Excersice 0
Install LearnBayes package in R/Rstudio and then access studentdata
#Install the LearnBayes package
#Keep in mind that R is case-sensitive
#install.packages('LearnBayes')
#You just need to install once and then you can directly use
#so long as you access the LearnBayes package
library(LearnBayes)
#Access studentdata from the LearnBayes package
data(studentdata)
attach(studentdata)
#show part of data
head(studentdata)## Student Height Gender Shoes Number Dvds ToSleep WakeUp Haircut Job Drink
## 1 1 67 female 10 5 10 -2.5 5.5 60 30.0 water
## 2 2 64 female 20 7 5 1.5 8.0 0 20.0 pop
## 3 3 61 female 12 2 6 -1.5 7.5 48 0.0 milk
## 4 4 61 female 3 6 40 2.0 8.5 10 0.0 water
## 5 5 70 male 4 5 6 0.0 9.0 15 17.5 pop
## 6 6 63 female NA 3 5 1.0 8.5 25 0.0 water\(\color{red}{\text{Q1}}\) The variable Dvds in the student dataset contains the number of movie DVDs owned by students in the class
- Construct a histogram of this variable using the hist command in R
# Histogram of Dvds
hist(studentdata$Dvds, prob=T)
- Summarize this variable using the summary command in R
# Histogram of Dvds
summary(studentdata)## Student Height Gender Shoes Number
## Min. : 1 Min. :54.0 female:435 Min. : 0.00 Min. : 1.00
## 1st Qu.:165 1st Qu.:64.0 male :222 1st Qu.: 6.00 1st Qu.: 4.00
## Median :329 Median :66.0 Median : 12.00 Median : 6.00
## Mean :329 Mean :66.7 Mean : 15.42 Mean : 5.67
## 3rd Qu.:493 3rd Qu.:70.0 3rd Qu.: 20.00 3rd Qu.: 7.00
## Max. :657 Max. :84.0 Max. :164.00 Max. :10.00
## NA's :10 NA's :22 NA's :2
## Dvds ToSleep WakeUp Haircut
## Min. : 0.00 Min. :-2.500 Min. : 1.000 Min. : 0.00
## 1st Qu.: 10.00 1st Qu.: 0.000 1st Qu.: 7.500 1st Qu.: 10.00
## Median : 20.00 Median : 1.000 Median : 8.500 Median : 16.00
## Mean : 30.93 Mean : 1.001 Mean : 8.383 Mean : 25.91
## 3rd Qu.: 30.00 3rd Qu.: 2.000 3rd Qu.: 9.000 3rd Qu.: 30.00
## Max. :1000.00 Max. : 6.000 Max. :13.000 Max. :180.00
## NA's :16 NA's :3 NA's :2 NA's :20
## Job Drink
## Min. : 0.00 milk :113
## 1st Qu.: 0.00 pop :178
## Median :10.50 water:355
## Mean :11.45 NA's : 11
## 3rd Qu.:17.50
## Max. :80.00
## NA's :32- Use the table command in R to construct a frequency table of the individual values of Dvds that were observed. If one constructs a barplot of these tabled values using the command
# Barplot of Dvds
barplot(table(Dvds),col='red') 
We observe from the barplot of Dvds (name of movie dvds owned) that the popular response values are 10 and 20. Is there any explanation for these popular values for the number of DVDs owned?
A: The students likely estimated the number of DVDs they own to the nearest 10.
\(\color{red}{\text{Q2}}\) The variable Height contains the height (in inches) of each student in the class.
- Construct parallel boxplots of the heights using the Gender variable. Hint: boxplot(Height~Gender)
# Barplot of Dvds
boxplot(Height~Gender)
- If one assigns the boxplot output to a variable, then output is a list that contains statistics used in constructing the boxplots. Print output to see the statistics that are stored.
# Assign boxplot to a variable named output
output=boxplot(Height~Gender)
print(output)## $stats
## [,1] [,2]
## [1,] 57.75 65
## [2,] 63.00 69
## [3,] 64.50 71
## [4,] 67.00 72
## [5,] 73.00 76
##
## $n
## [1] 428 219
##
## $conf
## [,1] [,2]
## [1,] 64.19451 70.6797
## [2,] 64.80549 71.3203
##
## $out
## [1] 56 76 55 56 76 54 54 84 78 77 56 63 77 79 62 62 61 79 59 61 78 62
##
## $group
## [1] 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2
##
## $names
## [1] "female" "male"- On average, how much taller are male students than female students?
# Method: using aggregate()
group_means <- aggregate(Height~Gender, data = studentdata, FUN = mean)
print(group_means)## Gender Height
## 1 female 64.75701
## 2 male 70.50767#Calculate the mean difference of heights between male and female students
# Using the results from aggregate()
mean_diff <- group_means[2,2] - group_means[1,2]
print(mean_diff) # Output: 5.750657## [1] 5.750657On average, the height of male students is 5.750657 inches taller than female students.
If you want to learn about R Markdown, please refer to https://bookdown.org/yihui/rmarkdown/.
\(\color{red}{\text{Q3}}\) The variables ToSleep and WakeUp contain, respectively, the time to bed and wake-up time for each student the previous evening. (The data are recorded as hours past midnight, so a value of −2 indicates 10 p.m.)
- Construct a scatterplot of ToSleep and WakeUp.
# Scatterplot of WakeUp vs ToSleep
plot(WakeUp~ToSleep)
- Find a least-squares fit to these data using the lm command and then place the least-squares fit on the scatterplot using the abline command.
# Least-squares fit
fit = lm(WakeUp~ToSleep)
summary(fit)##
## Call:
## lm(formula = WakeUp ~ ToSleep)
##
## Residuals:
## Min 1Q Median 3Q Max
## -4.4010 -0.9628 -0.0998 0.8249 4.6125
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 7.96276 0.06180 128.85 <2e-16 ***
## ToSleep 0.42472 0.03595 11.81 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.282 on 651 degrees of freedom
## (4 observations deleted due to missingness)
## Multiple R-squared: 0.1765, Adjusted R-squared: 0.1753
## F-statistic: 139.5 on 1 and 651 DF, p-value: < 2.2e-16# Plotting the best fit
plot(WakeUp~ToSleep)
abline(fit, col='blue', lwd=2)