Question 1196570: Because it is not practical to weigh bears in the field, researchers sought to develop a model to predict a bear's weight based on its length. Here are the results for a sample (this is the exact same data as the previous question):
Total Length (cm) Weight (kg)
139.0 110
138.0 60
139.0 90
120.5 60
149.0 85
141.0 100
141.0 95
150.0 85
166.0 155
151.5 140
129.5 105
150.0 110
The explanatory variable is .
If the bear length was instead measured in inches (there are 2.54 cm in one inch) and the weight was measured in pounds (there are 2.2 pounds in one kg), then the linear correlation coefficient would .
Found 2 solutions by ewatrrr, math_tutor2020:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
LENTH VS WEIGHT OF BEAR
First listed in cm & kg, then as listed in inches & pounds
139 110 353.06 242
138 60 350.52 132
139 90 353.06 198
120.5 60 306.07 132
149 85 378.46 187
141 100 358.14 220
141 95 358.14 209
150 858**Note typo 381 1887.6
166 155 421.64 341
151.5 140 384.81 308
129.5 105 328.93 231
150 110 381 242
Using Excel Function CORREL FOR BOTH LISTINGS to find
the linear correlation coefficient
Yes. Entered correctly in Excel...
r = 0.703903 r= 0.703903
As one would expect, this demonstrates that changing the units
has NO bearing on r.
Wish You the Best in your Studies.
Answer by math_tutor2020(3816)  (Show Source):
You can put this solution on YOUR website!
I don't know how the tutor @ewatrrr is getting r = 0.284846442 as that isn't correct.
The correct r value is approximately r = 0.703903176239897
which when rounded to four decimal places is roughly r = 0.7039
You can use any correlation coefficient calculator you like to confirm this.
Here is one such online calculator
http://www.alcula.com/calculators/statistics/correlation-coefficient/
Refer to this page to see an example how the r value is calculated
https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1196643.html
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However, @ewatrrr is correct in stating that the r value doesn't change when you scale the data sets consistently across the board.
Consider this example set A = {1,2,3,4}
The mean is 2.5 which is found by adding up the values and dividing by the sample size n = 4
Now multiply each item by 5 to get B = {5,10,15,20} and the mean of this is 12.5; this is a jump of "times 5" since 2.5*5 = 12.5
Therefore, the mean has scaled exactly the same way as the other values.
The standard deviation will also scale up by 5 through more complicated steps.
Apply that type of logic to this current data set and you'll find the mean and standard deviation are scaled the same amount.
If for instance we multiplied everything in the x data set by 5 then the old Zx = (x-xbar)/sigma will turn into the new Zx = (5x - 5*xbar)/(5*sigma), in which the 5's cancel
The 5 isn't that special and you can use any constant you like. The multiplier will cancel out anyway.
The Zx for the original x data set, and the Zx for the scaled x data set, are the same.
The Zx value doesn't change as long as we apply the same multiplier to each x.
Similar reasoning shows Zy doesn't change either. You don't have to use the same multiplier as you did with the x data set.
Hence the value
r = Sum(ZxZy)/(n-1)
will remain the same.
This fact is mentioned in the article here
https://en.wikipedia.org/wiki/Pearson_correlation_coefficient
Quote: "A key mathematical property of the Pearson correlation coefficient is that it is invariant under separate changes in location and scale in the two variables. That is, we may transform X to a + bX and transform Y to c + dY, where a, b, c, and d are constants with b, d > 0, without changing the correlation coefficient."
Also, this article mentions it as well
https://online.stat.psu.edu/stat509/lesson/18/18.1
scroll down to the portion that reads "The Pearson correlation coefficient is invariant to location and scale transformations"
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