#install libraries
library(tidyverse)── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
✔ ggplot2 3.3.6 ✔ purrr 0.3.5
✔ tibble 3.1.8 ✔ dplyr 1.0.10
✔ tidyr 1.2.1 ✔ stringr 1.4.1
✔ readr 2.1.3 ✔ forcats 0.5.2
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag() masks stats::lag()library(readxl)
library(ggplot2)Use the LungCapData to answer the following questions.
# read in data
lung_cap<- read_xls("_data/LungCapData.xls")
lung_cap# 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
# … with 715 more rows#plot histogram
hist(lung_cap$LungCap)
The histogram shows that the distribution is pretty close to a normal distribution, with an almost a bell shaped curve. Meaning that the data near the mean are more of a frequent occurrence which is true because there are fewer observations near the margins.
#Box plot
ggplot(lung_cap, aes (Gender,LungCap)) + geom_boxplot()
The box plot shows that male’s have a slightly higher lung capacity than females. Female’s have more values in the first quartile range and a lower minimum value than male’s. On the other hand male’s have a higher max value and more values in the 3rd quartile range.
lung_cap %>%
group_by(Smoke) %>%
summarise(avg_lung_cap = mean(LungCap))# A tibble: 2 × 2
Smoke avg_lung_cap
<chr> <dbl>
1 no 7.77
2 yes 8.65This does not make sense. Smokers have a higher sample mean than non-smokers which intuitively does not make sense because we would assume non-smokers would have a higher lung capacity.
#categorical variable of age_groups
df<- lung_cap %>%
group_by(Smoke,LungCap) %>%
summarise(age_group = case_when(Age<=13 ~ "13 & under",Age ==14 | Age == 15 ~ "14 to 15",Age== 16 | Age == 17 ~ "16 to 17", Age>=18 ~ "18 & older")) `summarise()` has grouped output by 'Smoke', 'LungCap'. You can override using
the `.groups` argument.#mean of lung capacity with new variable
df %>%
group_by(Smoke, age_group) %>%
summarise(avg_lung_cap = mean(LungCap)) %>%
arrange(desc(avg_lung_cap))`summarise()` has grouped output by 'Smoke'. You can override using the
`.groups` argument.# A tibble: 8 × 3
# Groups: Smoke [2]
Smoke age_group avg_lung_cap
<chr> <chr> <dbl>
1 no 18 & older 11.1
2 yes 18 & older 10.5
3 no 16 to 17 10.5
4 yes 16 to 17 9.38
5 no 14 to 15 9.14
6 yes 14 to 15 8.39
7 yes 13 & under 7.20
8 no 13 & under 6.36#histogram
ggplot(df, aes(x = LungCap)) + facet_grid(Smoke ~age_group) +geom_histogram()`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Using the package ggplot I used the function facet_grids to show a good visualization of the lung capacity between non-smokers and smokers within each age group. Looking at the histograms all age_groups that are non-smokers have a higher sample mean proving that non-smokers have
# Mean of non-smokers 13 & younger
df %>%
filter(Smoke == 'no' & age_group == '13 & under') %>%
pull(LungCap) %>%
mean()[1] 6.358746# Mean of smokers 13 & younger
df %>%
filter(Smoke == 'yes' & age_group == '13 &under') %>%
pull(LungCap) %>%
mean()[1] NaN#Mean of non-smokers 14 to 15
df %>%
filter(Smoke == 'no' & age_group == '14 to 15') %>%
pull(LungCap) %>%
mean()[1] 9.13881#Mean of smokers 14 to 15
df %>%
filter(Smoke == 'yes' & age_group == '14 to 15') %>%
pull(LungCap) %>%
mean()[1] 8.391667#Mean of non-smokers 16 to 17
df %>%
filter(Smoke == 'no' & age_group == '16 to 17') %>%
pull(LungCap) %>%
mean()[1] 10.46981#Mean of smokers 16 to 17
df %>%
filter(Smoke == 'yes' & age_group == '16 to 17') %>%
pull(LungCap) %>%
mean()[1] 9.38375#Mean of non-smokers 18 & older
df %>%
filter(Smoke == 'no' & age_group == '18 & older') %>%
pull(LungCap) %>%
mean()[1] 11.06885# Mean of smokers 18 & older
df %>%
filter(Smoke == 'yes' & age_group == '18 & older') %>%
pull(LungCap) %>%
mean()[1] 10.51333lung_cap# 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
# … with 715 more rows#Correlation
cor(lung_cap$LungCap, lung_cap$Age)[1] 0.8196749#Covariance
cov(lung_cap$LungCap, lung_cap$Age)[1] 8.738289The correlation is 0.81 which is pretty close to +1, meaning that the there is a positive relationship between lung capacity and age. The covariance is also high which shows that the two variables of lung capacity and age have a positive relationship. #2 Let X = number of prior convictions for prisoners at a state prison at which there are 810 prisoners.
# x= prior convictions
x<-c(0, 1, 2, 3, 4)
frequency<-c(128, 434, 160, 64, 24)
state_prison<- data_frame(x,frequency)Warning: `data_frame()` was deprecated in tibble 1.1.0.
ℹ Please use `tibble()` instead.state_prison# A tibble: 5 × 2
x frequency
<dbl> <dbl>
1 0 128
2 1 434
3 2 160
4 3 64
5 4 24# 2 prior convictions.P(2)/total
160/810[1] 0.1975309There is a 1.9% probability that a randomly selected inmate has exactly 2 prior convictions.
# less than 2 prior convictions. (P(0) + P(1))/total
(128+434)/810[1] 0.6938272There is a 6.9% probability that a randomly selected inmate has fewer than 2 prior convictions.
# 2 or fewer prior convictions. (P(0) + P(1) + P(2)) +total
(128+434+160)/810[1] 0.891358There is a 8.9% probability that a randomly selected inmate has 2 or fewer prior convictions.
# More than 2 prior convictions. (P(3) +P(4)) + total
(64+24)/810[1] 0.108642There is a 10.8% probability that a randomly selected inmate has more than 2 prior convictions.
#Prior convictions. ((0*P(0)) +(1*(P(1)) + (2*P(2)) + (3*P(3)) + (4*P(4)))
df<-((128*0/810) +(434*1/810) +(160*2/810) +(64*3/810) +(24*4/810))
mean(df)[1] 1.28642The expected value for number of prior convictions is 1.28
# variance
v<- ((0-1.28)^2) *(128/810) +((1-1.28)^2) * (434/810)+((2-1.28)^2) * (160/810)+((3-1.28)^2) *(64/810) +((4-1.28)^2) * (24/810)
v[1] 0.8562765# standard deviation
sd<- sqrt(v)
sd[1] 0.9253521The variance is 0.856 and the standard deviation is 0.925.