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Using Correlation to Describe Relationships between two Quantitative Variable.

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Using Correlation to Describe Relationships between two Quantitative Variable.

- When we describe the association between two variables, we can use a scatterplot to help our description.
- However, words like strong, moderate, and weak to determine the strength of the relationship can be very subjective. (Remember the saying. “Beauty is in the eye of the beholder”)
- So statisticians have another tool, a numeric measure, to help us clarify and be somewhat consistent when we describe these relationships.

- In the lesson on scatterplots, we indicated that a tighter oval around the data points indicated a stronger relationship. In some sense this must mean that the closer the points are to each other, the stronger the relationship.
- Pearson’s Correlation Coefficient helps us to numerically measure this “spread” of our data.

- We know that in describing a distribution that we think about both the “center” of the distribution, and the “spread” of the distribution.
- When looking at the relationship between two variables, we need to consider both the “center” and the “spread” of each, and how the combination of these two distributions interact.

- “r” is unitless, which allows us to change scales or calculate the relationship between two variables that are not the same units
- “r” measures the linear relationship between two quantitative variables.
- -1 ≤ r ≤ 1
- The sign of “r” indicates the direction of the relationship
- The closer “r” is to either +1 or -1, the stronger the relationship.
- The closer “r” is to 0, the weaker the relationship.

- Physical Sciences
“Hard Sciences”

≥ .80---Strong

.50 --.80—Moderate

≤ .50—Weak

- Social Sciences
“Soft Sciences”

≥ .50---Strong

.30 --.50—Moderate

≤ .30—Weak

Remember that these numbers are just guidelines. Each set of data is different and the context for the data must be considered.

Notice that the formula is adding terms together (we’ll talk about what those terms are shortly) and then dividing that sum by 1 less than the number of data points we have. So, it appears that we are looking for “an average” of sorts.

Now the terms that we are adding together are the product of z-scores.

Remember that a z-score is the number of standard deviations a piece of data is from the mean of the distribution.

So each term is the product of the z-scores in each direction (x and y) for each point.

So, how can we calculate this value?

- We can calculate “r” using this formula and the lists.
- L1X (amount of fat)
- L2Y (calories)
- L3Zx (x-xbar)/sx
- L4Zy (y-ybar)/sy
- L5L3*L4
- Once these lists are created, find the sum of L5 and then divide by n-1

Starting with our original formula

Now, the standard deviation of our x-values and the y-values are constants once our data has been collected, so they will be the same for each term in the summation.

This means that we can factor those out of the sum leaving:

Now, expanding the summation gives us:

Now, using the distributive property to multiply the binomials in each term gives:

Then, collapsing the sums gives:

Now, the ∑xiand the ∑yican be written as nxbar and nybar

But two of the last three terms cancel each other out, so we are left with:

- This formula is helpful to us because our calculator gives us each of the terms we see here
- With our data in the lists, L1(fat) and L2(calories) in this case, we calculate the 2-var stats to find these values.

- Now, calculate the 2-var stats for L1, L2
- STAT
CALC

- This gives us all the values we need to calculate “r”
- We can then describe numerically the relationship between amount of fat and calories in a burger.

Now, substituting the values for each of the variables we find that the correlation coefficient,

r=.96, indicating a strong, linear correlation in which as the amount of fat in the burger increases, so does the calories

r= .9606

- What about the point we determined was a long way away from the rest of our data----our possible “outlier”
- Since this burger appears to be somewhat different than the rest of our data, it would be wise to report the correlation both with and without it.

- If we calculate the correlation coefficient without this piece of data, we find that it drops to .8367. This change indicates that this piece of data is unusual

- In the next section, we will look at even another way to find the value of Pearson’s correlation coefficient.
- For now, either method used in this lesson is appropriate.

- The Practice of Statistics—YMM
- Pg 128 – 136

- The Practice of Statistics—YMS
- Pg 140-149

- The Basic Practice of Statistics—Moore
- Pg 88-94