anova.R

## Not assuming equal variances
oneway.test(extra ~ group, data = sleep, var.equal = T)

## 
##  One-way analysis of means
## 
## data:  extra and group
## F = 3.4626, num df = 1, denom df = 18, p-value = 0.07919

## Assuming equal variances
oneway.test(extra ~ group, data = sleep, var.equal = T)

## 
##  One-way analysis of means
## 
## data:  extra and group
## F = 3.4626, num df = 1, denom df = 18, p-value = 0.07919

## which gives the same result as
str(sleep)

## 'data.frame':    20 obs. of  3 variables:
##  $ extra: num  0.7 -1.6 -0.2 -1.2 -0.1 3.4 3.7 0.8 0 2 ...
##  $ group: Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 1 1 ...
##  $ ID   : Factor w/ 10 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8 9 10 ...

anova(lm(extra ~ group, data = sleep))

## Analysis of Variance Table
## 
## Response: extra
##           Df Sum Sq Mean Sq F value  Pr(>F)  
## group      1 12.482 12.4820  3.4626 0.07919 .
## Residuals 18 64.886  3.6048                  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

data_1 <- summary(aov(extra ~ group, data = sleep))
data_1

##             Df Sum Sq Mean Sq F value Pr(>F)  
## group        1  12.48  12.482   3.463 0.0792 .
## Residuals   18  64.89   3.605                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

t.test(sleep$extra[sleep$group ==1], sleep$extra[sleep$group ==2], var.equal = T)

## 
##  Two Sample t-test
## 
## data:  sleep$extra[sleep$group == 1] and sleep$extra[sleep$group == 2]
## t = -1.8608, df = 18, p-value = 0.07919
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -3.363874  0.203874
## sample estimates:
## mean of x mean of y 
##      0.75      2.33

#####################
class(data_1)

## [1] "summary.aov" "listof"

data_3 <- data_1[[1]]
data_3$`Pr(>F)`

## [1] 0.07918671         NA

data_2 <- unlist(data_1)
class(data_2)

## [1] "numeric"

data_2

##         Df1         Df2     Sum Sq1     Sum Sq2    Mean Sq1    Mean Sq2 
##  1.00000000 18.00000000 12.48200000 64.88600000 12.48200000  3.60477778 
##    F value1    F value2     Pr(>F)1     Pr(>F)2 
##  3.46262676          NA  0.07918671          NA

data_3 <- data.frame(data_2)
data_3 

##               data_2
## Df1       1.00000000
## Df2      18.00000000
## Sum Sq1  12.48200000
## Sum Sq2  64.88600000
## Mean Sq1 12.48200000
## Mean Sq2  3.60477778
## F value1  3.46262676
## F value2          NA
## Pr(>F)1   0.07918671
## Pr(>F)2           NA