1. Load the data:
setwd("/Users/traves/Dropbox/SM339/test1")
BT = read.csv("BaseballTimes.csv")
summary(BT)
## Game League Runs Margin Pitchers
## ARI-SD :1 AL:7 Min. : 3.0 Min. : 1.00 Min. : 4.0
## BOS-NYY:1 NL:8 1st Qu.: 5.5 1st Qu.: 1.00 1st Qu.: 5.0
## CHI-BAL:1 Median :10.0 Median : 4.00 Median : 7.0
## CHI-PIT:1 Mean :10.1 Mean : 3.87 Mean : 8.2
## CIN-HOU:1 3rd Qu.:13.0 3rd Qu.: 5.00 3rd Qu.:10.5
## CLE-DET:1 Max. :23.0 Max. :12.00 Max. :16.0
## (Other):9
## Attendance Time
## Min. :13478 Min. :133
## 1st Qu.:17734 1st Qu.:154
## Median :30395 Median :168
## Mean :29769 Mean :183
## 3rd Qu.:38102 3rd Qu.:194
## Max. :55058 Max. :317
##
attach(BT)
Get mean and median of Time:
mean(Time) #182.67 mins
## [1] 182.7
median(Time) # 168
## [1] 168
Regression Line:
TP = lm(Time ~ Pitchers)
summary(TP)
##
## Call:
## lm(formula = Time ~ Pitchers)
##
## Residuals:
## Min 1Q Median 3Q Max
## -37.94 -8.44 -3.10 9.75 50.79
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 94.84 13.39 7.08 8.2e-06 ***
## Pitchers 10.71 1.49 7.21 6.9e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21.5 on 13 degrees of freedom
## Multiple R-squared: 0.8, Adjusted R-squared: 0.784
## F-statistic: 51.9 on 1 and 13 DF, p-value: 6.88e-06
94.843 + 10.71 * 2 # = 116.263
## [1] 116.3
A game with only 2 pitchers lasts an average of 116.263 minutes and each extra pitcher increases this average time by 10.71 minutes.
Plotting:
plot(Time ~ Pitchers, pch = 19, col = "blue", xlab = "Number of Pitchers", ylab = "Length of Game (in minutes)",
main = "Length of Baseball Game versus Pitchers")
abline(TP, col = "green")
pdf(file = "plotoutput.pdf")
plot(Time ~ Pitchers, pch = 19, col = "blue", xlab = "Number of Pitchers", ylab = "Length of Game (in minutes)",
main = "Length of Baseball Game versus Pitchers")
abline(TP, col = "green")
dev.off()
## pdf
## 2
RSE:
summary(TP)
##
## Call:
## lm(formula = Time ~ Pitchers)
##
## Residuals:
## Min 1Q Median 3Q Max
## -37.94 -8.44 -3.10 9.75 50.79
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 94.84 13.39 7.08 8.2e-06 ***
## Pitchers 10.71 1.49 7.21 6.9e-06 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21.5 on 13 degrees of freedom
## Multiple R-squared: 0.8, Adjusted R-squared: 0.784
## F-statistic: 51.9 on 1 and 13 DF, p-value: 6.88e-06
# RSE = 21.46
Outliers:
plot(rstandard(TP) ~ TP$fitted, col = "blue", pch = 19)
abline(0, 0, col = "red")
# outlier at about (270,3)
which(rstandard(TP) > 2)
## 15
## 15
# Outlier is game 15
BT[15, ]
## Game League Runs Margin Pitchers Attendance Time
## 15 NYM-PHI NL 15 1 16 45204 317
rstandard(TP)[15]
## 15
## 2.956
Confidence Interval:
meantime = mean(Time)
sdtime = sd(Time)
n = length(Time)
conf = 0.95
tstat = qt(0.975, n - 1)
low = meantime - tstat * sdtime/sqrt(n)
high = meantime + tstat * sdtime/sqrt(n)
print(c(low, high))
## [1] 157.1 208.3
Alternatively:
t.test(Time, conf.level = 0.95)
##
## One Sample t-test
##
## data: Time
## t = 15.31, df = 14, p-value = 3.877e-10
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
## 157.1 208.3
## sample estimates:
## mean of x
## 182.7
We can only conclude that the confidence interval is valid for the length of games played on that day, Aug. 26, 2008. However, if that day is representative of all days on which games are played then we might be justified in making the stronger claim in part (f).
