Stat302_Assignment1_Solutions
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Stat 302 - Assignment 1 Solutions - Spring 2024 - (44 Points)
¶
In [1]:
### Call R package Stat2Data
#install.packages("Stat2Data") - For those who haven't installed the package
library(Stat2Data) # (1 point)
In [2]:
# Import "LongJumpOlympics2016" data
data("LongJumpOlympics2016") # (1 point)
# dimensions of the dataset
dim(LongJumpOlympics2016)
1. 28
2. 2
In [3]:
# First six rows of the dataset
head(LongJumpOlympics2016)
A data.frame: 6 × 2
Year
Gold
<int> <dbl>
1
1900
7.185
2
1904
7.340
3
1906
7.200
4
1908
7.480
5
1912
7.600
6
1920
7.150
In [4]:
# Regression output for predicting the "Olympic long jump length" (Response) from "Year" (explanatory variable)
reg_model1 <- lm(Gold~Year, data=LongJumpOlympics2016) # (1 point)
summary(reg_model1)
Call:
lm(formula = Gold ~ Year, data = LongJumpOlympics2016)
Residuals:
Min 1Q Median 3Q Max -0.39610 -0.15495 -0.00137 0.11606 0.75349 Coefficients:
Estimate Std. Error t value Pr(>|t|) (Intercept) -16.470194 2.666282 -6.177 1.56e-06 ***
Year 0.012508 0.001361 9.191 1.19e-09 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.2595 on 26 degrees of freedom
Multiple R-squared: 0.7646, Adjusted R-squared: 0.7556 F-statistic: 84.47 on 1 and 26 DF, p-value: 1.192e-09
Q1: 1.4 - (1 point)
¶
Slope of the Least Squares regression line:
¶
^β1=0.012508β1^=0.012508
Q2: 1.6 - (1 point)
¶
Intercept of the Least Squares regression line:
¶
^β0=−16.470194β0^=−16.470194
Q3: 1.8 - (1 point)
¶
Interpret the slope coefficient: Every year, the length of the gold-medal winning long jump distance has increased on average by 0.012508 meters.
¶
Q4: 1.10 - (1 point)
¶
Size of the typical error:
¶
^σ
=S=√
∈
SSEn−2=0.259522σ
^=S=SSEn−2=0.259522
∈
Q5: 1.12 - (1 point)
¶
Degrees of freedom of the regression standard error:
¶
28−2=2628−2=26
Q6: 1.14 - (2 points)
¶
^Y=78−0.5XY^=78−0.5X
^Y1=78−0.5(30)=63Y1^=78−0.5(30)=63
Residual: Y1−^Y1=60−63=−3Y1−Y1^=60−63=−3
Q7: 1.28 - (20 points)
¶
In [5]:
# Import datafile
data("SeaIce")
head(SeaIce)
A data.frame: 6 × 4
Year Extent Area
t
<int> <dbl> <dbl> <int>
1
1979
7.22
4.54
1
2
1980
7.86
4.83
2
3
1981
7.25
4.38
3
4
1982
7.45
4.38
4
5
1983
7.54
4.64
5
Year Extent Area
t
<int> <dbl> <dbl> <int>
6
1984
7.11
4.04
6
Part A - (3 points)
¶
In [6]:
# Scatterplot - 2 points
plot(SeaIce$t, SeaIce$Extent, pch=16, cex=1.2, col='blue', xlab='t', ylab='Extent', main='SeaIce Extent over time')
Pattern:Pattern: There is a strong, negative non-linear association between SeaIce Extent and time. (1 point)
Note to the marker: give full marks even if a student state that there is a strong, negative linear association since the non-
linearity is not clearly evident.
¶
Part B - (3 points)
¶
In [7]:
# Regression model for predicting Extent on time
reg_model2 <- lm(Extent~t, data=SeaIce)
# Residual vs. Fit plot
plot(reg_model2, 1)
Comment:Comment: The residuals versus fits graph too shows a curvature. In fact, it is somewhat easier to see in this plot. Given this amount of curvature, we should not fit a linear model to this data.
Note to the marker: give full marks even if a student state that liearity is statisfied since residuals are randomly distributed with no depatures from non-linearity. Furthermore, constant variance is also statisfied except for the point at the top of the plot.
¶
Part C - (3 points)
¶
In [8]:
# Scatterplot - 2 points
plot(SeaIce$t, SeaIce$Extent^2, pch=16, cex=1.2, col='blue', xlab='t', ylab='Extent^2', main='SeaIce Extent over time')
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