DATA 606 - LAB 3
Alexander Ng
March 2, 2019
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.
library(tidyverse) # Include all of the tidyverse libraries
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 1You’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 at http://www.openintro.org/stat/data/bdims.php, 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, sex == 1)
fdims <- subset(bdims, sex == 0)- 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? (ALEXANDER NG RESPONSE)
ggplot(data=mdims) + geom_histogram(mapping=aes(x=hgt ), binwidth=4, fill="lightblue") +
ggtitle("Male Heights")
ggplot(data=fdims) + geom_histogram(mapping=aes(x=hgt ), binwidth=4, fill="#FF9988") +
ggtitle("Female Heights")
ggplot(data=bdims) +
geom_histogram(mapping=aes(x=hgt, fill=sex) , binwidth=4, alpha=.5, position="identity" ) +
scale_fill_discrete(name="Gender", labels=c("Female", "Male")) + ggtitle("Overlapping Histogram\n of Males and Females")
Means and medians are almost identical for both male and females samples. We can see that distributions are quite symmetric.
Looking at the differences between males and females from the distributions, males are taller than females by an average of 12.87 centimeters. Male heights show larger dispersion with a higher standard deviation of 7.18 than females height standard deviation of 6.54.
We expect the distributions to be symmetric and bell-shaped. However, the distribution of female heights shows a data artifact (based on bin size selection) that suggests a bimodal distribution when using a binwidth of 2 to 3 cm. This is counterintuitive and likely to be caused by the small sample size. Hence we use binwidth=4 cm.
summary(mdims$hgt) # Key statistics of male heights## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 157.2 172.9 177.8 177.7 182.7 198.1summary(fdims$hgt) # Key statistics of female heights## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 147.2 160.0 164.5 164.9 169.5 182.9round( sd(mdims$hgt) , digits =2 ) # Males height standard deviation## [1] 7.18round( sd(fdims$hgt) , digits =2 ) # Females height standard deviation## [1] 6.54The 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$hgt, probability = TRUE)
x <- 140:190
y <- dnorm(x = x, mean = fhgtmean, sd = fhgtsd)
lines(x = x, y = y, col = "blue")
After plotting the density histogram with the first command, we create the x- and y-coordinates for the normal curve. We chose the x range as 140 to 190 in order to span the entire range of fheight. To create y, we use dnorm to calculate the density of each of those x-values in a distribution that is normal with mean fhgtmean and standard deviation fhgtsd. The final command draws a curve on the existing plot (the density histogram) by connecting each of the points specified by x and y. The argument col simply sets the color for the line to be drawn. If we left it out, the line would be drawn in black.
The top of the curve is cut off because the limits of the x- and y-axes are set to best fit the histogram. To adjust the y-axis you can add a third argument to the histogram function: ylim = c(0, 0.06).
- Based on the this plot, does it appear that the data follow a nearly normal distribution?
(Alexander Ng’s Response)
We show the histogram without the top of curve being cutoff but use ggplot2 instead of base graphics. Looking at the chart below, the normal distribution provides a good fit to the data. So my answer is yes.
ggplot(fdims, aes(hgt) ) +
geom_histogram(aes(y=..density..), fill="#FF9988", binwidth=4) +
stat_function(fun = dnorm, args=c(mean=fhgtmean, sd = fhgtsd), col = "blue" )
Evaluating the normal distribution
Eyeballing the shape of the histogram is one way to determine if the data appear to be nearly normally distributed, but it can be frustrating to decide just how close the histogram is to the curve. An alternative approach involves constructing a normal probability plot, also called a normal Q-Q plot for “quantile-quantile”.
qqnorm(fdims$hgt)
qqline(fdims$hgt)
A data set that is nearly normal will result in a probability plot where the points closely follow the line. Any deviations from normality leads to deviations of these points from the line. The plot for female heights shows points that tend to follow the line but with some errant points towards the tails. We’re left with the same problem that we encountered with the histogram above: how close is close enough?
A useful way to address this question is to rephrase it as: what do probability plots look like for data that I know came from a normal distribution? We can answer this by simulating data from a normal distribution using rnorm.
sim_norm <- rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)The first argument indicates how many numbers you’d like to generate, which we specify to be the same number of heights in the fdims data set using the length function. The last two arguments determine the mean and standard deviation of the normal distribution from which the simulated sample will be generated. We can take a look at the shape of our simulated data set, sim_norm, as well as its normal probability plot.
- 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?
