library(ggplot2)
Warning: package 'ggplot2' was built under R version 4.2.2
library(readxl)
Wilcoxon <- read_excel("D:/MARV BS MATH/4th year, 2nd sem/Nonparametric Statistics/Final Exam/Wilcoxon.xlsx")
Wilcoxon
# A tibble: 10 × 2
Patient Frequency
<dbl> <dbl>
1 1 9
2 2 10
3 3 8
4 4 4
5 5 8
6 6 3
7 7 0
8 8 10
9 9 15
10 10 9
To perform Wilcoxon Signed-Rank Test we have the following assumptions:
ks.test(Wilcoxon,"pnorm")
Warning in ks.test.default(Wilcoxon, "pnorm"): ties should not be present for
the Kolmogorov-Smirnov test
Asymptotic one-sample Kolmogorov-Smirnov test
data: Wilcoxon
D = 0.87725, p-value = 8.56e-14
alternative hypothesis: two-sided
Since the p-value 8.56e-14 is less than D = 0.87725, then we reject the null hypothesis and conclude that at least one value does not match the specifies distribution.
This observation is shown in the graph below.
ggplot(Wilcoxon, aes(Frequency)) +
geom_density()
ggplot(data=Wilcoxon) +
geom_histogram( aes(Frequency, ..density..) ) +
geom_density( aes(Frequency, ..density..) ) +
geom_rug( aes(Frequency) )
Warning: The dot-dot notation (`..density..`) was deprecated in ggplot2 3.4.0.
ℹ Please use `after_stat(density)` instead.
`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
By the above illustration and Kolmogorov-Smirnov test for normality, it shows that the data is skewed. Thus, we can perfom the Wilcoxon Signed-Rank Test.
Do the data support his contention that the median number of times he sees a patient is five?
summary(Wilcoxon$Frequency)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.00 5.00 8.50 7.60 9.75 15.00
library(ggpubr)
Warning: package 'ggpubr' was built under R version 4.2.2
ggboxplot(Wilcoxon$Frequency,
ylab = "Frequency", xlab = FALSE,
ggtheme = theme_minimal())
res <- wilcox.test(Wilcoxon$Frequency, mu = 5)
Warning in wilcox.test.default(Wilcoxon$Frequency, mu = 5): cannot compute exact
p-value with ties
res
Wilcoxon signed rank test with continuity correction
data: Wilcoxon$Frequency
V = 44, p-value = 0.1016
alternative hypothesis: true location is not equal to 5
res$p.value
[1] 0.1015756
The p-value of the test is 0.1015756, which is greater than the significance level alpha = 0.05. We do not reject the null hypothesis and conclude that The median number of times a physician sees each of his patients during the year is five significantly the same from a median of five with a p-value = 0.1015756.