Homework 1

Code

library(readxl)
library(dplyr)
library(magrittr)
knitr::opts_chunk$set(echo = TRUE)

Question 1

Loading the Data

Code

df <- read_excel("_data/LungCapData.xls")
df

# A tibble: 725 × 6
   LungCap   Age Height Smoke Gender Caesarean
     <dbl> <dbl>  <dbl> <chr> <chr>  <chr>    
 1    6.48     6   62.1 no    male   no       
 2   10.1     18   74.7 yes   female no       
 3    9.55    16   69.7 no    female yes      
 4   11.1     14   71   no    male   no       
 5    4.8      5   56.9 no    male   no       
 6    6.22    11   58.7 no    female no       
 7    4.95     8   63.3 no    male   yes      
 8    7.32    11   70.4 no    male   no       
 9    8.88    15   70.5 no    male   no       
10    6.8     11   59.2 no    male   no       
# ℹ 715 more rows

A

Code

summary(df)

    LungCap            Age            Height         Smoke          
 Min.   : 0.507   Min.   : 3.00   Min.   :45.30   Length:725        
 1st Qu.: 6.150   1st Qu.: 9.00   1st Qu.:59.90   Class :character  
 Median : 8.000   Median :13.00   Median :65.40   Mode  :character  
 Mean   : 7.863   Mean   :12.33   Mean   :64.84                     
 3rd Qu.: 9.800   3rd Qu.:15.00   3rd Qu.:70.30                     
 Max.   :14.675   Max.   :19.00   Max.   :81.80                     
    Gender           Caesarean        
 Length:725         Length:725        
 Class :character   Class :character  
 Mode  :character   Mode  :character  
                                      
                                      
                                      

Code

hist(df$LungCap, xlab = "Lung Capacity", main = "", freq = F)

The histogram shows that the distribution is almost a normal distribution with most of the values close to the mean. ## B

Code

boxplot(LungCap ~ Gender, df)

The minimum, first quartile, median, third quartile and maximum, all of them appear to be slightly higher for males than females.

C

Code

dff<- df |>
  group_by(Smoke) |>
  summarise(LungCap = mean(LungCap))

Lung capacity of non-smokers is higher than that of smokers which is against the expectation

D

Age less than or equal to 13

Code

df1 <- df |>
  filter(Age <= 13) |>
  group_by(Smoke)|>
  summarise(LungCap = mean(LungCap)) |>
  mutate(Age = "<=13")
df1

# A tibble: 2 × 3
  Smoke LungCap Age  
  <chr>   <dbl> <chr>
1 no       6.36 <=13 
2 yes      7.20 <=13 

Age 14 & 15

Code

df2 <- df |>
  filter(Age >= 14 & Age <= 15) |>
  group_by(Smoke)|>
  summarise(LungCap = mean(LungCap))|>
  mutate(Age = "14&15")
class(df2)

[1] "tbl_df"     "tbl"        "data.frame"

Age 16 to 17

Code

df3 <- df |>
  filter(Age >= 16 & Age <= 17) |>
  group_by(Smoke)|>
  summarise(LungCap = mean(LungCap))|>
  mutate(Age = "16&17")

Age greater than or equal to 18

Code

df4 <- df |>
  filter(Age >= 18) |>
  group_by(Smoke)|>
  summarise(LungCap = mean(LungCap))|>
  mutate(Age = ">=18")

E

Code

new_df <- rbind(df1,df2, df3,df4)
new_df

# A tibble: 8 × 3
  Smoke LungCap Age  
  <chr>   <dbl> <chr>
1 no       6.36 <=13 
2 yes      7.20 <=13 
3 no       9.14 14&15
4 yes      8.39 14&15
5 no      10.5  16&17
6 yes      9.38 16&17
7 no      11.1  >=18 
8 yes     10.5  >=18 

Code

ggplot(new_df, aes(fill=Smoke, y=LungCap, x=Age)) +
  geom_bar(position='dodge', stat='identity')

Error in ggplot(new_df, aes(fill = Smoke, y = LungCap, x = Age)): could not find function "ggplot"

Code

ggplot(dff, aes(y=LungCap, x=Smoke, fill = Smoke)) +
  geom_bar(position='dodge', stat='identity')  

Error in ggplot(dff, aes(y = LungCap, x = Smoke, fill = Smoke)): could not find function "ggplot"

Only age group that showed similar result that of step C is <=13. The analysis could go wrong if the lung capacity of smokers and non smokers are studied without considering the age.

Question 2

Code

df_prison <- tibble(X = c(0,1,2,3,4), Frequency = c(128,434,160,64,24))

df_prison

# A tibble: 5 × 2
      X Frequency
  <dbl>     <dbl>
1     0       128
2     1       434
3     2       160
4     3        64
5     4        24

Code

n <- sum(df_prison$Frequency)
n

[1] 810

Code

df_prison |>
  filter(X == 2) |>
  pull(Frequency) |>
  divide_by(n)

[1] 0.1975309

Code

#log(write) = 3.135 + 0.115 female + 0.0066 read + 0.0077 math



#write = - 99.164 + 5.389 female + 20.941 log(math) + 16.852 log(read)