Main goal of analysis
To test 3 different hypotheses about the data.
HW R 2_Despot
Dragana Despot
2023-01-12
Data
Import the data
data <- read.csv("./Fish.csv")
head(data)## Species Weight Length1 Length2 Length3 Height Width
## 1 Bream 242 23.2 25.4 30.0 11.5200 4.0200
## 2 Bream 290 24.0 26.3 31.2 12.4800 4.3056
## 3 Bream 340 23.9 26.5 31.1 12.3778 4.6961
## 4 Bream 363 26.3 29.0 33.5 12.7300 4.4555
## 5 Bream 430 26.5 29.0 34.0 12.4440 5.1340
## 6 Bream 450 26.8 29.7 34.7 13.6024 4.9274Explanations
unit of observation: 1 fish
sample size: 159
variables
– Species: species name of fish
– Weight: weight of fish in grams
– Length: there are three variables of length of the fish in centimeters. The source is explaining them as Length1=vertical, Length2=diagonal and Length3=cross length. Since these differences are unclear (I assume some are including the fins and others aren’t), and they are not detrimental to the analysis, it was decided to remove 2 of them. I will be keeping the middle length.
Cleaning the data
Removing columns and renaming variables
mydata <- data[c(-3, -5)]colnames(mydata) <- c("Species", "Weight", "Length", "Height", "Width")
mydata$Species <- factor (mydata$Species)
head(mydata)## Species Weight Length Height Width
## 1 Bream 242 25.4 11.5200 4.0200
## 2 Bream 290 26.3 12.4800 4.3056
## 3 Bream 340 26.5 12.3778 4.6961
## 4 Bream 363 29.0 12.7300 4.4555
## 5 Bream 430 29.0 12.4440 5.1340
## 6 Bream 450 29.7 13.6024 4.9274Replacing zero with NA and dropping NA
mydata <- mydata %>%
replace_with_na(replace = list(Weight = 0,
Length = 0,
Height = 0,
Width = 0))
mydata <- drop_na(mydata)1. statistical hypothesis test: t-test or Wicoxon Signed Rank Test
Testing the population arithmetic mean with the chosen parameter.
1.1. Adjust the data
mydata1 <- subset(mydata [c(-2, -4, -5)],
Species == "Bream" )
head(mydata1)## Species Length
## 1 Bream 25.4
## 2 Bream 26.3
## 3 Bream 26.5
## 4 Bream 29.0
## 5 Bream 29.0
## 6 Bream 29.71.2. Assumptions for parametrical test:
Variable is numeric.
Normality (variable on the population is normally distributed)
No outliers.
First one is met, the 2nd and 3rd will be checked bellow:
1.2.1. Check normality (Histogram and Shapiro)
ggplot(mydata1, aes(x=Length)) +
geom_histogram(binwidth = 2, colour = "black", fill = "lightblue")+
ylab("Frequency")+
xlab("Length of Bream")
Ho: variable (length of Bream) is normally distributed
H1: variable (length of Bream) is NOT normally distributed
shapiro.test(mydata1$Length)##
## Shapiro-Wilk normality test
##
## data: mydata1$Length
## W = 0.97961, p-value = 0.7463p = 0.75 At p-value of 75 % we cannot reject Ho (cannot reject that length is normally distributed).
Assumption of normality is met, therefore we can do the t-test (parametric).
1.2.2. Check if there are outliers
ggplot(mydata1, aes(y = Length)) +
geom_boxplot (fill = "lightblue") +
ggtitle("Bream length") +
ylab("Length [cm]") +
theme(axis.text.x=element_blank(),
axis.ticks.x=element_blank())
There are no outliers (no points), assumption is met, we can perform the parametrical test.
1.3. T-TEST
To be able to do the t-test, we need information to compare our data with. For the purpose of this test, we will say we are comparing the average length of Bream caught this year (the given data) and the last years average length of Bream, which was 35 cm.
Ho: Arithmetic mean of length of Bream caught this year = 35.
Average length of Bream caught this year is equal to average length of Bream caught last year.
H1: Arithmetic mean of length of Bream caught this year ≠ 35.
Average length of Bream caught this year is different to average length of Bream caught last year.
t.test(mydata1$Length,
mu = 35,
alternative = "two.sided")##
## One Sample t-test
##
## data: mydata1$Length
## t = -2.8604, df = 34, p-value = 0.007183
## alternative hypothesis: true mean is not equal to 35
## 95 percent confidence interval:
## 31.76478 34.45236
## sample estimates:
## mean of x
## 33.10857p-value is 0.008 therefore we can reject Ho, meaning we can say that there is statistical difference in the average length of Bream caught this year and average length of Bream caught last year.
