The Analysis of Variance (ANOVA) - Part 2

• Exam #3 opens today at 3:00 pm
  • Due date: November 2, 2:00 pm
  • You have 3 hours to complete the exam once opened
  • Chapters 9-12

Practice Problem #1

Many humans like the effect of caffeine, but it occurs in plants as a deterrent to herbivory by animals. Caffeine is also found in flower nectar, and nectar is meant as a reward for pollinators, not a deterrent. How does caffeine in nectar affect visitation by pollinators?

Practice Problem #1

Singaravelan et al. (2005) set up feeding stations where bees were offered a choice between a control solution with 20% sucrose or a caffeinated solution with 20% sucrose plus some quantity of caffeine. Over the course of the experiment, four different concentrations of caffeine were provided: 50, 100, 150, and 200 ppm. The response variable was the difference between the amount of nectar consumed from the caffeine feeders and that removed from the control feeders at the same station (grams).

Singaravelan et al. (2005) set up feeding stations where bees were offered a choice between a control solution with 20% sucrose or a caffeinated solution with 20% sucrose plus some quantity of caffeine. Over the course of the experiment, four different concentrations of caffeine were provided: 50, 100, 150, and 200 ppm. The response variable was the difference between the amount of nectar consumed from the caffeine feeders and that removed from the control feeders at the same station (grams).

Discuss: Describe the experimental design.

\[ t = \frac{\bar{Y}_{1}-\bar{Y}_{2}}{\mathrm{SE}_{\bar{Y}_{1}-\bar{Y}_{2}}} \]

Data: With \( i \) representing group \( i \), we have

\[ Y_{ij} - \bar{Y} = (\bar{Y}_{i} - \bar{Y}) + (Y_{ij} - \bar{Y}_{i}) \]

\[ \mathrm{SS}_{\mathrm{total}} = \mathrm{SS}_{\mathrm{groups}} + \mathrm{SS}_{\mathrm{error}} \]

Data: With \( i \) representing group \( i \), we have

\[ Y_{ij} - \bar{Y} = (\bar{Y}_{i} - \bar{Y}) + (Y_{ij} - \bar{Y}_{i}) \]

\[ \mathrm{SS}_{\mathrm{total}} = \sum_{i}\sum_{j}(Y_{ij}-\bar{Y})^2 \]

Data: With \( i \) representing group \( i \), we have

\[ Y_{ij} - \bar{Y} = (\bar{Y}_{i} - \bar{Y}) + (Y_{ij} - \bar{Y}_{i}) \]

\[ \scriptsize{\mathrm{SS}_{\mathrm{total}} = \sum_{i}\sum_{j}(Y_{ij}-\bar{Y})^2 = \sum_{i}n_{i}(\bar{Y}_{i}-\bar{Y})^2 + \sum_{i}\sum_{j}(Y_{ij}-\bar{Y}_{i})^2} \]

Data: With \( i \) representing group \( i \), we have

\[ Y_{ij} - \bar{Y} = (\bar{Y}_{i} - \bar{Y}) + (Y_{ij} - \bar{Y}_{i}) \]

\[ \scriptsize{ \begin{eqnarray*} \mathrm{SS}_{\mathrm{total}} = \sum_{i}\sum_{j}(Y_{ij}-\bar{Y})^2 & = & \sum_{i}n_{i}(\bar{Y}_{i}-\bar{Y})^2 + \sum_{i}\sum_{j}(Y_{ij}-\bar{Y}_{i})^2 \\ & = & \mathrm{SS}_{\mathrm{groups}} + \mathrm{SS}_{\mathrm{error}} \end{eqnarray*} } \]

