Duryea Homework 1

hw1

descriptive statistics

probability

Homework 1 for DACSS 603

Author

Emily Duryea

Published

October 3, 2022

Question 1

Code

library(ggplot2)
library(dplyr)


Attaching package: 'dplyr'

The following objects are masked from 'package:stats':

    filter, lag

The following objects are masked from 'package:base':

    intersect, setdiff, setequal, union

Code

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

hist(df$LungCap)

Part A: Plotting probability density histogram

Code

hist(df$LungCap, 
     col="yellow",
     border="black",
     prob = TRUE,
     xlab = "LungCap",
     main = "Density Plot")

lines(density(df$LungCap),
      lwd = 2,
      col = "chocolate3")

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

Code

ggplot(df, aes(y = dnorm(LungCap), color = Gender)) +
  geom_boxplot() +
  labs(title = "LungCap Probability Distribution for Males and Females", y = "Probability density")

Part C: Compare the mean lung capacities for smokers and non-smokers

Code

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

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

The means of smokers vs non smokers does not make sense since non smokers have a lower mean lung cap, when one would think it would be the other way around. However, limited data is provided on the sample, so there could be other factors in play.

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

df %>%
  ggplot(aes(y = LungCap, color = Smoke)) +
  geom_histogram(bins = 25) +
  facet_wrap(vars(AgeGroup)) +
  theme_classic() + 
  labs(title = "LungCap and Smoke based on age groups", y = "Lung Capacity", x = "Frequency")

Based on the histograms, Part D seems to contrast with Part C, since the plots seem to demonstrate non-smokers having higher lung capacity than smokers in all age groups. Additionally, lung capacity appears to decrease with age based on the graph.

Part E: Compare the lung capacities for smokers and non-smokers within each age group

Code

df %>%
  ggplot(aes(x = Age, y = LungCap, color = Smoke)) +
  geom_line() +
  facet_wrap(vars(Smoke)) +
  labs(title = "LungCap and Smoke based on age and smoker vs nonsmoker", y = "Lung Capacity", x = "Age")

Based on information gained in Part D and Part E, it appears that lung capacity decreases with age, and, despite the means in Part C, lung capacity is higher for non-smokers.

Part F: Calculate the correlation and covariance between Lung Capacity and Age

Code

Cov_lungcapage <- cov(df$LungCap, df$Age)
Cor_lungcapeage <- cor(df$LungCap, df$Age)
Cov_lungcapage

[1] 8.738289

Code

Cor_lungcapeage

[1] 0.8196749

Because both the covariance and correlation are positive numbers, the relationship between lung capacity and age are positively related, meaning as one increases, the other also increases in a proportional manner.

Question 2

Code

Prior_Convictions <- c(0:4)
Inmate_Number <- c(128, 434, 160, 64, 24)
ip <- tibble(Prior_Convictions, Inmate_Number)

ip <- mutate(ip, Probability = Inmate_Number/sum(Inmate_Number))
ip

# A tibble: 5 × 3
  Prior_Convictions Inmate_Number Probability
              <int>         <dbl>       <dbl>
1                 0           128      0.158 
2                 1           434      0.536 
3                 2           160      0.198 
4                 3            64      0.0790
5                 4            24      0.0296

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

Code

ip %>%
  filter(Prior_Convictions == 2) %>%
  select(Probability)

# A tibble: 1 × 1
  Probability
        <dbl>
1       0.198

The probability that a randomly selected inmate has exactly two prior convictions is 0.1975309.

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

Code

partb <- ip %>%
  filter(Prior_Convictions < 2)
sum(partb$Probability)

[1] 0.6938272

The probability that a randomly selected inmate has fewer than two prior convictions is 0.6938272.

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

Code

partc <- ip %>%
  filter(Prior_Convictions <= 2)
sum(partc$Probability)

[1] 0.891358

The probability that a randomly selected inmate has two or fewer prior convictions is 0.891358.

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

Code

partd <- ip %>%
  filter(Prior_Convictions > 2)
sum(partd$Probability)

[1] 0.108642

The probability that a randomly selected inmate has more than two prior convictions is 0.108642.

Part E: What is the expected value for the number of prior convictions?

Code

ip <- mutate(ip, vl = Prior_Convictions*Probability)
parte <- sum(ip$vl)
parte

[1] 1.28642

The expected value for the number of prior convictions is 1.28642.

Part F: Calculate the variance and the standard deviation for the Prior Convictions

Code

ip_var <-sum(((ip$Prior_Convictions-parte)^2)*ip$Probability)
ip_var

[1] 0.8562353

Code

sqrt(ip_var)

[1] 0.9253298

The variance for prior convictions is 0.8562353 and the standard deviation is 0.9253298.