OpenIntro Lab 3 - Distributions

Lab 3 - Distributions

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In this lab we’ll investigate the probability distribution that is most central to statistics: the normal distribution. If we are confident that our data are nearly normal, that opens the door to many powerful statistical methods. Here we’ll use the graphical tools of R to assess the normality of our data and also learn how to generate random numbers from a normal distribution.

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(url("http://www.openintro.org/stat/data/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

You’ll see that for every observation we have 25 measurements, many of which are either diameters or girths. A key to the variable names can be found here, but we’ll be focusing on just three columns to get started: weight in kg wgt, height in cm hgt, and sex (1 indicates male, 0 indicates female).

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, bdims$sex == 1)
fdims = subset(bdims, bdims$sex == 0)

Exercise

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?

hist(mdims$hgt)

hist(fdims$hgt)

# draw both at the same plot
# from http://stackoverflow.com/a/3557042/861423

p1 = hist(mdims$hgt)

p2 = hist(fdims$hgt)

xlim = c(min(fdims$hgt), max(mdims$hgt)) + c(-5, 5)
plot( p1, col=rgb(0,0,1,1/4), xlim=xlim)
plot( p2, col=rgb(1,0,0,1/4), add=T)

Each is symmetric and has one mode: most likely they are normal

Normal Distribution

In your description of the distributions, did you use words like bell-shaped or normal? It’s tempting to say so when faced with a unimodal symmetric distribution.

To see how accurate that description is, we can plot a normal distribution curve on top of a histogram to see how closely the data follow a normal distribution. This normal curve should have the same mean and standard deviation as the data. We’ll be working with women’s heights, so let’s store them as a separate object and then calculate some statistics that will be referenced later.

fhgtmean = mean(fdims$hgt)
fhgtsd = sd(fdims$hgt)

Next we make a density histogram to use as the backdrop and use the lines function to overlay a normal probability curve. The difference between a frequency histogram and a density histogram is that while in a frequency histogram the heights of the bars add up to the total number of observations, in a density histogram the areas of the bars add up to 1. The area of each bar can be calculated as simply the height \(\times\) the width of the bar. Using a density histogram allows us to properly overlay a normal distribution curve over the histogram since the curve is a normal probability density function. Frequency and density histograms both display the same exact shape; they only differ in their y-axis. You can verify this by comparing the frequency histogram you constructed earlier and the density histogram created by the commands below.

hist(fdims$hgt, probability = TRUE, ylim=c(0, 0.07))
x = 140:190
y = dnorm(x = x, mean = fhgtmean, sd = fhgtsd)
lines(x = x, y = y, col = "blue")

Exercise

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

Yes

Evaluating the normal distribution

Eyeballing the shape of the histogram is one way to determine if the data appear to be nearly normally distributed, but it can be frustrating to decide just how close the histogram is to the curve. An alternative approach involves constructing a normal probability plot, also called a normal Q-Q plot for quantile-quantile.

qqnorm(fdims$hgt, col="orange", pch=19)
qqline(fdims$hgt, lwd=2)

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 by simulating data from a normal distribution using rnorm.

sim.norm = rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)

Exercise

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?

qqnorm(sim.norm, col="orange", pch=19)
qqline(sim.norm, lwd=2)

Looks similar - also not perfect

Even better than comparing the original plot to a single plot generated from a normal distribution is to compare it to many more plots using the following function. It may be helpful to click the “zoom” button in the plot window.

qqnormsim(fdims$hgt)

Exercise

Does the normal probability plot for fdims$hlstd look similar to the plots created for the simulated data? That is, do plots provide evidence that the female heights are nearly normal?

Yes, similar. Looks like it

Exercise

Using the same technique, determine whether or not female weights appear to come from a normal distribution.

wgtm = mean(fdims$wgt)
wgts = sd(fdims$wgt)

xlim = c(min(fdims$wgt), max(fdims$wgt)) + c(-5, +5)
hist(fdims$wgt, probability=T, xlim=xlim)

x = seq(xlim[1], xlim[2], 0.5)
y = dnorm(x=x, mean=wgtm, sd=wgts)
lines(x=x, y=y, col="blue")

looks a bit skewed

qqnorm(fdims$wgt, col="orange", pch=19)
qqline(fdims$wgt, lwd=2)

sim.norm = rnorm(n=length(fdims$wgt), mean=wgtm, sd=wgts)
qqnorm(sim.norm, col="orange", pch=19)
qqline(sim.norm, lwd=2)

qqnormsim(fdims$wgt)

most likely not normal

Normal probabilities

Okay, so now you have a slew of tools to judge whether or not a variable is normally distributed. Why should we care?

It turns out that statisticians know a lot about the normal distribution. Once we decide that a random variable is approximately normal, we can answer all sorts of questions about that variable related to probability. Take, for example, the question of, “What is the probability that a randomly chosen young adult female is taller than 182 cm” (The study that published this data set is clear to point out that the sample was not random and therefore inference to a general population is not suggested. We do so here only as an exercise.)

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] 0.004434

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] 0.003846

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.

Excercise

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?

Question 1

the prob of having height between 160 and 170

# drawing normal curve with shades: 
# http://stackoverflow.com/a/3494767/861423

x = 140:190
y = dnorm(x=x, mean=fhgtmean, sd=fhgtsd)
plot(x=x, y=y, type="l")

x1 = min(which(x >= 160))  
x2 = max(which(x <= 170))

polygon(x=x[c(x1, x1:x2, x2)], y=c(0, y[x1:x2], 0), col="skyblue")

pers.170 = pnorm(q=170, mean=fhgtmean, sd=fhgtsd)
pers.160 = pnorm(q=160, mean=fhgtmean, sd=fhgtsd)

# prop of being smaller than 170
pers.170

## [1] 0.7833

# prob of being smaller than 160
pers.160

## [1] 0.2283

# prop of being smaller than 170 and higher than 160
prob.theor = pers.170 - pers.160

prob.theor

## [1] 0.555

So, the theoretical probability is 0.555

Now let’s calculate the empirical one:

prob.emp = sum(fdims$hgt >= 160 & fdims$hgt <= 170) / length(fdims$hgt)
prob.emp

## [1] 0.5846

the result is 0.584 - not that far away from the theoretical

abs(prob.emp - prob.theor)

## [1] 0.02958

This was quite expected since we knew that the height follows the normal distribution

Question 2:

What’s the probability of having weight greater than 57?

xlim = c(min(fdims$wgt), max(fdims$wgt)) + c(-5, +5)
hist(fdims$wgt, probability=T, xlim=xlim)

x = seq(xlim[1], xlim[2], 0.5)
y = dnorm(x=x, mean=wgtm, sd=wgts)
lines(x=x, y=y, col="blue")

x1 = min(which(x >= 57))
x2 = max(which(x >= 57))

polygon(x=x[c(x1, x1:x2, x2)], y=c(0, y[x1:x2], 0), col=rgb(0,0.5,1,0.5))

Theoretical:

pers.57 = pnorm(q=57, mean=wgtm, sd=wgts)
prob.theor = 1 - pers.57
prob.theor

## [1] 0.646

Emperical

prob.emp = sum(fdims$wgt >= 57) / length(fdims$wgt)
prob.emp

## [1] 0.5846

the result is 0.584 - quite far away from the theoretical

abs(prob.emp - prob.theor)

## [1] 0.06134

This is expectable, because we know that most likely weight doesn’t follow the normal distribution