Regression (Part 2)

• Homework #11: Due: Friday, April 28, 11:59 pm
  • Whitlock & Schluter, Chapter 16: #20, 26
  • Whitlock & Schluter, Chapter 17: #30, 34-36
  • Whitlock & Schluter, Chapter 18: #10, 14, 16, 18
• Reading Assignment #18: Due: Friday, April 14, 11:00 am
  • Whitlock & Schluter, Chapter 18: Multiple explanatory variables
• Lab this week: Projects and Homework #18

There are two other types of “confidence intervals” of interest:

• Confidence bands: Confidence intervals of mean \( Y \) values for fixed \( X \) (i.e. confidence intervals for specific points on regression line)
• Prediction intervals: \( (1-\alpha)\% \) of the probability distribution of \( Y \) values for fixed \( X \).

(biteAnova <- anova(biteRegression))

Analysis of Variance Table

Response: territory.area
          Df Sum Sq Mean Sq F value  Pr(>F)  
bite       1 194.37 194.374  5.8012 0.03934 *
Residuals  9 301.55  33.506                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Tests null hypothesis \( H_{0}: \beta = 0 \).

Definition: If the assumptions of linear regression are met, then the sampling distribution of \( b \) is a normal distribution having a mean equal to \( \beta \) and a standard error estimated from data as \( \mathrm{SE}_{b} \).

\[ H_{0}: \beta = \beta_{0}\\ H_{A}: \beta \neq \beta_{0} \]

\( F \)-test using ANOVA tests \( \beta_{0}=0 \) only, but a \( t \)-test can be used when \( \beta_{0}\neq 0 \)!!!

Test statistic: \[ t = \frac{b-\beta_{0}}{\mathrm{SE}_{b}} \]

Degree of freedom: \[ df = n-2 \]

You can get the results of a \( t \)-test on slope in R for \( \beta_{0}=0 \) by using the summary command on a regression results from lm:

summary(biteRegression)

Call:
lm(formula = territory.area ~ bite, data = biteData)

Residuals:
    Min      1Q  Median      3Q     Max 
-7.0472 -4.5101 -0.5504  3.6689 10.2237 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)  
(Intercept)  -31.539     23.513  -1.341   0.2127  
bite          11.677      4.848   2.409   0.0393 *
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 5.788 on 9 degrees of freedom
Multiple R-squared:  0.3919,    Adjusted R-squared:  0.3244 
F-statistic: 5.801 on 1 and 9 DF,  p-value: 0.03934

For a \( t \)-test with   \( H_{0}: \beta_{0} \neq 0 \), go manual.

First, get estimate.

biteRegression$coefficients

(Intercept)        bite 
  -31.53929    11.67730 

(b <- biteRegression$coefficients['bite'])

   bite 
11.6773 

For a \( t \)-test with   \( H_{0}: \beta_{0} \neq 0 \), go manual.

Second, get standard error.

(coef <- summary(biteRegression)$coefficients)

             Estimate Std. Error   t value  Pr(>|t|)
(Intercept) -31.53929  23.512646 -1.341375 0.2126627
bite         11.67730   4.848226  2.408571 0.0393411

(sderr <- coef['bite','Std. Error'])

[1] 4.848226

For a \( t \)-test with   \( H_{0}: \beta_{0} \neq 0 \), go manual.

Third, compute test statistic.

\[ \begin{align} t & = \frac{b - \beta_{0}}{\mathrm{SE}_{b}} \\ & = \frac{11.6772959 - \beta_{0}}{4.8482259} \end{align} \]

Then continue with usual hypothesis test steps by computing \( P \)-value of test statistic with \( df = n-2 \).

• Linearity
• \( Y \)-values are normally distributed
• Variance of \( Y \)-values the same for all \( X \)
• \( Y \)-values a random sample from sub-population with corresponding \( X \)-values.

We discussed using residual plots to check the second and third point above.

Outliers can also have large effects on linear fits.

Visualizing the data will help spot nonlinearities.

Transformations can make some nonlinear relationships linear.

Transformations can also correct for non-equal variances (just like in ANOVA).

In this example, the response variable for the untransformed data on the left was transformed to the right using a square-root transformation.

• Measurement error in \( Y \) increases variance of residuals, but has no effect on the estimate of the slope (does effect precision of estimate, however).
• Measurement error in \( X \) affects both residual error and estimate of slope (underestimates).

The problem with nonlinear regression? An infinite number of possible nonlinear functions to fit!

Book covers:

• Asymptotic curve fitting
• Quadratic curve fitting
• Formula-free curve fitting

We will just discuss asymptotic curve fitting.

Consider the relationship between population growth rate of a phytoplankton in culture and the concentration of iron in the medium.

• Left panel is overfitting
• Left panel fit will perform poorly for prediction
• Left panel fit has poor biological motivation

The right panel shows a fit with the Michaelis-Menten curve

\[ Y = \frac{aX}{b+X}, \]
with \( a \) and \( b \) parameters.

phytoplankton <- read.csv(url("http://www.zoology.ubc.ca/~schluter/WhitlockSchluter/wp-content/data/chapter17/chap17f8_1IronAndPhytoplanktonGrowth.csv"))
head(phytoplankton)

  ironConcentration phytoGrowthRate
1        0.01096478            0.00
2        0.02398833            0.50
3        0.02818383            0.35
4        0.07244360            1.04
5        0.07585776            0.97
6        0.11481536            1.19

One can perform a nonlinear regression using nonlinear least squares. This can be accomplished in R as follows with the package nls.

phytoCurve <- nls(phytoGrowthRate ~ a*ironConcentration / ( b+ironConcentration), 
                  data = phytoplankton, 
                  list(a = 1, b = 1))

The last argument is a list containing initial values for the parameters (initial guesses at the best fit).