2. Load Data:
MR = read.csv("MetabolicRate.csv")
summary(MR)
## Computer BodySize LogBodySize Instar
## Min. :1.0 Min. :0.002 Min. :-2.6778 Min. :1.00
## 1st Qu.:1.0 1st Qu.:0.033 1st Qu.:-1.4815 1st Qu.:2.00
## Median :1.0 Median :0.167 Median :-0.7765 Median :3.00
## Mean :1.5 Mean :0.684 Mean :-0.7389 Mean :3.16
## 3rd Qu.:2.0 3rd Qu.:1.030 3rd Qu.: 0.0128 3rd Qu.:4.00
## Max. :3.0 Max. :4.803 Max. : 0.6815 Max. :5.00
## CO2ppm Mrate LogMrate
## Min. : 0.4 Min. : 0.03 Min. :-1.5477
## 1st Qu.: 12.8 1st Qu.: 0.94 1st Qu.:-0.0265
## Median : 57.3 Median : 4.51 Median : 0.6541
## Mean : 177.1 Mean : 14.34 Mean : 0.6294
## 3rd Qu.: 205.9 3rd Qu.: 16.49 3rd Qu.: 1.2173
## Max. :1467.3 Max. :100.57 Max. : 2.0025
attach(MR)
Plot 4 graphs:
plot(Mrate ~ BodySize, pch = 19, col = "blue") # kind of linear but a lot of spread
plot(Mrate ~ LogBodySize, pch = 19, col = "blue") # not linear
plot(LogMrate ~ BodySize, pch = 19, col = "blue") # not linear
plot(LogMrate ~ LogBodySize, pch = 19, col = "blue") # good linear fit
The strongest linear fit is in the log-log model: LogMrate ~ LogBodySize. The next strongest linear fit is in the original variables: Mrate ~ BodySize.
Regression:
MB = lm(LogMrate ~ LogBodySize)
summary(MB)
##
## Call:
## lm(formula = LogMrate ~ LogBodySize)
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.5920 -0.1119 0.0036 0.1212 0.4729
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 1.3066 0.0136 96.3 <2e-16 ***
## LogBodySize 0.9164 0.0124 74.2 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.175 on 303 degrees of freedom
## Multiple R-squared: 0.948, Adjusted R-squared: 0.948
## F-statistic: 5.51e+03 on 1 and 303 DF, p-value: <2e-16
REGRESSION LINE: LogMrate = 1.30655 + 0.91641*LogBodySize
10^(1.30655 + 0.01641 * log(base = 10, 1)) # = 20.25583
## [1] 20.26
The model predicts that the metabolic rate of a 1 gram caterpillar should be 20.25583.
Sleep = c(5.5, 5, 6, 6.5, 6, 6.5, 7, 10, 6.5, 8, 7)
Class = c(4, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1)
Time = 4 - Class
Mean and dtandard deviation:
mean(Sleep)
## [1] 6.727
sd(Sleep)
## [1] 1.348
80% CI:
t.test(Sleep, conf.level = 0.8)
##
## One Sample t-test
##
## data: Sleep
## t = 16.55, df = 10, p-value = 1.357e-08
## alternative hypothesis: true mean is not equal to 0
## 80 percent confidence interval:
## 6.169 7.285
## sample estimates:
## mean of x
## 6.727
H0: \( \mu \) = 7.5 hours
versus
H1 (or HA): \( \mu \) < 7.5 hours
To actually do the test we look at the statistic \( (\bar{Sleep} - 7.5)/(sd(Sleep)/sqrt(11)) \) which is distributed as a T-distribution with 10 degrees of freedom.
