DACSS 603 Homework 1

hw1

desriptive statistics

probability

This is Homework 1 for DASS 603.

Author

Laura Collazo

Published

February 19, 2023

Libraries

Code

library(readxl)
library(tidyverse)

Question 1

Read in data for question 1.

Code

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

a

The distribution of LungCap looks as follows:

Code

hist(df$LungCap)

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

The distribution of LungCap by Gender looks as follows:

Code

boxplot(df$LungCap ~ df$Gender)

The distribution shows that males, on average, have a higher lung capacity than females.

c

The mean lung capacities for smokers and non-smokers is shown below.

Code

df %>% 
  group_by(Smoke) %>% 
  summarise(mean = mean(LungCap))

# A tibble: 2 × 2
  Smoke  mean
  <chr> <dbl>
1 no     7.77
2 yes    8.65

These means are not what I would have expected to see. This shows smokers have a higher mean lung capacity than non-smokers.

d

The mean lung capacities for smokers and non-smokers by age group is shown below.

Code

df_age_group <- df %>% 
  mutate(
    age_group = case_when(
      Age < 14 ~ "less than or equal to 13",
      Age == 14 ~ "14 to 15",
      Age == 15 ~ "14 to 15",
      Age == 16 ~ "16 to 17",
      Age == 17 ~ "16 to 17",
      Age > 17 ~ "greater than or equal to 18")
  ) 

df_age_group %>%
  group_by(age_group, Smoke) %>% 
  summarise(mean = mean(LungCap))

`summarise()` has grouped output by 'age_group'. You can override using the
`.groups` argument.

# A tibble: 8 × 3
# Groups:   age_group [4]
  age_group                   Smoke  mean
  <chr>                       <chr> <dbl>
1 14 to 15                    no     9.14
2 14 to 15                    yes    8.39
3 16 to 17                    no    10.5 
4 16 to 17                    yes    9.38
5 greater than or equal to 18 no    11.1 
6 greater than or equal to 18 yes   10.5 
7 less than or equal to 13    no     6.36
8 less than or equal to 13    yes    7.20

e

The distribution of LungCap by age_group and smoker looks as follows:

Code

ggplot(df_age_group, aes(x=age_group, y=LungCap, color = Smoke)) +
  geom_boxplot()

The mean lung capacity for non-smokers is higher than smokers for all age groups except for those age 13 or younger. This is interesting as it was observed earlier that overall smokers have a higher mean lung capacity than non-smokers. This led me to wonder if there are more individuals age 13 or younger in this sample than other age groups, and also what the count of smokers versus non smokers is for this sample.

The count by age group and smoker type is shown below.

Code

df_age_group %>% 
  group_by(age_group, Smoke) %>% 
  summarise(total_count = n())

`summarise()` has grouped output by 'age_group'. You can override using the
`.groups` argument.

# A tibble: 8 × 3
# Groups:   age_group [4]
  age_group                   Smoke total_count
  <chr>                       <chr>       <int>
1 14 to 15                    no            105
2 14 to 15                    yes            15
3 16 to 17                    no             77
4 16 to 17                    yes            20
5 greater than or equal to 18 no             65
6 greater than or equal to 18 yes            15
7 less than or equal to 13    no            401
8 less than or equal to 13    yes            27

The overall count by smoker type is shown below.

Code

df %>% 
  group_by(Smoke) %>% 
  summarise(total_count = n())

# A tibble: 2 × 2
  Smoke total_count
  <chr>       <int>
1 no            648
2 yes            77

Question 2

Data for question 2.

Code

convictions <- data.frame(number_convictions = 0:4, freq = c(128, 434, 160, 64, 24)) %>% 
  mutate(prop = freq/810)

print(convictions)

  number_convictions freq       prop
1                  0  128 0.15802469
2                  1  434 0.53580247
3                  2  160 0.19753086
4                  3   64 0.07901235
5                  4   24 0.02962963

a

The probability that a randomly selected inmate has exactly 2 prior convictions is as follows:

Code

dbinom(x= 1, size= 1, prob= 160/810)

[1] 0.1975309

b

The probability that a randomly selected inmate has fewer than 2 prior convictions is as follows:

Code

dbinom(x= 1, size= 1, prob= sum(128/810 + 434/810))

[1] 0.6938272

c

The probability that a randomly selected inmate has 2 or fewer prior convictions is as follows:

Code

dbinom(x= 1, size= 1, prob= sum(128/810 + 434/810 + 160/810))

[1] 0.891358

d

The probability that a randomly selected inmate has more than 2 prior convictions is as follows:

Code

dbinom(x= 1, size= 1, prob= sum(64/810 + 24/810))

[1] 0.108642

e

The expected value for the number of prior convictions is as follows:

Code

ev <- sum(convictions$number_convictions * convictions$prop)

print(ev)

[1] 1.28642

f

The variance for prior convictions is as follows:

Code

var <- sum((convictions$number - ev) ^ 2 * convictions$prop)

print(var)

[1] 0.8562353

The standard deviation for prior convictions is as follows:

Code

sd <- sqrt(var)

print(sd)

[1] 0.9253298