You should know now:

• the command t.test() and the different kinds of usage
• that y ~ x in R mean y dependent on x (formula syntax)
• that R can roll dice for you (as one example for random number generator)

For the babies data set, the variable age contains the recorded mom's age and dage contains the dad's age for several different cases in the sample. Do a significance test of the null hypothesis of equal ages against a one-sided alternative that the dads are older in the sampled population.

• install the package UsingR (you may check if the package is already installed)
• do the respective t-test
• redo the test assuming equal variances! How do the results differ? Would this assumption be appropriate?

• a technique we use when all explanatory variables are categorical (factor)
• if there is one factor with three or more levels we use one-way ANOVA (only two levels: t-test should be preferred, would give exactly the same answer since with 2 levels F = t²⁾
• for more factors there there is two-way, three-way anova
• central idea is to compare two or more means by comparing variances

A data frame with 14 observations on 2 variables.

Source: M. Crawley, The R-Book

• read in the respective data file anova.txt using the read.table() command

gardens <- read.table("session5dat/gardens2.txt", header = T)
head(gardens)

##   ozone garden index
## 1     6      a    10
## 2     8      a     7
## 3     5      a     1
## 4     9      a     4
## 5     7      a     5
## 6     8      a     6

• first we plot the values

• there is a lot of scatter, indicating that the variance in ozone is large
• to get a feel for the overall variance we plot the overall mean (8.5) and indicate each of the residuals by a vertical line

• we refer to this overall variation as the total sum of squares, SSY or TSS

\[SSY = \sum(y-\bar{y})^2\]

• here we have \[ SSY = 55.5 \]

• now instead of fitting the overall mean, let us fit the individual garden means

• now instead of fitting the overall mean, let us fit the individual garden means

• Garden A: black
• Garden B: red

• now we see that the mean ozone concentration is substantially higher in garden B
• the aim of ANOVA is to determine
  • whether it is significantly higher or
  • whether this kind of difference could come by chance alone

When the means are significantly different then the sum of squares computed from the individual garden means will be smaller than the sum of squares computed from the overall mean.

• we define the new sum of squares as the error sum of squares (error in the sense of redidual) \[ SSE = \sum(y_{garden A}-\bar{y}_{garden A})^2+\sum(y_{garden B}-\bar{y}_{garden B})^2 \]

• here we have \[ SSE = 24.0 \]

• then the component of the variation that is explained by the difference (i.e. explained by the predictor) of the means is called the treatment sum of sqares SSA
• analysis of variance is based on the notion that we break down the total sum of squares into useful and informative components

\[ SSY = SSE + SSA \]

where

• SSA = explained variation
• SSE = unexplained variation

• now we need to test whether an F ratio of 15.75 is large or small
  • we can use a table or software package
  • I use here R to calculate the cumulative probability

1 - pf(15.75,1,12)

## [1] 0.001864103

• in R we use the lm() command and
• the formula syntax a ~ b
• we assign the result to a variable

mm <- lm(ozone ~ garden, data = gardens)
mm

## 
## Call:
## lm(formula = ozone ~ garden, data = gardens)
## 
## Coefficients:
## (Intercept)      gardenb  
##           7            3

summary(mm)

## 
## Call:
## lm(formula = ozone ~ garden, data = gardens)
## 
## Residuals:
##    Min     1Q Median     3Q    Max 
##     -2     -1      0      1      2 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   7.0000     0.5345  13.096 1.82e-08 ***
## gardenb       3.0000     0.7559   3.969  0.00186 ** 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 1.414 on 12 degrees of freedom
## Multiple R-squared:  0.5676, Adjusted R-squared:  0.5315 
## F-statistic: 15.75 on 1 and 12 DF,  p-value: 0.001864

anova(mm)

## Analysis of Variance Table
## 
## Response: ozone
##           Df Sum Sq Mean Sq F value   Pr(>F)   
## garden     1   31.5    31.5   15.75 0.001864 **
## Residuals 12   24.0     2.0                    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

• there are two traditional ways of plotting the results of ANOVA
• box-and-whisker plots
• barplots with error bars