Name: Kirk Swanson

Collaborators: None

Stat 202 HW 1

1) Sparrows: Complete exercise 1.2 in the text. The slope of the sparrows regression line is 0.467 grams per mm, which indicates that when we observe a sparrow which has a winglength longer than some other observed sparrow by one mm, we expect its weight to also be larger by 0.467 grams compared to that other observed sparrow.

2) Sparrows: Complete exercise 1.4 in the text. The regression standard error estimate is labeled S in Minitab output, so we see that the size of a typical error is 1.39959 grams.

# Use R as your calculator.

3) Cereal: Complete exercise 1.8 in the text.

load("Cereal.RData")
plot(Cereal$Sugar, Cereal$Calories)

cor(Cereal$Sugar, Cereal$Calories)

## [1] 0.5154

There seems to be a positive linear trend with sugar as the predictor variable and calories as the response variable in the scatter plot. The correlation coefficient is approximately .5, which supports this claim. So, we will choose a linear model.

m <- lm(Cereal$Calories ~ Cereal$Sugar)
abline(m, col = "Red")

## Error: plot.new has not been called yet

m

## 
## Call:
## lm(formula = Cereal$Calories ~ Cereal$Sugar)
## 
## Coefficients:
##  (Intercept)  Cereal$Sugar  
##        87.43          2.48

Fill in values for a and b below. \[ \widehat{\text{calories}} = 87.428 + 2.481 \cdot \text{sugar} \] The slope is 2.481 calories per serving per number of grams of sugar per serving. When we find a cereal with one more gram of sugar per serving, we expect its number of calories per serving to be larger by 2.481.

4) Caterpillar waste (fun!): Complete exercise 1.12 in the text.

load("Caterpillars.RData")
plot(Caterpillars$Mass, Caterpillars$WetFrass)

plot(Caterpillars$LogMass, Caterpillars$LogWetFrass)

There seems to be a general positive trend, but the scatter plot shows a lot of curvature as mass increases.

• There seems to be a positive linear trend with increasing log of mass, although there are four groups of lower errors.
  
• I would prefer the log-log plot, as the relationship appears to be more linear. The fitted equation is \( \widehat{Log(WetFrass)}=-0.7386+1.0536\text{Log(Mass)} \).

abline(lm(Caterpillars$LogWetFrass ~ Caterpillars$LogMass), col = "Red")

## Error: plot.new has not been called yet

lm(Caterpillars$LogWetFrass ~ Caterpillars$LogMass)

## 
## Call:
## lm(formula = Caterpillars$LogWetFrass ~ Caterpillars$LogMass)
## 
## Coefficients:
##          (Intercept)  Caterpillars$LogMass  
##               -0.739                 1.054

symb <- 1:5
plot(Caterpillars$LogMass, Caterpillars$LogWetFrass, pch = symb[Caterpillars$Instar])

Based on the symbol groupings, it appears that stage five is the most linear for logWetFrass versus logMass in base 10 log.

symb <- 1:2
plot(Caterpillars$LogMass, Caterpillars$LogWetFrass, pch = symb[Caterpillars$Fgp])

Based on the symbol groupings, it appears that growth period two is the most linear for logWetFrass versus logMass in base 10 log.

5) Baseball: Complete exercise 1.27 in the text.

load("BaseballTimes.RData")
hist(BaseballTimes$Time, breaks = 10)

summary(BaseballTimes$Time)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##     133     154     168     183     194     317

sd(BaseballTimes$Time)

## [1] 46.21

The histogram shows a right-skewed distribution with a large outlier at 317 minutes. The mean is 182.7 minutes with a standard deviation of 46.20554 minutes. Looking at the data table, the largest time game was also the game with the largest number of pitchers - perhaps this hints at a possible explanation, although I don't know much about baseball. It also had the second largest attendence and one of the smallest margins.

par(mfrow = c(2, 2))
plot(BaseballTimes$Runs, BaseballTimes$Time)
plot(BaseballTimes$Margin, BaseballTimes$Time)
plot(BaseballTimes$Pitchers, BaseballTimes$Time)
plot(BaseballTimes$Attendance, BaseballTimes$Time)

• Margin and Attendance do not appear to be good linear predictors of game time. Runs and pitchers do look somewhat linear, but since the runs data seems to fan out as the number of runs increases, I will choose pitchers to be my explanatory variable. We need to be wary of the outlier at the end, of course.

summary(lm(BaseballTimes$Time ~ BaseballTimes$Pitchers))

## 
## Call:
## lm(formula = BaseballTimes$Time ~ BaseballTimes$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 ***
## BaseballTimes$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

plot(BaseballTimes$Time ~ BaseballTimes$Pitchers)
abline(lm(BaseballTimes$Time ~ BaseballTimes$Pitchers), col = "Red")

Fill in values for a, b, and the predictor below. \[ \widehat{\text{Time}} = 94.84 + 10.71\cdot \text{Pitchers} \]

• The estimated parameters give us a model that explains 79.98 percent of the variability in game time, from the R-squared value. The slope is 10.71 minutes per number of pitchers. If a game has one more pitcher than some other game, we expect this game to be 10.71 mintues longer.
  

resid <- BaseballTimes$Time - (94.84 + 10.71 * BaseballTimes$Pitchers)
qqnorm(resid)
qqline(resid)

plot(BaseballTimes$Pitchers, resid)
abline(0, 0)

• There do not appear to be significant trends in the Normal QQ plot of the residuals, other than three points that might be outliers. This plot indicates that the residuals are distributed approximately normally. There also do not appear to be any significant trends in the plot of residuals against pitchers, which indicates that the error obeys the constant variance condition. Based on these residual plots, it seems that the linear model is valid for inference. These observations are in addition to the fact that the original scatter plot appeared linear and that the observations as separate games are probably independent data points.
  

6) Caterpillar Metablic Rates: Complete exercise 1.30 in the text.

load("Metabolic.RData")
plot(MetabolicRate$Computer, MetabolicRate$Mrate)

plot(MetabolicRate$BodySize, MetabolicRate$Mrate)

plot(MetabolicRate$LogBodySize, MetabolicRate$Mrate)

plot(MetabolicRate$LogBodySize, MetabolicRate$LogMrate)

plot(MetabolicRate$BodySize, MetabolicRate$LogMrate)

• The log-log plot for LogMrate as a function of LogBodySize looks very linear, so I suggest using LogMrate versus LogBodySize.
  

m <- lm(MetabolicRate$LogMrate ~ MetabolicRate$LogBodySize)
summary(m)

## 
## Call:
## lm(formula = MetabolicRate$LogMrate ~ MetabolicRate$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 ***
## MetabolicRate$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

plot(MetabolicRate$LogBodySize, MetabolicRate$LogMrate)
abline(m, col = "Red")

m

## 
## Call:
## lm(formula = MetabolicRate$LogMrate ~ MetabolicRate$LogBodySize)
## 
## Coefficients:
##               (Intercept)  MetabolicRate$LogBodySize  
##                     1.307                      0.916

For a caterpillar that has a body size of 1 gram, we predict that LogMrate = 1.3066+0.9164LogBodySize = .3066, or that Mrate = 1.3587 metabolic rate units.