CS Week 6 : Exercise

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.

data.imelda <- read.csv("treatment.csv")
print(data.imelda)

##      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
## 11   11 not-treated     improved
## 12   12 not-treated not-improved
## 13   13 not-treated not-improved
## 14   14 not-treated not-improved
## 15   15 not-treated     improved
## 16   16 not-treated     improved
## 17   17     treated     improved
## 18   18     treated     improved
## 19   19 not-treated not-improved
## 20   20 not-treated not-improved
## 21   21     treated not-improved
## 22   22 not-treated not-improved
## 23   23     treated not-improved
## 24   24 not-treated     improved
## 25   25     treated     improved
## 26   26     treated     improved
## 27   27 not-treated not-improved
## 28   28 not-treated     improved
## 29   29     treated not-improved
## 30   30     treated     improved
## 31   31 not-treated not-improved
## 32   32 not-treated not-improved
## 33   33     treated     improved
## 34   34 not-treated     improved
## 35   35     treated not-improved
## 36   36 not-treated     improved
## 37   37     treated     improved
## 38   38 not-treated not-improved
## 39   39 not-treated     improved
## 40   40     treated     improved
## 41   41 not-treated     improved
## 42   42 not-treated     improved
## 43   43 not-treated not-improved
## 44   44 not-treated     improved
## 45   45 not-treated     improved
## 46   46     treated     improved
## 47   47     treated not-improved
## 48   48 not-treated not-improved
## 49   49     treated     improved
## 50   50     treated     improved
## 51   51 not-treated not-improved
## 52   52     treated     improved
## 53   53 not-treated     improved
## 54   54     treated     improved
## 55   55     treated     improved
## 56   56 not-treated     improved
## 57   57     treated     improved
## 58   58 not-treated not-improved
## 59   59     treated     improved
## 60   60     treated     improved
## 61   61     treated     improved
## 62   62 not-treated     improved
## 63   63     treated not-improved
## 64   64     treated not-improved
## 65   65 not-treated     improved
## 66   66 not-treated     improved
## 67   67 not-treated     improved
## 68   68 not-treated not-improved
## 69   69 not-treated not-improved
## 70   70     treated     improved
## 71   71     treated not-improved
## 72   72 not-treated not-improved
## 73   73     treated not-improved
## 74   74 not-treated     improved
## 75   75 not-treated not-improved
## 76   76 not-treated not-improved
## 77   77     treated not-improved
## 78   78 not-treated     improved
## 79   79     treated     improved
## 80   80     treated     improved
## 81   81     treated     improved
## 82   82 not-treated not-improved
## 83   83     treated     improved
## 84   84 not-treated not-improved
## 85   85     treated     improved
## 86   86 not-treated     improved
## 87   87 not-treated not-improved
## 88   88     treated     improved
## 89   89 not-treated not-improved
## 90   90     treated     improved
## 91   91 not-treated not-improved
## 92   92 not-treated     improved
## 93   93     treated not-improved
## 94   94     treated not-improved
## 95   95 not-treated not-improved
## 96   96     treated     improved
## 97   97 not-treated     improved
## 98   98     treated     improved
## 99   99 not-treated not-improved
## 100 100 not-treated     improved
## 101 101     treated     improved
## 102 102     treated     improved
## 103 103 not-treated not-improved
## 104 104     treated     improved
## 105 105 not-treated not-improved

table(data.imelda$treatment, data.imelda$improvement)

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

chisq.test(data.imelda$treatment, data.imelda$improvement, correct = FALSE)

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

After we calculate the data, We have a chi-squared value of 5.55. Since we get a p-Value less than the significance level of 0.05.

1.2 Exercise 2

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

data(mtcars)
print(mtcars)

##                      mpg cyl  disp  hp drat    wt  qsec vs am gear carb
## Mazda RX4           21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
## Mazda RX4 Wag       21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
## Datsun 710          22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
## Hornet 4 Drive      21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
## Hornet Sportabout   18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
## Valiant             18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
## Duster 360          14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
## Merc 240D           24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
## Merc 230            22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
## Merc 280            19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4
## Merc 280C           17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4
## Merc 450SE          16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3
## Merc 450SL          17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3
## Merc 450SLC         15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3
## Cadillac Fleetwood  10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4
## Lincoln Continental 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4
## Chrysler Imperial   14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
## Fiat 128            32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
## Honda Civic         30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2
## Toyota Corolla      33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1
## Toyota Corona       21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1
## Dodge Challenger    15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2
## AMC Javelin         15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2
## Camaro Z28          13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4
## Pontiac Firebird    19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2
## Fiat X1-9           27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
## Porsche 914-2       26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2
## Lotus Europa        30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2
## Ford Pantera L      15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4
## Ferrari Dino        19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6
## Maserati Bora       15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8
## Volvo 142E          21.4   4 121.0 109 4.11 2.780 18.60  1  1    4    2

table(mtcars$cyl, mtcars$carb)

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

chisq.test(mtcars$cyl, mtcars$carb, correct = FALSE)

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

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

Of the calculation that have been made, we get high chi-squared value and a p-value of less that 0.05 significance level. When \(a > p-value\) we reject the null hypothesis and conclude that \(carb\) and \(cyl\) have a significant relationship (independent).

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)

category <- c("Aries",
              "Taurus",
              "Gemini",
              "Cancer",
              "Leo",
              "Virgo",
              "Libra",
              "Scorpio",
              "Sagittarius",
              "Capricorn",
              "Aquarius",
              "Pisces")

observed <- c(29,24,22,19,21,18,19,20,23,18,20,23)

expected <- c(256/12)

residual <- c(observed - expected)

obs_exp <- c((observed-expected)^2)

component <- c(obs_exp/expected)


zodiac_sign <- (data.frame(category, 
                        observed, 
                        expected, 
                        residual, 
                        obs_exp, 
                        component))
print(zodiac_sign)

##       category observed expected   residual    obs_exp   component
## 1        Aries       29 21.33333  7.6666667 58.7777778 2.755208333
## 2       Taurus       24 21.33333  2.6666667  7.1111111 0.333333333
## 3       Gemini       22 21.33333  0.6666667  0.4444444 0.020833333
## 4       Cancer       19 21.33333 -2.3333333  5.4444444 0.255208333
## 5          Leo       21 21.33333 -0.3333333  0.1111111 0.005208333
## 6        Virgo       18 21.33333 -3.3333333 11.1111111 0.520833333
## 7        Libra       19 21.33333 -2.3333333  5.4444444 0.255208333
## 8      Scorpio       20 21.33333 -1.3333333  1.7777778 0.083333333
## 9  Sagittarius       23 21.33333  1.6666667  2.7777778 0.130208333
## 10   Capricorn       18 21.33333 -3.3333333 11.1111111 0.520833333
## 11    Aquarius       20 21.33333 -1.3333333  1.7777778 0.083333333
## 12      Pisces       23 21.33333  1.6666667  2.7777778 0.130208333

sum(zodiac_sign$component)

## [1] 5.09375

If we calculate the data manual, we’re going through a lot of steps. The above steps are manual steps by calculating one by one the existing data. Then, add the total of the component variabel as the value of its Chi-squared.

A faster way is to use the \(chisq.test\) function to find out the value of its Chi-squared without having to go through a long step, because we only give 1 variable to be researched.

chisq.test(zodiac_sign$observed, correct = FALSE)

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

Of the calculation that have been made, we get chi-squared value of 5.09 and a p-value is greater than 0.05 significance level. When \(a < p-value\) we accepted the null hypothesis and conclude that zodiac signs are evenly distributed across visual artists.