\[ \scriptsize{ \begin{eqnarray*} \mathrm{SS}_{\mathrm{total}} & = & \sum_{i}\sum_{j}(Y_{ij}-\bar{Y})^2 \\ & = & \sum_{i}\sum_{j}\left[(\bar{Y}_{i} - \bar{Y}) + (Y_{ij} - \bar{Y}_{i})\right]^2 \\ & = & \sum_{i}\sum_{j}\left[(\bar{Y}_{i} - \bar{Y})^2 + (Y_{ij} - \bar{Y}_{i})^2 + 2(\bar{Y}_{i} - \bar{Y})(Y_{ij} - \bar{Y}_{i})\right] \\ & = & \sum_{i}\sum_{j}(\bar{Y}_{i} - \bar{Y})^2 + \sum_{i}\sum_{j}(Y_{ij} - \bar{Y}_{i})^2 + \sum_{i}\sum_{j}2(\bar{Y}_{i} - \bar{Y})(Y_{ij} - \bar{Y}_{i}) \\ & = & \sum_{i}n_{i}(\bar{Y}_{i} - \bar{Y})^2 + \sum_{i}\sum_{j}(Y_{ij} - \bar{Y}_{i})^2 + \sum_{i}\sum_{j}2(\bar{Y}_{i} - \bar{Y})(Y_{ij} - \bar{Y}_{i}) \\ & = & \mathrm{SS}_{\mathrm{groups}} + \mathrm{SS}_{\mathrm{error}} + \sum_{i}\sum_{j}2(\bar{Y}_{i} - \bar{Y})(Y_{ij} - \bar{Y}_{i}) \end{eqnarray*} } \]

Can show:

\[ \sum_{i}\sum_{j}2(\bar{Y}_{i} - \bar{Y})(Y_{ij} - \bar{Y}_{i}) = 0, \]

and thus

\[ \mathrm{SS}_{\mathrm{total}} = \mathrm{SS}_{\mathrm{groups}} + \mathrm{SS}_{\mathrm{error}}. \]

Data: With \( i \) representing group \( i \), we have

\[ Y_{ij} - \bar{Y} = (\bar{Y}_{i} - \bar{Y}) + (Y_{ij} - \bar{Y}_{i}) \]

\[ \scriptsize{\mathrm{SS}_{\mathrm{total}} = \sum_{i}\sum_{j}(Y_{ij}-\bar{Y})^2 = \sum_{i}n_{i}(\bar{Y}_{i}-\bar{Y})^2 + \sum_{i}\sum_{j}(Y_{ij}-\bar{Y}_{i})^2} \]

Definition: The group mean square is given by

\[ \mathrm{MS}_{\mathrm{groups}} = \frac{\mathrm{SS}_{\mathrm{groups}}}{df_{\mathrm{groups}}}, \] with \( df_{\mathrm{groups}} = k-1 \).

Definition: The error mean square is given by

\[ \mathrm{MS}_{\mathrm{error}} = \frac{\mathrm{SS}_{\mathrm{error}}}{df_{\mathrm{error}}}, \] with \( df_{\mathrm{error}} = \sum (n_{i}-1) = N-k \).

str(strungOutBees)

'data.frame':   20 obs. of  2 variables:
 $ ppmCaffeine                     : Factor w/ 4 levels "ppm50","ppm100",..: 1 2 3 4 1 2 3 4 1 2 ...
 $ consumptionDifferenceFromControl: num  -0.4 0.01 0.65 0.24 0.34 -0.39 0.53 0.44 0.19 -0.08 ...

Discuss: Is this data tidy or messy?

Definition: Tidy!

stripchart(consumptionDifferenceFromControl ~ ppmCaffeine, 
           data = strungOutBees, 
           vertical = TRUE, 
           method = "jitter", 
           xlab="Caffeine (ppm)",
           col="red")

Discuss: State the null and alternative hypotheses appropriate for this question.

\[ \begin{eqnarray*} H_{0} & : & \mu_{50} = \mu_{100} = \mu_{150} = \mu_{200} \\ H_{A} & : & \mathrm{At \ least \ one \ of \ the \ means \ is \ different} \end{eqnarray*} \]

Short cut using R

caffResults <- lm(consumptionDifferenceFromControl ~ ppmCaffeine, data=strungOutBees)
anova(caffResults)

Analysis of Variance Table

Response: consumptionDifferenceFromControl
            Df Sum Sq Mean Sq F value  Pr(>F)  
ppmCaffeine  3 1.1344 0.37814  4.1779 0.02308 *
Residuals   16 1.4482 0.09051                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Definition: The \( R^{2} \) value in ANOVA is the “fraction of the variation explained by groups” and is given by

\[ R^{2} = \frac{\mathrm{SS}_{\mathrm{groups}}}{\mathrm{SS}_{\mathrm{total}}}. \] Note: \( 0 \leq R^2 \leq 1 \).

