.65 , -1.39 ]
## var(d) = 0.1
## p-value(d) = 0
## U3(d) = 2.17 %
## CLES(d) = 7.66 %
## Cliff's Delta = -0.85
##
## g [ 95 %CI] = -1.99 [ -2.61 , -1.37 ]
## var(g) = 0.1
## p-value(g) = 0
## U3(g) = 2.31 %
## CLES(g) = 7.93 %
##
## Correlation ES:
##
## r [ 95 %CI] = -0.72 [ -0.82 , -0.56 ]
## var(r) = 0
## p-value(r) = 0
##
## z [ 95 %CI] = -0.9 [ -1.16 , -0.64 ]
## var(z) = 0.02
## p-value(z) = 0
##
## Odds Ratio ES:
##
## OR [ 95 %CI] = 0.03 [ 0.01 , 0.08 ]
## p-value(OR) = 0
##
## Log OR [ 95 %CI] = -3.66 [ -4.8 , -2.53 ]
## var(lOR) = 0.34
## p-value(Log OR) = 0
##
## Other:
##
## NNT = -5.05
## Total N = 59Effect size (more than or equal to) < = 0.1 is small, 0.25 is medium, 0.4 is large (Cohen, 1988)
Summary of ANOVA:
Observations from the study were analyzed by conducting a one-way analysis of variance using R version 3.6.1. First, all assumptions are met and there is no adjustment made. Results suggest that the task conditions (predictor) has a significant effect on performace score (outcome) (F(2, 86) = 29.89, p < .001). [We can not consider age group as blocks because it also has a significant effect on outcomes F(1, 87) = 60.21, p < .001.] Continuing the discussion with specifically which task condition produced the signiificantaly differed measures of the performance score, a Tukey’s hoc test was established. The result suggested that there is a significant difference between task condition 1 and 2, condition 2 and 3, and condition 1 and 3 (All p-value < 0.001) in terms of the performance score. The effect were large, Cohen’ D = 0.95, 1.05, and 2.02.