Homework 1

DACSS 603

Load Libraries

Code

library(tidyverse)

-- Attaching packages --------------------------------------- tidyverse 1.3.2 --
v ggplot2 3.4.0      v purrr   0.3.5 
v tibble  3.1.8      v dplyr   1.0.10
v tidyr   1.2.1      v stringr 1.5.0 
v readr   2.1.3      v forcats 0.5.2 
-- Conflicts ------------------------------------------ tidyverse_conflicts() --
x dplyr::filter() masks stats::filter()
x dplyr::lag()    masks stats::lag()

Code

library(ggplot2)
library(formattable)

Question 1

a

First, let’s read in the data from the Excel file:

Code

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

The distribution of LungCap looks as follows:

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hist(df$LungCap, xlab= 'LungCap', main = 'Histogram of LungCap Distribution')

The histogram suggests that the distribution is close to a normal distribution. Most of the observations are close to the mean. Very few observations are close to the margins (0 and 15).

##b

Compare the probability distribution of the LungCap with respect to Males and Females:

Code

boxplot(df$LungCap~df$Gender, xlab = 'Gender', ylab = 'LungCap')

##c

Compare the mean lung capacities for smokers and non-smokers. Does it make sense?

Code

LungCap_mean <- df %>% 
  group_by(Smoke) %>%
  summarise(Lungcap_Mean = mean(LungCap))

LungCap_mean

# A tibble: 2 x 2
  Smoke Lungcap_Mean
  <chr>        <dbl>
1 no            7.77
2 yes           8.65

These results show the lung capacity for non-smokers is lower than the lung capacity for smokers. With no background knowledge of lung capacity, I would think these are not standard results.

##d)

Examine the relationship between Smoking and Lung Capacity within age groups: “less than or equal to 13”, “14 to 15”, “16 to 17”, and “greater than or equal to 18”.

Code

df_Agegroup <- df %>% 
  mutate(
    Age_group = dplyr::case_when(
      Age <= 13            ~ "0-13",
      Age > 13 & Age <= 15 ~ "14-15",
      Age > 15 & Age <= 17 ~ "16-17",
      Age >= 18             ~ "18+"))


ggplot(data = df_Agegroup, aes(x=Age_group, y=LungCap)) + 
  geom_boxplot(aes(fill=Gender))

##e)

Compare the lung capacities for smokers and non-smokers within each age group. Is your answer different from the one in part c. What could possibly be going on here?

Code

ggplot(data = df_Agegroup, aes(x=Age_group, y=LungCap)) + 
  geom_boxplot(aes(fill=Smoke))

This data is different from what was seen in part C. The only age group displaying greater lung capacity for smokers is the group 0-13. This would suggest this sample of the population is influencing the overall lung capacity mean. This by far is the largest sample of age groups with 428 participants

Code

count(df_Agegroup, Age_group, Smoke)

# A tibble: 8 x 3
  Age_group Smoke     n
  <chr>     <chr> <int>
1 0-13      no      401
2 0-13      yes      27
3 14-15     no      105
4 14-15     yes      15
5 16-17     no       77
6 16-17     yes      20
7 18+       no       65
8 18+       yes      15

Question 2

First create the dataframe:

Code

stateprison <- data.frame (X  = c("0", "1", "2", "3", "4"),
                  Frequency = c("128", "434", "160", "64", "24")
                  )
stateprison$Frequency <- as.integer(stateprison$Frequency)
stateprison$X <- as.integer(stateprison$X)
stateprison

  X Frequency
1 0       128
2 1       434
3 2       160
4 3        64
5 4        24

##a

What is the probability that a randomly selected inmate has exactly 2 prior convictions?

Code

formattable::percent(160/summarise(stateprison, sum(Frequency)))

[1] 19.75%

##b

What is the probability that a randomly selected inmate has fewer than 2 prior convictions?

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formattable::percent((128+434)/summarise(stateprison, sum(Frequency)))

[1] 69.38%

##c

What is the probability that a randomly selected inmate has 2 or fewer prior convictions?

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formattable::percent((128+434+160)/summarise(stateprison, sum(Frequency)))

[1] 89.14%

##d

What is the probability that a randomly selected inmate has more than 2 prior convictions?

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formattable::percent((64+24)/summarise(stateprison, sum(Frequency)))

[1] 10.86%

##e

What is the expected value for the number of prior convictions?

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stateprison_PRB <- transform(stateprison, PROB = (Frequency/810))

df_prison_ev <- transform(stateprison_PRB, XP = X*PROB )


EV <- sum(df_prison_ev$XP)
print(EV)

[1] 1.28642

##f

Calculate the variance and the standard deviation for the Prior Convictions.

Variance:

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varX <-sum(((df_prison_ev$X-EV)^2)*df_prison_ev$PROB)
print(varX)

[1] 0.8562353

Standard deviation:

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sdX <- sqrt(varX)
print(sdX)

[1] 0.9253298