beeAnovaSummary <- summary(caffResults)
beeAnovaSummary$r.squared

[1] 0.4392573

Long way

Question: Calculate the following summary statistics for each group: \( n_{i} \), \( \bar{Y}_{i} \), and \( s_{i} \).

library(dplyr)
beeStats <- strungOutBees %>% 
  group_by(ppmCaffeine) %>% 
  summarise(n = n(), 
            mean = mean(consumptionDifferenceFromControl),
            sd = sd(consumptionDifferenceFromControl))
knitr::kable(beeStats)

Compute sum-of-squares

\[ \scriptsize{ \begin{eqnarray*} \mathrm{SS}_{\mathrm{groups}} & = & \sum_{i}n_{i}(\bar{Y}_{i}-\bar{Y})^2 \end{eqnarray*} } \]

grandMean <- mean(strungOutBees$consumptionDifferenceFromControl)
(SS_groups <- sum(beeStats$n*(beeStats$mean - grandMean)^2))

[1] 1.134415

Compute sum-of-squares

\[ \scriptsize{ \begin{eqnarray*} \mathrm{SS}_{\mathrm{error}} & = & \sum_{i}\sum_{j}(Y_{ij}-\bar{Y}_{i})^2 = \sum_{i}(n_{i}-1)s_{i}^2 \end{eqnarray*} } \]

(SS_error <- sum((beeStats$n-1)*beeStats$sd^2))

[1] 1.44816

Compute degree of freedom

\[ \begin{equation} \mathrm{df}_{\mathrm{groups}} = k-1 \end{equation} \]

where \( k \) is number of groups.

(df_groups <- 4-1)

[1] 3

(df_error <- nrow(strungOutBees)-4)

[1] 16

Compute degree of freedom

\[ \begin{equation} \mathrm{df}_{\mathrm{error}} = N-k \end{equation} \]

where \( N \) is number of total observations in data.

(df_error <- nrow(strungOutBees)-4)

[1] 16

Compute mean squares

\[ \begin{equation} \mathrm{MS}_{\mathrm{groups}} = \frac{\mathrm{SS}_{\mathrm{groups}}}{\mathrm{df}_{\mathrm{groups}}} \end{equation} \]

(MS_groups <- SS_groups/df_groups)

[1] 0.3781383

Compute mean squares

\[ \begin{equation} \mathrm{MS}_{\mathrm{error}} = \frac{\mathrm{SS}_{\mathrm{error}}}{\mathrm{df}_{\mathrm{error}}} \end{equation} \]

(MS_error <- SS_error/df_error)

[1] 0.09051

Compute \( F \)-statistic and \( P \)-value

\[ \begin{equation} F = \frac{\mathrm{MS}_{\mathrm{groups}}}{\mathrm{MS}_{\mathrm{error}}} \end{equation} \]

(F_ratio <- MS_groups/MS_error)

[1] 4.177862

(pval <- pf(F_ratio, df_groups, df_error, lower.tail=FALSE))

[1] 0.02307757

Using Statistical F Table

Using Statistical F Table

Assumptions (same as 2-sample \( t \)-test)

• Measurements in each group represent a random sample from corresponding population.
• Variable is normally distributed in each of the \( k \) populations.
• Variance is the same in all \( k \) populations.

Robustness (same as 2-sample \( t \)-test)

• Robust to deviations in normality.
• Somewhat robust to deviations in equal variances when:
  • Sample sizes are “large”, about the same size in each group
  • Sample sizes are about the same (balanced)
  • Standard deviations are within about a 3-fold difference

Definition: The Kruskal-Wallis test is a nonparametric method for mulutiple groups based on ranks.

The Kruskal-Wallis test is similar to the Mann-Whitney \( U \)-test and has the same assumptions:

• Group samples are random samples.
• To use as a test of difference between means or medians, the distributions must have the same shape in every population.

Power of Kruskal-Wallis test is nearly as powerful as ANOVA when sample sizes are large, but has smaller power than ANOVA for small sample sizes.