Stat 111 – Final – Supplement with R Output

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) ## ## ## ## ## ## ## 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 ## Action Adventure ## 20 15 Thriller 6 favstats( ~ earnings, data = movies) ## ## 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 2 ● earnings 300 ● ● 200 ● ● ● ● 100 ● ● ● ● ● 0 ● ● ● ●● ●●● ●●● ●● ● ●● ● ● 0 50 ● ● ● ● ●● 100 ● ● 150 budget bwplot(earnings ~ genre, data = movies) ● earnings 300 ● ● 200 ● ● 100 ● ● ● ● 0 Action Adventure Thriller STAT 111 3 ## A model fit earnings_model <- lm(earnings ~ genre + budget, data = movies) summary(earnings_model) ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## ## 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) ## ## ## ## ## ## ## ## ## 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) ## ## ## ## ## 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 4 0.008 Density 0.006 0.004 0.002 ●●● ●● ● ● ● ●● ● ● ● ● ●● ●● ●● ● ●●● ● ●● ● ●● ●●● 0.000 −100 0 ● 100 200 300 resid(earnings_model) xyplot(resid(earnings_model) ~ fitted(earnings_model)) ● resid(earnings_model) 200 100 ● ● 0 ● ● ● ● ● ● ● ● ●●● ● ● ● ● ● ● ● ●● ●● ● ● ● ●●● ● ● ● ● ● ● ● −100 0 50 100 ● ● 150 ● 200 fitted(earnings_model) STAT 111 5 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$edu, levels = c("Primary school", "Lower\nmiddle school", "Upper\nmiddle school", "Technical\nor vocational", "College")) bwplot(child_care ~ edu, data = china) STAT 111 6 ● ● ● ● 150 ● ● ● ● ● ● ● ● ● child_care 100 ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● 50 ● ● ● ● ● 0 Primary school Lower middle school Upper middle school Technical or vocational densityplot(~ child_care, group = edu, data = china) College STAT 111 7 0.03 Density 0.02 0.01 ● ●● ● ● ● ● ● ● ● ● ●●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●●● ● ●● ● ●● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ● ●● ● ● ●● ●● ●● ● ● ● ●● ● ● ●● ●●● ● ● ● ● ● ● ● ●●● ● ● ● ● ● ● ● ●● ●●● ● ● ● ● ● ● ● ●● ● ● ● ● ●● ● ● 0.00 0 50 ● ● ● 100 ● ● ● ● ● 150 child_care Here’s an ANOVA table: xtable(anova(lm(china$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 ● ● ● ●