Test that the mean high temperature of city 1 is greater than the mean high temperature of city 2.
temp_city1 = c(83,91,89,89,94,96,91,92,90) # X
temp_city2 = c(78,82,81,77,79,81,80,81) # Y
wilcox.test(temp_city1,
temp_city2,
alternative = "g",
exact=FALSE)
Wilcoxon rank sum test with continuity correction
data: temp_city1 and temp_city2
W = 72, p-value = 0.0003033
alternative hypothesis: true location shift is greater than 0x = c(0.80, 0.83, 1.89, 1.04, 1.45, 1.38, 1.91, 1.64, 0.73, 1.46)
y = c(1.15, 0.88, 0.90, 0.74, 1.21)
wilcox.test(x, y, alternative = "g")
Wilcoxon rank sum exact test
data: x and y
W = 35, p-value = 0.1272
alternative hypothesis: true location shift is greater than 0Test if the first born twin is more aggressive than the second born twin
FBT = c(86,71,77,68,91,72,77,91,70,71,88,87) # X
SBT = c(88,77,76,64,96,72,65,90,65,80,81,72) # Y
#in R D= X-Y
wilcox.test(SBT,FBT,
paired=TRUE,
alternative="l",
exact = FALSE,
conf.int=TRUE)
Wilcoxon signed rank test with continuity correction
data: SBT and FBT
V = 24.5, p-value = 0.2382
alternative hypothesis: true location shift is less than 0
95 percent confidence interval:
-Inf 2.000018
sample estimates:
(pseudo)median
-1.823501 The cake batter is mixed until it reaches a specified level of consistency. The time required using two mixers A and B.
Find 95 confidence interval of the difference between two means.
# Time required
A=c(7.3,6.9,7.2,7.8,7.2)
B=c(7.4,6.8,6.9,6.7,7.1)
wilcox.test(A,B,alternative = "t",conf.int = TRUE,exact=FALSE)
Wilcoxon rank sum test with continuity correction
data: A and B
W = 19.5, p-value = 0.1719
alternative hypothesis: true location shift is not equal to 0
95 percent confidence interval:
-0.1999974 0.8999827
sample estimates:
difference in location
0.3000632 library(agricolae)
data(corn)
head(corn) method observation rx
1 1 83 11.0
2 1 91 23.0
3 1 94 28.5
4 1 89 17.0
5 1 89 17.0
6 1 96 31.5kruskal.test(observation~method,data=corn)
Kruskal-Wallis rank sum test
data: observation by method
Kruskal-Wallis chi-squared = 25.629, df = 3, p-value = 1.141e-05Median.test(corn$observation,corn$method)
The Median Test for corn$observation ~ corn$method
Chi Square = 17.54306 DF = 3 P.Value 0.00054637
Median = 89
Median r Min Max Q25 Q75
1 91.0 9 83 96 89.00 92.00
2 86.0 10 81 91 83.25 89.75
3 95.0 7 91 101 93.50 98.00
4 80.5 8 77 82 78.75 81.00
Post Hoc Analysis
Groups according to probability of treatment differences and alpha level.
Treatments with the same letter are not significantly different.
corn$observation groups
3 95.0 a
1 91.0 b
2 86.0 b
4 80.5 c## KW test is reject --> post hoc analysis
# install.packages(c("PMCMRplus", "PMCMR"))# install package is needed
library(PMCMRplus)
kwAllPairsConoverTest(corn$observation,corn$method,
p.adjust.method = "bon")Warning in kwAllPairsConoverTest.default(corn$observation, corn$method, : Ties
are present. Quantiles were corrected for ties.
Pairwise comparisons using Conover's all-pairs testdata: corn$observation and corn$method 1 2 3
2 0.04257 - -
3 0.02382 1.2e-05 -
4 3.9e-07 0.00058 5.3e-10
P value adjustment method: bonferronikwAllPairsConoverTest(corn$observation,
corn$method,p.adjust.method = "none")Warning in kwAllPairsConoverTest.default(corn$observation, corn$method, : Ties
are present. Quantiles were corrected for ties.
Pairwise comparisons using Conover's all-pairs test
data: corn$observation and corn$method 1 2 3
2 0.0071 - -
3 0.0040 1.9e-06 -
4 6.4e-08 9.7e-05 8.8e-11
P value adjustment method: none3 instructors compared grades they assigned over the past semester:
| Grades | I 1 | I 2 | I 3 |
|---|---|---|---|
| A | 4 | 10 | 6 |
| B | 14 | 6 | 7 |
| C | 17 | 9 | 8 |
| D | 6 | 7 | 6 |
| F | 2 | 6 | 1 |
Do some instructors tend to give higher grades than other instructors?
