Study Guide
Stat 111 – Final – Supplement with R Output
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University:
Amherst College -
Course:
STAT 111 | Introduction to Statistics Academic year:
2016
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Views:
90
Pages:
7
Author:
sammerito22923g0
STAT 111
1
Stat 111 – Final – Supplement with R Output
Name:
I
Word Problems
1. Here is R output relevant to word problem number 1:
library(mosaic)
## Read in the data and look at just a subset of it for this problem
movies <- read.csv("Movie_budgets.csv")[, c(1, 3, 4, 7)]
names(movies) <- c("title", "budget", "earnings", "genre")
genres_to_keep <- c("Action", "Adventure", "Thriller")
movies <- filter(movies, genre %in% genres_to_keep) %>%
mutate(genre = factor(as.character(genre, levels = genres_to_keep)))
## A first look
head(movies)
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title budget earnings
genre
1
Elektra
65.0 24.409722
Action
2 Assault on Precinct 13
30.0 20.040895
Action
3 Pooh's Heffalump Movie
20.0 18.081626 Adventure
4
Constantine
75.0 75.976178
Action
5
The Jacket
28.5 6.301131 Thriller
6
Hostage
75.0 34.636443
Action
nrow(movies)
## [1] 41
tally( ~ genre, data = movies)
## genre
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Action Adventure
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20
15
Thriller
6
favstats( ~ earnings, data = movies)
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min
Q1
median
Q3
max
mean
sd n missing
4.476235 18.78223 34.63644 74.09886 380.2706 69.1629 83.91076 41
0
xyplot(earnings ~ budget, data = movies) STAT 111
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earnings
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budget
bwplot(earnings ~ genre, data = movies)
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Action
Adventure
Thriller STAT 111
3
## A model fit
earnings_model <- lm(earnings ~ genre + budget, data = movies)
summary(earnings_model)
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Call:
lm(formula = earnings ~ genre + budget, data = movies)
Residuals:
Min
1Q
-117.469 -29.143
Median
-0.013
3Q
29.620
Max
225.128
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-23.3434
21.4441 -1.089
0.283
genreAdventure 34.3790
21.6282
1.590
0.120
genreThriller
2.5180
30.3570
0.083
0.934
budget
1.2531
0.2464
5.087 1.08e-05 ***
--Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 63.13 on 37 degrees of freedom
Multiple R-squared: 0.4764,Adjusted R-squared: 0.434
F-statistic: 11.22 on 3 and 37 DF, p-value: 2.221e-05
anova(earnings_model)
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Analysis of Variance Table
Response: earnings
Df Sum Sq Mean Sq F value
Pr(>F)
genre
2 31066
15533 3.8975
0.02911 *
budget
1 103115 103115 25.8731 1.079e-05 ***
Residuals 37 147460
3985
--Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
confint(earnings_model)
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2.5 %
97.5 %
(Intercept)
-66.7933244 20.106536
genreAdventure -9.4438126 78.201777
genreThriller -58.9911322 64.027055
budget
0.7539425 1.752274
## A plot of residuals
densityplot( ~ resid(earnings_model)) STAT 111
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0.008
Density
0.006
0.004
0.002
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resid(earnings_model)
xyplot(resid(earnings_model) ~ fitted(earnings_model))
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fitted(earnings_model) STAT 111
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2. Here is R output relevant to word problem 2:
# load data --------------------------------------------------------china <- read.csv("china.csv")
# subset and clean data --------------------------------------------china <- china[!is.na(china$gender) & !is.na(china$child_care)
& !is.na(china$edu) & china$child_care != -99 & china$edu != 9,]
china$edu[china$edu == 1] <- "Primary school"
china$edu[china$edu == 2] <- "Lower\nmiddle school"
china$edu[china$edu == 3] <- "Upper\nmiddle school"
china$edu[china$edu == 4] <- "Technical\nor vocational"
china$edu[china$edu == 5] <- "College"
china$edu <- factor(china$edu,
levels = c("Primary school", "Lower\nmiddle school",
"Upper\nmiddle school", "Technical\nor vocational",
"College"))
bwplot(child_care ~ edu, data = china) STAT 111
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Primary school
Lower
middle school
Upper
middle school
Technical
or vocational
densityplot(~ child_care, group = edu, data = china)
College STAT 111
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0.03
Density
0.02
0.01
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child_care
Here’s an ANOVA table:
xtable(anova(lm(china$child_care ~ china$edu)), digits = 2)
china$edu
Residuals
Df
4.00
794.00
Sum Sq
4142.09
653047.83
Mean Sq
1035.52
822.48
F value
1.26
Pr(>F)
0.28
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Stat 111 – Final – Supplement with R Output
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