(Alexander Ng’s Response)
We fix the seed to the pseudo-random number generator for ease of analysis. The sequence of points clearly show deviations from normality in the randomly generated data at the both left and right tails. Thus, normally distributed data can also show departures from linearity in the QQ plot.
set.seed(117) # Fix the seed
sim_norm <- rnorm(n = length(fdims$hgt), mean = fhgtmean, sd = fhgtsd)
qqnorm(sim_norm)
qqline(sim_norm)
Even better than comparing the original plot to a single plot generated from a normal distribution is to compare it to many more plots using the following function. It may be helpful to click the zoom button in the plot window.
qqnormsim(fdims$hgt)
- Does the normal probability plot for
fdims$hgtlook similar to the plots created for the simulated data? That is, do plots provide evidence that the female heights are nearly normal?
(Alexander Ng’s Response)
The distribution of females heights appears to deviate slightly from the qqplot simulations. The tails of the empirical distribution appear to bend away from the \(x=y\) line slightly more.
- Using the same technique, determine whether or not female weights appear to come from a normal distribution.
(Alexander Ng’s Response)
The distribution of female heights is mostly normally distributed within the first 1.7 standard deviations from the mean of 0. However, the sample distribution appears to have skinnier tails than the normal distribution. This makes biological sense. Left tail corresponds to dwarfs and right tails correspond to giantesses. There are likely physical limits to the size of human bodies reflected in the tail distribution properties. We conclude the female weights don’t come from a normal distribution but another distribution with slightly tighter tails.
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 6 feet (about 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] 0.004434387Note 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] 0.003846154Although 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.
- 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?
(Alexander Ng’s Response)
We consider two empirical question for female heights and weights.
What is the probability that female heights are below 5 feet? This is equivalent to 152.4 cm.
What is the probability that female weights exceed 190 pounds? This is equivalent to 86.18 kg.
We estimate the probability that heights are less than 152.4 cm empirically. Then we compare the probability that heights are below 152.4 cm using a normal distribution.
sum( fdims$hgt < 152.4) /length(fdims$hgt) #empirical estimate## [1] 0.02692308pnorm( 152.4, mean = fhgtmean, sd = fhgtsd) #normal distribution prob estimate## [1] 0.028342We estimate the probability that weights exceed 86.18 kg empirically.
sum( fdims$wgt > 86.18 ) /length(fdims$wgt)## [1] 0.01538462we estimate the same probability using the normal distribution below.
1 - pnorm( 86.18 , mean= mean(fdims$wgt), sd = sd(fdims$wgt) )## [1] 0.003904793qqnorm(fdims$wgt)
qqline(fdims$wgt)
qqnorm(fdims$hgt)
qqline(fdims$hgt)
Looking at the above QQ plots, it is clear the weights are more right skewed. Two outliers in weight suggest somes survey participants are obese. The curve of the QQ-plot of the weights has a U-shape consistent with a right skew.
By contrast, the height QQ plot has fewer outliers which are closer to normality. The curve shape suggests narrower tails below and above the mean.
I conclude that the height data better fits the normal distribution.
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 ____. Bb. The histogram for female elbow diameter (
elb.di) belongs to normal probability plot letter ____. Cc. The histogram for general age (
age) belongs to normal probability plot letter ____. Dd. The histogram for female chest depth (
che.de) belongs to normal probability plot letter ____. A
(Alexander Ng’s Additional Responses)
Note that to guess the distribution associated with each histogram, it suffices to identify two key traits: The skewness of the histogram: left or right. The Z-score of the max and min values from the histogram.
One can translate each key trait to the scale or the shape of the curve in the Q-Q plot.
histQQmatch
- Note that normal probability plots C and D have a slight stepwise pattern.
Why do you think this is the case?
(Alexander Ng’s Response)
The probability plots C and D have a stepwise pattern because their data values are more discretized. This suggests either the underlying data uses more discrete values. We see that age and elbow diameter are the most discretized data.
age assume only 37 values.
elbow diameter assumes only 42 values.
- 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.
(Alexander Ng’s response)
Based on the normal probability plot below, female knee diameter distribution is skewed right. Some knee diameters are much larger than predicted.
The histogram below the QQ-plot affirms the right skew distribution.
qqnorm( fdims$kne.di)
qqline( fdims$kne.di)
ggplot(data=fdims) + geom_histogram(mapping=aes(x=kne.di ), bins = 20 , fill="#FF9988") +
ggtitle("Female Knee Diameters Are Right Skewed")