Video Transcript
The table represents the power
output and rotor diameter of several helicopters. Find the Spearman’s rank
correlation coefficient, and round your answer to four decimal places.
We’re given a set of paired data
for the power output and rotor diameter of some helicopters. We use the term paired data because
each pair of data is unique to one helicopter. So, for example, the helicopter
with a power output of 1,218 kilowatts has a rotor diameter of 10.2 meters. And to calculate Spearman’s rank
correlation coefficient, we’ll use the formula given. In this formula, 𝑛 corresponds to
the number of data pairs. 𝑑 𝑖 corresponds to the difference
in ranks for each pair, where 𝑖 takes value from one to 𝑛, and we calculate the
sum of the differences squared.
The first thing we need to do then
is to rank each of our two data sets. And to do this, let’s make some
room. If we begin by ranking the power
output, we could start at either the lowest or the highest power output. It should make no difference to the
Spearman’s correlation coefficient, provided we stick to the same direction for the
rotor diameter rankings. So let’s start with the last power
output, which is 944, which we rank as one. And to avoid confusion later on,
let’s strike this out. Our next lowest power output is
1,218, so we can strike this out and rank this two. And the next lowest is 1,864, which
we can rank third. 3,324 can be ranked fourth, 3,552
is ranked fifth, 3,758 is ranked sixth, and our highest power output is 4,698, which
is ranked seventh.
And now for our rotor diameters,
our lowest value is 10.2 meters. But this occurs twice, so
effectively we have tied ranks for the first place. How this works statistically,
however, is we take the average of the places that these data points would
occupy. That is first and second places so
that the ranks of the two data points with values 10.2 or one plus two over two. That is the first place and the
second place over two, which is 1.5, so that both of our instances of a rotor
diameter of 10.2 meters are ranked 1.5. And we can strike these two
out.
Now, our third lowest value is 14,
so we can strike this out. And since first and second places
have already been taken by the 10.2 values, we must rank this third. Our next lowest value is 16.2,
which we rank fourth, followed by 16.3, which is ranked fifth, followed by 17.7,
which is ranked sixth, and finally 18.59, which is ranked seventh.
Now, to use our formula, we need
the differences in ranks squared for each data pair. So let’s first take the differences
in ranks. To do this, we subtract the
diameter rank from the power rank for each pair so that, in our first data column,
we have two minus 1.5, which is 0.5, for our next column, three minus three, which
is zero, one minus 1.5, which is negative 0.5. We have seven minus seven is zero,
five minus four is one, four minus five is negative one, and six minus six is
zero.
Our next step is to work out the
differences squared. In our first column, 0.5 squared is
0.25. In our second data column, zero
squared is zero. In our third column, negative 0.5
squared is 0.25. In our fourth column, zero squared
is zero. In our fifth column, one squared is
one. In our sixth column, negative one
squared is one. And in our final column, zero
squared is zero.
Now for a formula, we want the sum
of the differences squared. That is 0.25 plus zero plus 0.25
plus zero plus one plus one plus zero, which is 2.5. Now, before we use the formula,
let’s just check that the sum of the differences is equal to zero as it should
be. We have 0.5 plus zero plus negative
0.5 plus zero plus one plus negative one plus zero, and that is indeed equal to
zero.
Now we have seven pairs of data so
that our 𝑛 is equal to seven. And so Spearman’s rank correlation
coefficient is one minus six times 2.5 over seven times seven squared minus one. That is one minus 15 over 336. If you do this on your calculator,
it’s very important at this point to separate the one from the fraction. And to do this, we calculate 15
divided by 336; that’s 0.04464. And so Spearman’s rank correlation
coefficient for this data is 0.9554 to four decimal places.
