download.file("http://www.openintro.org/stat/data/bdims.RData", destfile = "bdims.RData")
load("bdims.RData")
head(bdims)
## bia.di bii.di bit.di che.de che.di elb.di wri.di kne.di ank.di sho.gi
## 1 42.9 26.0 31.5 17.7 28.0 13.1 10.4 18.8 14.1 106.2
## 2 43.7 28.5 33.5 16.9 30.8 14.0 11.8 20.6 15.1 110.5
## 3 40.1 28.2 33.3 20.9 31.7 13.9 10.9 19.7 14.1 115.1
## 4 44.3 29.9 34.0 18.4 28.2 13.9 11.2 20.9 15.0 104.5
## 5 42.5 29.9 34.0 21.5 29.4 15.2 11.6 20.7 14.9 107.5
## 6 43.3 27.0 31.5 19.6 31.3 14.0 11.5 18.8 13.9 119.8
## che.gi wai.gi nav.gi hip.gi thi.gi bic.gi for.gi kne.gi cal.gi ank.gi
## 1 89.5 71.5 74.5 93.5 51.5 32.5 26.0 34.5 36.5 23.5
## 2 97.0 79.0 86.5 94.8 51.5 34.4 28.0 36.5 37.5 24.5
## 3 97.5 83.2 82.9 95.0 57.3 33.4 28.8 37.0 37.3 21.9
## 4 97.0 77.8 78.8 94.0 53.0 31.0 26.2 37.0 34.8 23.0
## 5 97.5 80.0 82.5 98.5 55.4 32.0 28.4 37.7 38.6 24.4
## 6 99.9 82.5 80.1 95.3 57.5 33.0 28.0 36.6 36.1 23.5
## wri.gi age wgt hgt sex
## 1 16.5 21 65.6 174.0 1
## 2 17.0 23 71.8 175.3 1
## 3 16.9 28 80.7 193.5 1
## 4 16.6 23 72.6 186.5 1
## 5 18.0 22 78.8 187.2 1
## 6 16.9 21 74.8 181.5 1
mdims <- subset(bdims, sex == 1)
fdims <- subset(bdims, sex == 0)
#Exercise 1. Make a histogram of men's heights and a histogram of women's heights.
#How would you compare the various aspects of the two distributions?
fhgtmean <- mean(fdims$hgt)
fhgtsd <- sd(fdims$hgt)
hist(fdims$hgt, probability = TRUE)
x <- 140:190
y <- dnorm(x = x, mean = fhgtmean, sd = fhgtsd)
lines(x = x, y = y, col = "blue")
#Exercise 2. Based on the this plot, does it appear that the data follow a nearly normal
#distribution?
qqnorm(fdims$hgt)
qqline(fdims$hgt)
sim_norm <- rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)
#Exercise 3. Make a normal probability plot of sim_norm. Do all of the points fall on the line?
#How does this plot compare to the probability plot for the real data?
qqnormsim(fdims$hgt)
#Exercise 4. Does the normal probability plot for fdims$hgt look similar to the plots created for
#the simulated data? That is, do plots provide evidence that the female heights are nearly
#normal?
#Exercise 5. Using the same technique, determine whether or not female weights appear to come
#from a normal distribution.
1 - pnorm(q = 182, mean = fhgtmean, sd = fhgtsd)
## [1] 0.004434387
sum(fdims$hgt > 182) / length(fdims$hgt)
## [1] 0.003846154
#Exercise 6. Write out two probability questions that you would like to answer; one regarding
#female heights and one regarding female weights. Calculate the those probabilities using both
#the theoretical normal distribution as well as the empirical distribution (four probabilities
#in all). Which variable, height or weight, had a closer agreement between the two methods?
#On Your Own
# Now let's consider some of the other variables in the body dimensions data set.
#Using the figures at the end of the exercises, match the histogram to its normal probability plot.
#All of the variables have been standardized (first subtract the mean, then divide by the standard deviation), so the units won't be of any help. If you are uncertain based on these figures, generate the plots in R to check.
#a. The histogram for female biiliac (pelvic) diameter (bii.di) belongs to normal probability
#plot letter __B__.
#b. The histogram for female elbow diameter (elb.di) belongs to normal probability plot
#letter __C__.
#c. The histogram for general age (age) belongs to normal probability plot letter __D__.
##d. The histogram for female chest depth (che.de) belongs to normal probability plot letter __A__.
#2. ote that normal probability plots C and D have a slight stepwise pattern.
#Why do you think this is the case?
# The data set is not perfect normal distribution.
#3. As you can see, normal probability plots can be used both to assess normality and
#visualize skewness. Make a normal probability plot for female knee diameter (kne.di).
#Based on this normal probability plot, is this variable left skewed, symmetric,
#or right skewed? Use a histogram to confirm your findings.
qqnorm(fdims$kne.di)
qqline(fdims$kne.di)
hist(fdims$kne.di, probability = TRUE)
#Based on the plot, the variable is right skewed.