Measuring Central Tendancy

Measures of Central Tendency (Averages): The mean and median both attempt to measure the center of a dataset.

• The mean of a set of observations is the traditional average. We typically denote the mean by \(\bar{x}\) (or \(\mu\) in the case of population-level data) and it is computed as follows: \(\bar{x} = \frac{\displaystyle{\sum_{i=1}^{n}{x_i}}}{n} = \frac{x_1+x_2+x_3+...+x_n}{n}\)

• The median is the middle value for a set of observations. To compute the median, list the numbers in ascending order and find the number or number(s) in the middle of the list. In the case that there is a single middle number, that is the median. In the case where there are two middle numbers, we take the mean of those two.

1. Use the code block below to compute the mean of Sample_One

In R we can easily compute the means and medians for our samples or for the entire dataset! Remember from our most recent tutorial that the $ operator can be used to access an entire column of a data frame. I’ve stored the samples in a data frame called samples. R includes a function mean() for computing the mean of a list of numbers and a function median() for computing the median. This means that we could compute the mean of Sample_Two using mean(samples$Sample_Two).

2. Use the code block below so that it computes the mean of Sample_Three using the $ operator to access the Sample_Three column of the samples data frame.

mean(samples$Sample_Three)

3. Use the median() function and the code block below to compute the median of each of the samples and then answer the question that follows.

4. In our first tutorial we saw that we can use sample data to make generalizations about populations for which the sample is representative. Answer the following questions with this in mind.
   Quiz

Aside: Defining my own data

For data which is not already known to R (ie. data which is not part of a data frame), we can still use R to quickly perform compuations. Consider the distributions of doors knocked on by two political campaign workers last week (Monday - Friday): \(\begin{array}{lcl} \text{Worker A} & : & 23,~24,~25,~26,~27\\ \text{Worker B:} & : & 0,~15,~25,~35,~50\end{array}\). We do this below with the help of the c() function in R, which can be used to create lists of values.

The following code block finds the mean and median for Worker A – execute the code block to find the mean and median. Once you’ve done this for Worker A, add two lines to the bottom of the code block so that it also finds the mean and median for Worker B.

mean(c(23, 24, 25, 26, 27))
median(c(23, 24, 25, 26, 27))

mean(c(23, 24, 25, 26, 27))
median(c(23, 24, 25, 26, 27))
mean(c(0, 15, 25, 35, 50))
median(c(0, 15, 25, 35, 50))

5. Use your explorations of the means and medians for the poll workers to answer the following question.

The standard deviation of a set of observations is denoted by \(s\) (or \(\sigma\) in the case of population-level data) and is computed as follows: \[s = \sqrt{\frac{\displaystyle{\sum_{i=1}^{n}{\left(x_i-\bar{x}\right)^2}}}{n-1}}\]

We should also note that if you are certain that you are working with population-level data, then the denominator used to compute the standard deviation should be changed to \(N\) (the population size). We can do this because there is no uncertainty in estimating the population standard deviation if we have records from every element of the population.

Explaining the Standard Deviation Formula: The standard deviation seeks to measure an “average deviation” from the mean.

• If we don’t look too closely at the formula, we can see the summation symbol \(\left(\sum\right)\) as well as division (by just about the number of values we’ve added up). That’s almost like an average!
• What are we averaging? The quantity \(\left(x - \bar{x}\right)\) denotes an observed value’s deviation from the mean. We shouldn’t average these values though, since the mean sits in center of the data and we would have deviations above the mean (positive) “cancelling out” deviations below the mean (negative).
  • We square the deviations which has two effects: (1) all of the squared deviations are now non-negative, so that no cancellation can occur, and (2) large deviations from the mean carry a larger weight in measuring the standard deviation.
• Since we squared the deviations before computing the “average”, the units of measure are no longer comparable to the original units that the variable was measured in – the units are square units now. This is why we see the large square root as the last piece of the formula – taking the square root brings us back to the original units.

The inter-quartile range (IQR) of a set of observations measures the spread of the “middle-50-percent” of the observations. The IQR is the distance between \(Q1\) (the 25th percentile) and \(Q3\) (the 75th percentile).
* The median of a set of observations splits the set into two halves: an upper half and a lower half. The median of the lower half is called the first quartile (\(Q_1\)) while the median of the upper half is called the third quartile (\(Q_3\)). The interquartile range is the distance between \(Q_1\) and \(Q_3\). That is, \[IQR = Q_3-Q_1\]

6. Check your intuition about the standard deviation and interquartile range by answering the questions below.
   Quiz

The two plots below are a histogram (left) and a boxplot (right), each showing the distribution of carat-weights for the diamonds in our population.

• The histogram does a nice job showing the true shape of the data, but does not always do a good job showing the presence of outliers. The purple line has been added to the histogram to show the true mean carat weight.
• The boxplot doesn’t show the detailed shape that the histogram does, but it does a great job showing the IQR, median, and any outliers present.
  • The lone dots in the boxplot show any outliers (extending more than 1.5 times the IQR, the distance from \(Q1\) to \(Q3\)).
  • The box in the boxplot shows the IQR – the left edge of the box is at \(Q1\) and the right edge of the box is at \(Q3\).
  • The line through the middle of the box denotes the location of the median.
  • I’ve added a faded purple line to our boxplot, showing the location of the mean carat weight. We can really see the impact of those outliers on the mean here.

A Note on Skew: It is common to refer to data as skewed if the presence of outliers cause the mean and median to disagree with one another on the location of the “center” of our data. In this case, we say that the data is skewed in the direction that those outliers have pulled the mean. For example, we would say that the carat weight data (from above) is skewed right.

In R we can easily compute the standard deviation with the function sd(), and IQR with the function quantiles() or IQR(), for our samples or for the entire dataset! Recall that our diamond samples are stored in a data frame called samples. The code block below is preset to compute the standard deviation, Q1, Q3, and IQR for Sample_One. Note that in the quantiles() function the 0.25 identifies the 25th percentile (\(Q1\)) and the 0.75 identifies the 75th percentile (\(Q3\)). Run the code to find the standard deviation, Q1, Q3, and the interquartile range for Sample_One. Once you’ve done that, edit the existing code to compute these metrics for the other two samples.

sd(samples$Sample_One)
quantile(samples$Sample_One, c(0.25, 0.75))
IQR(samples$Sample_One)

Remark: Our third sample of diamond carat sizes contained an outlier. The presence of this outlier drastically impacted the computed mean and standard deviation, but didn’t have much (if any) effect on the median or \(IQR\). Because of this, we say that the median and \(IQR\) are robust statistics in the presence of outliers.

In R we can also easily explore these measures of spread for our campaign workers from earlier. Recall their door-knocking data: \(\begin{array}{lcl} \text{Worker A} & : & 23,~24,~25,~26,~27\\ \text{Worker B:} & : & 0,~15,~25,~35,~50\end{array}\)

Use the code blocks below to find the standard deviation and IQR for the doors visited by the campaign workers.

Below, we can see the distributions of diamond cut from Sample_Two (left) and from our entire population (right) below. Even with a sample of 8 diamonds, we gain “some” insight as to the most and least common diamond cuts. You may also notice that the frequency and relative frequency plots look identical aside from the scale on the vertical axis – this will be the case in general.