(mean(Sleep) - 7.5)/(sd(Sleep)/sqrt(length(Sleep)))
## [1] -1.901
p = pt(-1.900658, df = 10)
p
## [1] 0.04326
The p-value is 0.04326425. This is less than the level of significance, 0.05, so we reject the null hypothesis. The data supports the alternative hypothesis. MIDN Dozy's claim is supported by the evidence.
However, MIDN Dozy should not try to extrapolate his infrerence to all college students since midshipmen are very special kinds of college students. As well, the date that MIDN Dozy chose to do his sampling on is actually a very special day (the first day of x-week) and so the responses will be different from students at USNA than at most other colleges, which don't have x-week exams.
MIDN Dozy's study was an observational study. He did not randomly assign midshipmen to a control and response group.
Sleep is a quantitative variable and Class is a qualitative variable.
Plots and linear models:
plot(Sleep ~ Time, col = "blue", pch = 19, xlab = "Time (in full years) at USNA",
ylab = "Amount of Sleep in hours", main = "Sleep versus Time")
ST = lm(Sleep ~ Time)
abline(ST, col = "red", lwd = 3)
summary(ST)
##
## Call:
## lm(formula = Sleep ~ Time)
##
## Residuals:
## Min 1Q Median 3Q Max
## -1.0435 -0.5326 -0.1304 0.0652 2.7609
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.630 0.517 10.89 1.8e-06 ***
## Time 0.804 0.299 2.69 0.025 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.06 on 9 degrees of freedom
## Multiple R-squared: 0.446, Adjusted R-squared: 0.385
## F-statistic: 7.26 on 1 and 9 DF, p-value: 0.0246
The general pattern revealed in the data is that the amount of sleep midshipmen get increases as they spend more time at USNA.
Residuals:
require(lattice)
## Loading required package: lattice
require(latticeExtra)
## Loading required package: latticeExtra
## Loading required package: RColorBrewer
densityplot(ST$residuals)
The extra-long right tail means that the residuals are right-skewed.
Outlier:
plot(rstandard(ST) ~ ST$fitted)
which(rstandard(ST) > 2)
## 8
## 8
Sleep[8]
## [1] 10
Time[8]
## [1] 2
Sleep[8] - (5.6304 + 0.8043 * 2) # residual is 2.761
## [1] 2.761
The outlier is the 2nd class that slept 10 hours. The residual is 2.761 hours.
The regression line with all data is: Sleep = 5.6304+0.8043*(TimeAtUSNA)
STo = lm(Sleep[-8] ~ Time[-8])
summary(STo)
##
## Call:
## lm(formula = Sleep[-8] ~ Time[-8])
##
## Residuals:
## Min 1Q Median 3Q Max
## -0.5620 -0.3151 0.0434 0.2934 0.5041
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.562 0.203 27.44 3.4e-09 ***
## Time[-8] 0.645 0.119 5.41 0.00063 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 0.414 on 8 degrees of freedom
## Multiple R-squared: 0.786, Adjusted R-squared: 0.759
## F-statistic: 29.3 on 1 and 8 DF, p-value: 0.000635
The regression line computed without the outlier is: Sleep = 5.5620 + 0.64446*(TimeAtUSNA).
The regression line doesn't change that much because the outlier's explanatory variable (TimeAtUSNA) is close to its mean. This outlier does not have a lot of leverage. We can see this intuitively when looking at the two regression lines:
plot(Sleep ~ Time, col = "blue", pch = 19, xlab = "Time (in full years) at USNA",
ylab = "Amount of Sleep in hours", main = "Sleep versus Time")
legend(0.25, 9.5, c("Regression with all data", "Regression without outlier"),
lty = c(1, 1), lwd = c(3, 3), col = c("red", "green"))
abline(ST, col = "red", lwd = 3)
abline(STo, col = "green", lwd = 3)
