Excersice 0
Sofia Hernandez
2026-01-24
Install LearnBayes package in R/Rstudio and then access studentdata
library(LearnBayes)
data(studentdata)
attach(studentdata)
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,
main = "Histogram of Student Owned DVDs",
xlab = "# of DVDs",
col = "lightblue",
border = "white")
- Summarize this variable using the summary command in R
# Summary of student owned 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),
main = "Barplot of DVDs Owned by Students",
col='red',
xlab = "# of DVDs",
ylab = "Frequency") 
We observe from the barplot of Dvds (name of movie dvds owned) that the popular response values are 10 and 20. This is most likely due to respondents rounding dvd counts providing an estimate, not an exact count.
\(\color{red}{\text{Q2}}\) 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)
# Boxplot of Heights by Gender
boxplot(Height~Gender,
main = "Distribution of Heights by Gender",
col = c("lightpink", "lightblue"))
- 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 output var
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?
# Aggregate()
group_means <- aggregate(Height~Gender, data = studentdata, FUN = mean)
print(group_means)## Gender Height
## 1 female 64.75701
## 2 male 70.50767#Mean height difference between males and females
mean_diff <- group_means[2,2] - group_means[1,2]
print(mean_diff)## [1] 5.750657On average, the height of male students is 5.750657 inches taller than female students.
\(\color{red}{\text{Q3}}\) 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.
plot(ToSleep, WakeUp,
xlab = "Time to Sleep (neg = hours before midnight)",
ylab = "Wake-up Time (hours past midnight)",
main = "Scatterplot of Bedtime and Wake-up Times")
b. 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.
plot(ToSleep, WakeUp,
xlab = "Time to Sleep (neg = hours before midnight)",
ylab = "Wake-up Time (hours past midnight)",
main = "Scatterplot of Bedtime and Wake-up Times")
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-16abline(fit, col='blue', lwd=2)
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