The Data

This week we’ll be working with measurements of body dimensions. This data set contains measurements from 247 men and 260 women, most of whom were considered healthy young adults.

load("more/bdims.RData")

Let’s take a quick peek at the first few rows of the data.

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

Since males and females tend to have different body dimensions, it will be useful to create two additional data sets: one with only men and another with only women.

mdims <- subset(bdims, sex == 1)
fdims <- subset(bdims, sex == 0)
  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?

Answer: The average of the heights of man (177.74) is 10cm taller than women (164.87) in this sample. But the standard deviation of men (7.1836) is widerthan women’s (6.5446).

Men’s heights

mhgtmean <- mean(mdims$hgt)
mhgtsd   <- sd(mdims$hgt)
mhgtmean
## [1] 177.7453
mhgtsd 
## [1] 7.183629
hist(mdims$hgt, probability = TRUE)
mx <- 140:250
my <- dnorm(x = mx, mean = mhgtmean, sd = mhgtsd)
lines(x = mx, y = my, col = "blue")

###Women’s heights

fhgtmean <- mean(fdims$hgt)
fhgtsd   <- sd(fdims$hgt)
fhgtmean
## [1] 164.8723
fhgtsd 
## [1] 6.544602
hist(fdims$hgt, probability = TRUE)
fx <- 140:190
fy <- dnorm(x = fx, mean = fhgtmean, sd = fhgtsd)
lines(x = fx, y = fy, ylim = c(0, 0.15), col = "blue")

Women’s and men’s heights(blue->men, yellow->women)

Comparing men’s heights data and women’s, they are normal distributions and men’s cure shifts to right since the average of men is taller than women’s.

hist(rnorm(247, mean=mhgtmean, sd=mhgtsd), col='light blue',
 xlim=c(140, 250))
hist(rnorm(260, mean=fhgtmean, sd=fhgtsd), col='light yellow',
 add=T)

Evaluating the normal distribution

An alternative approach involves constructing a normal probability plot, also called a normal Q-Q plot for “quantile-quantile”.

  1. Based on the this plot, does it appear that the data follow a nearly normal distribution?

Answer: Yes, it follows a nearly normal since the points closes to the line.

Men’s heights

qqnorm(mdims$hgt)
qqline(mdims$hgt)

Women’s heights

qqnorm(fdims$hgt)
qqline(fdims$hgt)

A data set that is nearly normal will result in a probability plot where the points closely follow the line. Any deviations from normality leads to deviations of these points from the line. The plot for female heights shows points that tend to follow the line but with some errant points towards the tails. We’re left with the same problem that we encountered with the histogram above: how close is close enough?

A useful way to address this question is to rephrase it as: what do probability plots look like for data that I know came from a normal distribution? We can answer this

  1. 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?

Answer: Almost all simulation data fall on the line. They are more close to the line than the real data. by simulating data from a normal distribution using rnorm.

Simulation of women’s heights

sim_fnorm <- rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)

qqnorm(sim_fnorm)
qqline(sim_fnorm)

  1. 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?

Answer: Yes. The female heights are nearly normal since mean (164.87) and standard deviation (6.5446) of the real data are very close to the simulation data’s mean (165.51) and standard deviation (6.1027).

sim_fhgtmean <- mean(sim_fnorm)
sim_fhgtsd   <- sd(sim_fnorm)
sim_fhgtmean
## [1] 164.4472
sim_fhgtsd 
## [1] 6.088038
  1. Using the same technique, determine whether or not female weights appear to come from a normal distribution.

Answer:The female weights appear to come from a normal distribution with small positive skew.However, since there are some women weights bigger than the mean and lead to deviations of these points from the line, the distribution of female weights has a long tail on rigth side. The QQplot shows this deviattion that there are some points at low weights and high weights aren’t close to the line.

