Lab6: Goodness of Fit

1 Exercise

1.1 Exercise 1

Please work out in R by doing a chi-squared test on the treatment (X) and improvement (Y) columns in treatment.csv.

Answer

data   <- read.csv("C:/Users/user/Downloads/treatment.csv")  # Read Data
head(data, 10)                                               # Shows 10 from 105 data of treatment

##    id   treatment  improvement
## 1   1     treated     improved
## 2   2     treated     improved
## 3   3 not-treated     improved
## 4   4     treated     improved
## 5   5     treated not-improved
## 6   6     treated not-improved
## 7   7 not-treated not-improved
## 8   8     treated not-improved
## 9   9 not-treated     improved
## 10 10     treated     improved

• \(H_0\) : The two variables are independent.
• \(H_a\) : The two variables are dependent.

table(data$treatment, data$improvement)

##              
##               improved not-improved
##   not-treated       26           29
##   treated           35           15

chisq.test(data$treatment, data$improvement, correct = FALSE)    # Chi-Square test

## 
##  Pearson's Chi-squared test
## 
## data:  data$treatment and data$improvement
## X-squared = 5.5569, df = 1, p-value = 0.01841

So, the chi-squared value is 5.5569 and p-value is 0.01841 (less than the significance level of 0.05). Then, we reject the null hypothesis and conclude that the two variables are dependent.

1.2 Exercise 2

Find out if the cyl and carb variables in mtcars dataset are dependent or not.

Answer

• \(H_0\) : The two variables are independent.
• \(H_a\) : The two variables are dependent.

data("mtcars")                           # Data of `mtcars`
table(mtcars$carb, mtcars$cyl)

##    
##     4 6 8
##   1 5 2 0
##   2 6 0 4
##   3 0 0 3
##   4 0 4 6
##   6 0 1 0
##   8 0 0 1

chisq.test(mtcars$carb, mtcars$cyl)

## Warning in chisq.test(mtcars$carb, mtcars$cyl): Chi-squared approximation may be
## incorrect

## 
##  Pearson's Chi-squared test
## 
## data:  mtcars$carb and mtcars$cyl
## X-squared = 24.389, df = 10, p-value = 0.006632

So, the chi-squared value is 24.389 and p-value is 0.006632 (less than the significance level of 0.05). Then, we reject the null hypothesis and conclude that the two variables are dependent.

1.3 Exercise 3

256 visual artists were surveyed to find out their zodiac sign. The results were: Aries (29), Taurus (24), Gemini (22), Cancer (19), Leo (21), Virgo (18), Libra (19), Scorpio (20), Sagittarius (23), Capricorn (18), Aquarius (20), Pisces (23). Test the hypothesis that zodiac signs are evenly distributed across visual artists. (Reference)

Answer

• \(H_0\) : Zodiac signs are evenly distributed across visual artists.
• \(H_a\) : Zodiac signs are not evenly distributed across visual artists.

observed <- c(29, 24, 22, 19, 21, 18, 19, 20, 23, 18, 20, 23)
n        <- 256  
expected <- c(1/12) * n
alpha    <- .05
r        <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)
df       <- 12 - 1
(chisq   <- sum((observed - expected)^2 / expected))

## [1] 5.09375

(p_value <- pchisq(q = chisq, df = df, lower.tail = F))

## [1] 0.9265414

chisq.test(observed)                                          # Chi-square Result

## 
##  Chi-squared test for given probabilities
## 
## data:  observed
## X-squared = 5.0938, df = 11, p-value = 0.9265

So, the chi-squared value is 5.0938 and p-value is 0.9265 (more than the significance level of 0.05). Then, we do not reject the null hypothesis and that means the Zodiac signs are evenly distributed across visual artists.