1.4. EFFECT SIZE
effectsize::cohens_d(mydata1$Length, mu = 35)## Cohen's d | 95% CI
## --------------------------
## -0.48 | [-0.83, -0.13]
##
## - Deviation from a difference of 35.interpret_cohens_d(0.48, rules = "sawilowsky2009")## [1] "small"
## (Rules: sawilowsky2009)Based on the sample data, we found that the average length of Bream caught this year (is 33.11 cm and) has decreased compared to average length of Bream caught last year (p = 0.008, d = 0.48 - small sized effect).
2. statistical hypothesis test: independent t-test or Wicoxon Rank Sum Test
Hypothesis about the difference between two population arithmetic means.
2.1. Adjust the data
mydata2 <- subset(mydata[c(-2, -4, -5)],
Species == "Bream" | Species == "Perch" )
head(mydata2)## Species Length
## 1 Bream 25.4
## 2 Bream 26.3
## 3 Bream 26.5
## 4 Bream 29.0
## 5 Bream 29.0
## 6 Bream 29.72.2. Assumptions for parametrical test
Variable is numeric.
The distribution of the variable is normal in both populations.
Variable has the same variance in both populations (if not: apply Welch correction)
First one is met, the 2nd and 3rd will be checked bellow:
2.2.1. Check normality (Histogram and Shapiro)
ggplot(mydata2, aes(x=Length)) +
geom_histogram(binwidth = 5, colour="black", fill="lightblue") +
facet_wrap(~Species, ncol = 1) +
ylab("Frequency")
Ho: Distribution is normal.
H1: Distribution is not normal.
mydata2 %>%
group_by(Species)%>%
shapiro_test(Length)## # A tibble: 2 × 4
## Species variable statistic p
## <fct> <chr> <dbl> <dbl>
## 1 Bream Length 0.980 0.746
## 2 Perch Length 0.938 0.00624Bream: Cannot reject Ho (p=0.747).
Perch: Reject Ho at p=0.007. Length of Perch is not normally distributed.
Assumption of normal distribution of variable in both populations is violated.The non-parametrical test will be done.
2.3. WILCOXON RANK SUM TEST
Ho: Distribuion locations are the same
H1: Distribution locations are different.
wilcox.test(mydata2$Length ~ mydata2$Species,
paired = FALSE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided")##
## Wilcoxon rank sum test
##
## data: mydata2$Length by mydata2$Species
## W = 1348, p-value = 0.002672
## alternative hypothesis: true location shift is not equal to 0We can reject Ho at p=0.003, meaning the distribution locations for the 2 fish are different.
2.4. EFFECT SIZE
effectsize(wilcox.test(mydata2$Length ~ mydata2$Species,
paired = FALSE,
correct = FALSE,
exact = FALSE,
alternative = "two.sided"))## r (rank biserial) | 95% CI
## --------------------------------
## 0.38 | [0.15, 0.56]interpret_rank_biserial(0.38)## [1] "large"
## (Rules: funder2019)Based on the sample data, we found that length of Bream and Perch differ (p=0.003). The difference in distribution location is large (r=0.38).
3. statistical hypothesis test: the population proportion
3.1. Adjust the data
mydata3 <- mydata[c(-2, -3, -4, -5)]
head(mydata3) ## Species
## 1 Bream
## 2 Bream
## 3 Bream
## 4 Bream
## 5 Bream
## 6 BreamWe take the data set with all available species of fish and will look if the Perch represent 30 % or more of the sample.
3.2. Assumptions for parametrical test:
n * π > 5 and n(1 - π) > 5
plyr::count(mydata3, "Species")## Species freq
## 1 Bream 35
## 2 Parkki 11
## 3 Perch 56
## 4 Pike 17
## 5 Roach 19
## 6 Smelt 14
## 7 Whitefish 6nrow(mydata3)## [1] 158158 * 0.3 > 5 This assumption is met. 158 * (1 - 0.3) > 5 This assumption is met.
3.3. PROP-TEST
Ho: π = 0.3
H1: π > 0.3
prop.test(x = 56,
n = 158,
p = 0.3,
correct = FALSE,
alternative = "greater")##
## 1-sample proportions test without continuity correction
##
## data: 56 out of 158, null probability 0.3
## X-squared = 2.2291, df = 1, p-value = 0.06772
## alternative hypothesis: true p is greater than 0.3
## 95 percent confidence interval:
## 0.2947674 1.0000000
## sample estimates:
## p
## 0.3544304The p-value of 0.999 is too high, therefore we cannot reject Ho, therefore we can conclude that Perch represents more than 30 % of the sample.