Women’s weights

fwgtmean <- mean(fdims$wgt)
fwgtsd   <- sd(fdims$wgt)

hist(fdims$wgt, probability = TRUE)
fx <- 30:120
fy <- dnorm(x = fx, mean = fwgtmean, sd = fwgtsd)
lines(x = fx, y = fy, ylim = c(0, 0.10), col = "blue")

Women’s weights

qqnorm(fdims$wgt)
qqline(fdims$wgt)

Simulation of women’s weights

sim_fnorm <- rnorm(n = length(fdims$wgt), mean = fwgtmean, sd = fwgtsd)
qqnorm(sim_fnorm)
qqline(sim_fnorm)

Normal probabilities

If we assume that female heights are normally distributed (a very close approximation is also okay), we can find this probability by calculating a Z score and consulting a Z table (also called a normal probability table). In R, this is done in one step with the function pnorm.

1 - pnorm(q = 182, mean = fhgtmean, sd = fhgtsd)
## [1] 0.004434387

Note that the function pnorm gives the area under the normal curve below a given value, q, with a given mean and standard deviation. Since we’re interested in the probability that someone is taller than 182 cm, we have to take one minus that probability.

Assuming a normal distribution has allowed us to calculate a theoretical probability. If we want to calculate the probability empirically, we simply need to determine how many observations fall above 182 then divide this number by the total sample size.

sum(fdims$hgt > 182) / length(fdims$hgt)
## [1] 0.003846154

Although the probabilities are not exactly the same, they are reasonably close. The closer that your distribution is to being normal, the more accurate the theoretical probabilities will be.

  1. 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?

Answer:Regarding female heights, the approximate probability of female who are taller than 182 (0.004434387) is close to its theoretical probability (0.003846154) since the distribution of female heights is small.

However, for female weights, the approximate probability of female who are heavier than 90 (0.001116107) is significan smaller to its theoretical probability (0.007692308) since the distribution of female heights is big.

The approximate probability of female who are heavier than 90

1 - pnorm(q = 90, mean = fwgtmean, sd = fwgtsd)
## [1] 0.001116107

The theoretical probability of female who are heavier than 90

sum(fdims$wgt > 90) / length(fdims$wgt)
## [1] 0.007692308

On Your Own

qqnorm(fdims$bii.di)
qqline(fdims$bii.di)

**b.** The histogram for female elbow diameter (`elb.di`) belongs to normal 
probability plot letter __C__.
qqnorm(fdims$elb.di)
qqline(fdims$elb.di)

**c.** The histogram for general age (`age`) belongs to normal probability 
plot letter __D__.
qqnorm(fdims$age)
qqline(fdims$age)

**d.** The histogram for female chest depth (`che.de`) belongs to normal 
probability plot letter __A__.
qqnorm(fdims$che.de)
qqline(fdims$che.de)

Answer: Both distributions of female’s age and chest depth are positive (right) skewness since they have long tail on the right sides of the normal curves.

Women’s age

fagemean <- mean(fdims$age)
fagesd   <- sd(fdims$age)

hist(fdims$age, probability = TRUE)
fx <- 10:80
fy <- dnorm(x = fx, mean = fagemean, sd = fagesd)
lines(x = fx, y = fy, ylim = c(0, 0.10), col = "red")

Women’s chest depth

fche.demean <- mean(fdims$che.de)
fche.desd   <- sd(fdims$che.de)

hist(fdims$che.de, probability = TRUE)
fx <- 0:80
fy <- dnorm(x = fx, mean = fche.demean, sd = fche.desd)
lines(x = fx, y = fy, ylim = c(0, 0.10), col = "orange")

  • 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.

Answer: It is right skewed since it has long tail on right side of the curve. The qq plot also shows the deviations on the higher value of the line.

Women’s knee diameter normal probability plot

fkne.dimean <- mean(fdims$kne.di)
fkne.disd   <- sd(fdims$kne.di)

hist(fdims$kne.di, probability = TRUE)
fx <- 0:40
fy <- dnorm(x = fx, mean = fkne.dimean, sd = fkne.disd)
lines(x = fx, y = fy, ylim = c(0, 0.50), col = "green")

Women’s knee diameter histogram

hist(rnorm(260, mean=fkne.dimean, sd=fkne.disd), col='green')

qqnorm(fdims$kne.di)
qqline(fdims$kne.di)

histQQmatch

histQQmatch