exercise0

#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

Question 1 a) Construct a histogram of this variable using the hist command in R.

hist(studentdata$Dvds, main="Histogram of DVDs Owned", xlab="Number of DVDs", ylab="Frequency", col="yellow", border="black")

2. Summarize this variable using the summary command in R.

summary(studentdata$Dvds)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.    NA's 
   0.00   10.00   20.00   30.93   30.00 1000.00      16 

3. 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(table(Dvds),col=‘red’) one will see that particular response values are very popular. Is there any explanation for these popular values for the number of DVDs owned?

# Construct a frequency table for Dvds
dvd_table <- table(studentdata$Dvds)

# Print the frequency table
print(dvd_table)

   0    1    2  2.5    3    4    5    6    7    8    9   10   11   12   13   14 
  26   10   13    1   18    9   27   14   12   12    7   78    3   20    7    4 
  15   16   17 17.5   18   20   21   22 22.5   23   24   25 27.5   28   29   30 
  46    1    3    1    4   83    3    3    1    3    2   31    3    1    1   45 
  31   33   35   36   37   40   41   42   45   46   48   50   52   53   55   60 
   1    1   12    4    1   26    1    1    5    1    2   26    1    2    1    7 
  62   65   67   70   73   75   80   83   85   90   97  100  120  122  130  137 
   1    2    1    4    1    3    4    1    1    1    1   10    2    1    2    1 
 150  152  157  175  200  250  500  900 1000 
   6    1    1    1    8    1    1    1    1 

# Create a barplot of the frequency table
barplot(dvd_table, col='yellow', main='Frequency of DVDs Owneds', xlab='Number of DVDs', ylab='Frequency')

1C Income might be a reason why people own different amounts of dvds. Maybe some people dont have space to have a large collection.

Question 2 a) Construct parallel boxplots of the heights using the Gender variable. Hint: boxplot(Height~Gender)

boxplot(Height ~ Gender, data = studentdata, main = "Boxplot of Heights by Gender", xlab = "Gender", ylab = "Height (inches)", col = c("yellow", "blue"))

2. If one assigns the boxplot output to a variable output=boxplot(Height~Gender) then output is a list that contains statistics used in constructing the boxplots. Print output to see the statistics that are stored.

output <- boxplot(Height ~ Gender, data = studentdata, main = "Boxplot of Heights by Gender", xlab = "Gender", ylab = "Height (inches)", col = c("blue", "red"))

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"  

3. On average, how much taller are male students than female students? 3

# Calculate the mean height for each gender
mean_height_male <- mean(Height[Gender == "male"], na.rm = TRUE)
mean_height_female <- mean(Height[Gender == "female"], na.rm = TRUE)

# Calculate the difference between the mean heights
difference <- mean_height_male - mean_height_female

# Print the results
mean_height_male

[1] 70.50767

mean_height_female

[1] 64.75701

difference

[1] 5.750657

On average male students are 5.750657 taller.

Question 3 a) Construct a scatterplot of ToSleep and WakeUp.

plot(ToSleep, WakeUp, 
     xlab = "Time to Sleep", 
     ylab = "Time to WakeUp", 
     main = "Scatterplot of ToSleep and WakeUp",
     pch = 19, # plot points as solid circles
     col = "blue") # color of the points

2. 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) fit = lm(WakeUp~ToSleep) summary(fit) abline(fit, col=‘blue’, lwd=2)

# Construct a scatterplot of ToSleep and WakeUp
plot(ToSleep, WakeUp, 
     xlab = "Time to Sleep", 
     ylab = "Time to WakeUp", 
     main = "Scatterplot of ToSleep and WakeUp with Least-Squares Fit",
     pch = 19, # plot points as solid circles
     col = "red") # color of the points

# Find a least-squares fit to the data
fit = lm(WakeUp ~ ToSleep)

# Display the summary of the fit
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

# Add the least-squares fit line to the scatterplot
abline(fit, col = "blue", lwd = 2) # Use blue color for the line and set line width to 2