Definition: Logistic regression is nonlinear regression developed for a categorical response variable with two levels.

Caption: Mortality of guppies in relation to duration of exposure to a temperature of 5\( ^{\circ} \) C. (Estimating probability of an event)

Properties:

• Outcomes at every \( X \) have a binomial distribution (not a normal distribution).
• Probability of an event is given by corresponding predicted value on the regression curve.
• To correct for differences in variance of residuals at different \( X \) values, residuals are weighted by estimated variance obtained from binomial distribution.

The following curve is fit in logistic regression:

\[ \textrm{log-odds}(Y) = \alpha + \beta X, \]

where

\[ \textrm{log-odds}(Y) = \ln\left[\frac{Y}{1-Y}\right] \]

and \( \alpha, \beta \) are population parameters.

Note: \( \textrm{Log-odds}(Y) \) is called the logit link function.

The estimates \( a \) and \( b \) for \( \alpha \) and \( \beta \), respectively, describe the linear relationship between \( X \) and the predicted log odds of \( Y \), i.e.

\[ \textrm{log-odds}(\hat{Y}) = a + b X. \]

To obtain the original predicted values, we need to solve for \( \hat{Y} \), obtaining

\[ \hat{Y} = \frac{e^{a+bX}}{1+e^{a+bX}} = \frac{1}{1 + e^{-(a + bX)}}. \]

This is a sigmoidal (S-shaped) function!! When \( b > 0 \), function is increasing. When \( b < 0 \), function is decreasing. When \( b = 0 \), function is constant.

The threat of bioterrorism makes it necessary to quantify the risk of exposure to infectious agents such as anthrax (Bacillus anthracis). Hans (2002) measured the mortality of rhesus monkeys in an exposure chamber to aerosolized anthrax spores of varying concentrations.

Look at data

anthrax <- read.csv("http://whitlockschluter.zoology.ubc.ca/wp-content/data/chapter17/chap17q35AnthraxMortality.csv")
str(anthrax)

'data.frame':   9 obs. of  3 variables:
 $ anthraxConcentration: int  29300 32100 45300 57300 64800 67000 100000 125000 166000
 $ survived            : int  7 4 3 2 3 5 0 1 0
 $ died                : int  1 4 5 6 5 3 8 7 8

Discuss: What format is this data in, tidy or messy?

From wide table to long

# Create survived data frame
survived <- with(anthrax, 
                 data.frame(anthraxConcentration = rep(anthraxConcentration, survived), 
                            mortality = rep(1, sum(survived))))

# Create died data frame
died <- with(anthrax, 
             data.frame(anthraxConcentration = rep(anthraxConcentration, died), 
                        mortality = rep(0, sum(died))))

# Bind them together
anthrax_long <- rbind(died, survived)

From wide table to long

str(anthrax_long)

'data.frame':   72 obs. of  2 variables:
 $ anthraxConcentration: int  29300 32100 32100 32100 32100 45300 45300 45300 45300 45300 ...
 $ mortality           : num  0 0 0 0 0 0 0 0 0 0 ...

Show data

plot(jitter(mortality, 
            factor = 0.1) ~ 
     jitter(anthraxConcentration, 
            factor = 6), 
     data = anthrax_long, 
     pch=16, 
     cex=1.5, 
     cex.lab=1.5, 
     col="firebrick", 
     xlab="Anthrax concentration (spores/l)", 
     ylab="Mortality")

Show data

Generalized linear models

anthraxGlm <- glm(mortality ~ anthraxConcentration, 
                  data = anthrax_long, 
                  family = binomial(link = logit))

glm stands for generalized linear model.

Notice the family = binomial(link = logit) argument.

Add regression curve

# Plot data
plot(jitter(mortality, factor = 0.1) ~ jitter(anthraxConcentration, factor = 6), data = anthrax_long, pch=16, cex=1.5, cex.lab=1.5, col="firebrick", xlab="Anthrax concentration (spores/l)", ylab="Mortality")

# Plot fit
xpts <- seq(min(anthrax_long$anthraxConcentration), 
            max(anthrax_long$anthraxConcentration), 
            length.out = 100)
ypts <- predict(anthraxGlm, 
                newdata = data.frame(anthraxConcentration = xpts), 
                type = "response")
lines(xpts, ypts)

Add regression curve

Table of parameter estimates

summary(anthraxGlm)

Table of parameter estimates

Call:
glm(formula = mortality ~ anthraxConcentration, family = binomial(link = logit), 
    data = anthrax_long)

Deviance Residuals: 
    Min       1Q   Median       3Q      Max  
-1.4750  -0.9292  -0.3420   0.9161   2.3951  

Coefficients:
                       Estimate Std. Error z value Pr(>|z|)   
(Intercept)           1.745e+00  6.921e-01   2.521  0.01171 * 
anthraxConcentration -3.643e-05  1.119e-05  -3.255  0.00113 **
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

(Dispersion parameter for binomial family taken to be 1)

    Null deviance: 92.982  on 71  degrees of freedom
Residual deviance: 73.962  on 70  degrees of freedom
AIC: 77.962

Number of Fisher Scoring iterations: 5

Note: \( P \)-values are not to be trusted!!! These report Wald statistics, which we know are inaccurate.

Better to create an analysis of deviance table.

anova(anthraxGlm, test = "Chi")

This will test the null hypothesis that anthrax concentration has no effect on mortality (i.e. \( \beta_{0} = 0 \)). See the book for details.