Question 1208428: Good afternoon, please help me solve this problem.
During the baking of a certain type of bread roll, each bread goes through a series of heat process. The length of time spent under this heat treatment is related to the lifespan of the bread rolls. A sample of eight bread rolls that underwent different baking times was selected and the life span (hours) of each roll was recorded as follows:
Length of time 18 13 18 15 10 12 8 4
Life span 23 20 18 16 14 11 10 7
a) Calculate the Pearson’s correlation coefficient and explain the result [10 Marks
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Answer: 0.8703
This value is approximate.
Explanation
This value can be quickly calculated using technology.
Here are some options:- Use the LinReg command on a TI83 (or similar). Make sure that Diagnostic is set to "on".
- Use the the CORREL command in a spreadsheet.
- Input the data into the spreadsheet mode of GeoGebra. Refer to this video for an example.
- Search out "pearson correlation coefficient calculator" to find many free tools online. Two search results I found were this and this (among many other results).
- Use a programming language such as R or Python.
- There are many other technology options. I encourage the student to explore his/her favorite.
If you aren't allowed to use a specialized correlation coefficient calculator, then check out any of the three methods shown below.
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Method 1
Define the following
x = length of baking time (in minutes)
y = life span (in hours)
The given table is
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 |
There are n = 8 columns of values.
In other words, there are 8 pairs of (x,y) values.
Add up the x values and divide by n = 8 to find the sample mean of the x values.
xbar = sample mean of x values = x with a horizontal bar over top
xbar = (sum of x values)/(number of x values)
xbar = (18+13+18+15+10+12+8+4)/8
xbar = 98/8
xbar = 12.25
If you are using spreadsheet software, then the function you can use is called Average
Apply the same idea to the y values.
ybar = (sum of y values)/(number of y values)
ybar = (23+20+18+16+14+11+10+7)/8
ybar = 119/8
ybar = 14.875
Introduce a new row labeled xy. This is the product of the x and y values.
For example, we'll have x*y = 18*23 = 414 as the first entry of this row.
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 | | xy | 414 | 260 | 324 | 240 | 140 | 132 | 80 | 28 |
Add up the items in this new row to get Sum(xy) = 414+260+324+240+140+132+80+28 = 1618
Spreadsheet software is strongly recommended.
The last two things we need are:
SD(x)
SD(y)
which represent the sample standard deviations of the x and y values respectively.
You could find these by hand (see part d of this page) but I'll use technology to speed things up.
SD(x) = 4.862392
SD(y) = 5.409978
These values are approximate. I rounded each to 6 decimal places.
Make sure you use the sample standard deviation and NOT the population standard deviation.
Let's calculate the correlation coefficient.
 - \text{n}*\overline{\text{x}}*\overline{\text{y}}}{(\text{n}-1)*SD(\text{x})*SD(\text{y})})
 - 8*12.25*14.875}{(8-1)*4.862392*5.409978})
Many textbooks will round to 4 decimal places. If your teacher instructs otherwise, then be sure to follow said instructions.
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Method 2
Return back to this table.
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 | | xy | 414 | 260 | 324 | 240 | 140 | 132 | 80 | 28 |
Let's introduce rows labeled x^2 and y^2.
As their name suggests, we'll square each value.
Eg: x^2 = 18^2 = 324
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 | | xy | 414 | 260 | 324 | 240 | 140 | 132 | 80 | 28 | | x^2 | 324 | 169 | 324 | 225 | 100 | 144 | 64 | 16 | | y^2 | 529 | 400 | 324 | 256 | 196 | 121 | 100 | 49 |
Define the following
A = sum of the x values = 98
B = sum of the y values = 119
C = sum of the xy values = 1618
D = sum of the x^2 values = 1366
E = sum of the y^2 values = 1975
Spreadsheet software can quickly compute each of these.
We can now compute the correlation coefficient.
*(n*E - B^2) } })
*(8*1975 - 119^2) } })
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Method 3
Let's return back to the original table.
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 |
Recall these previous calculations
xbar = 12.25
SD(x) = 4.862392
We'll use them to determine the z score of each x value.
For instance, if x = 18, then,
z = (x-xbar)/( SD(x) )
z = (18-12.25)/( 4.862392 )
z = 1.182545545485
z = 1.182546
Repeat similar calculations for the other x values. This new row will be labeled Zx to indicate the z scores of the x values.
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 | | Zx | 1.182546 | 0.154245 | 1.182546 | 0.565565 | -0.462735 | -0.051415 | -0.874055 | -1.696696 |
For the y values we have
ybar = 14.875
SD(y) = 5.409978 (approximate)
Here's what it looks like to introduce the z scores for the y values.
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 | | Zx | 1.182546 | 0.154245 | 1.182546 | 0.565565 | -0.462735 | -0.051415 | -0.874055 | -1.696696 | | Zy | 1.501855 | 0.947324 | 0.577636 | 0.207949 | -0.161738 | -0.716269 | -0.901113 | -1.455644 |
As mentioned earlier, spreadsheet software is strongly recommended.
The next thing we'll do is multiply each Zx and Zy value to form a new row.
Example: 1.182546*1.501855 = 1.776013 approximately
| x | 18 | 13 | 18 | 15 | 10 | 12 | 8 | 4 | | y | 23 | 20 | 18 | 16 | 14 | 11 | 10 | 7 | | Zx | 1.182546 | 0.154245 | 1.182546 | 0.565565 | -0.462735 | -0.051415 | -0.874055 | -1.696696 | | Zy | 1.501855 | 0.947324 | 0.577636 | 0.207949 | -0.161738 | -0.716269 | -0.901113 | -1.455644 | | Zx*Zy | 1.776013 | 0.14612 | 0.683081 | 0.117609 | 0.074842 | 0.036827 | 0.787622 | 2.469785 |
The decimal values are approximate.
The values in that bottom row add to 6.091899 which is also approximate.
We can finally calculate the r value.
r = Sum(Zx*Zy)/(n-1)
r = 6.091899/(8-1)
r = 0.870271285714
r = 0.8703
Other methods are possible.
I'll let the student explore them.
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Further Reading
https://www.statology.org/pearson-correlation-coefficient/
https://statisticsbyjim.com/basics/correlation-coefficient-formula/
https://www.statisticshowto.com/probability-and-statistics/correlation-coefficient-formula/
https://www.howtoexcel.org/correlation-coefficient/
More Practice
https://www.algebra.com/algebra/homework/word/evaluation/Evaluation_Word_Problems.faq.question.1196643.html
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