The data here is ordinal (A to F). We can code A to be 5 and F=1
I1= c(rep(5,4),rep(4,14),rep(3,17),rep(2,6),rep(1,2))
I2= c(rep(5,10),rep(4,6),rep(3,9),rep(2,7),rep(1,6))
I3= c(rep(5,6),rep(4,7),rep(3,8),rep(2,6),rep(1,1))
grades = c(I1,I2,I3)
Instr = c(rep("I1",43),rep("I2",38),rep("I3",28))
kruskal.test(grades,Instr)
Kruskal-Wallis rank sum test
data: grades and Instr
Kruskal-Wallis chi-squared = 0.32093, df = 2, p-value = 0.851712 homeowners are randomly selected to participate in an experiment with a plant nursery. Each homeowner was asked to select four fairly identical areas in his yard and to plant 4 different types of grasses.
At the end of the experiment, each homeowner was asked to rank the grass types in order of preference. The rank 1 was assigned to the least preferred.
The hypothesis was that there is no difference in preferences of the grass types.
library(tidyverse)
hw = (1:12) # homeowner id
grass1 = c(4,4,3,3,4,2,1,2,3.5,4,4,3.5) # preference for grass type 1
grass2 = c(3,2,1.5,1,2,2,3,4,1,1,2,1)# preference for grass type 2
grass3 = c(2,3,1.5,2,1,2,2,1,2,3,3,2)# preference for grass type 3
grass4 = c(1,1,4,4,3,4,4,3,3.5,2,1,3.5)# preference for grass type 4
df = data.frame(hw,grass1,grass2,grass3,grass4)
df1 = df %>%
pivot_longer(cols = -hw, names_to = "GrassType", values_to = "preference")
head(df1)# A tibble: 6 × 3
hw GrassType preference
<int> <chr> <dbl>
1 1 grass1 4
2 1 grass2 3
3 1 grass3 2
4 1 grass4 1
5 2 grass1 4
6 2 grass2 2friedman.test(y=df1$preference,df1$GrassType,blocks=df1$hw)
Friedman rank sum test
data: df1$preference, df1$GrassType and df1$hw
Friedman chi-squared = 8.0973, df = 3, p-value = 0.04404Results: Reject the null hypothesis. There is evidence to suggest that preference of grasses types are different among at least two homeowners.
frdcomp = frdAllPairsConoverTest(y=df1$preference,
df1$GrassType,
blocks=df1$hw,
p.adjust.method = "bon")
frdcomp
Pairwise comparisons using Conover's all-pairs test for a two-way balanced complete block designdata: y, groups and blocks grass1 grass2 grass3
grass2 0.15 - -
grass3 0.22 1.00 -
grass4 1.00 0.60 0.81
P value adjustment method: bonferroniboxplot(preference~GrassType, data= df1)
Multiple comparison with the Bonferroni correction shows no significance difference. We may conclude the significance results from the Friedman test may be a false positive.
12 MBA students are studied to measure the strength of the relationship between their GMAT and GPA.
gmat=c(710,610,640,580,545,560,610,530,560,540,570,560)
gpa=c(4,4,3.9,3.8,3.7,3.6,3.5,3.5,3.5,3.3,3.2,3.2)
cor.test(gmat,gpa,method = "s",alternative = "t")Warning in cor.test.default(gmat, gpa, method = "s", alternative = "t"): Cannot
compute exact p-value with ties
Spearman's rank correlation rho
data: gmat and gpa
S = 117.25, p-value = 0.04344
alternative hypothesis: true rho is not equal to 0
sample estimates:
rho
0.5900188 cor.test(gmat,gpa,method = "k",alternative = "t")Warning in cor.test.default(gmat, gpa, method = "k", alternative = "t"): Cannot
compute exact p-value with ties
Kendall's rank correlation tau
data: gmat and gpa
z = 1.8967, p-value = 0.05787
alternative hypothesis: true tau is not equal to 0
sample estimates:
tau
0.4390389 cor.test(gmat,gpa,method = "p",alternative = "t")
Pearson's product-moment correlation
data: gmat and gpa
t = 2.8004, df = 10, p-value = 0.01878
alternative hypothesis: true correlation is not equal to 0
95 percent confidence interval:
0.1437660 0.8959716
sample estimates:
cor
0.6629678