bivariate data – scatter plots and correlation coefficient…… section 3.1 and 3.2
TRANSCRIPT
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Bivariate Data – Scatter Plots and Correlation Coefficient……
Section 3.1 and 3.2
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2 Quantitative Variables…… We represent 2 variables that are quantitative
by using a scatter plot.
Scatter Plot – a plot of ordered pairs (x,y) of bivariate data on a coordinate axis system. It is a visual or pictoral way to describe the nature of the relationship between 2 variables.
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Input and Output Variables…… X:
a. Input Variable
b. Independent Var
c. Controlled Var
Y:
a. Output Variable
b. Dependent Var
c. Results from the Controlled
variable
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Example…… When dealing with
height and weight, which variable would you use as the input variable and why?
Answer:
Height would be used as the input variable because weight is often predicted based on a person’s height.
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Constructing a scatter plot…… Do a scatter plot of the
following data:
Independent Dependent
Variable Variable
Age Blood Pressure
43 128
48 120
56 135
61 143
67 141
70 152
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What do we look for?...... A. Is it a positive correlation, negative
correlation, or no correlation?
B. Is it a strong or weak correlation?
C. What is the shape of the graph?
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Answer……With TI
Age Blood Pressure
43 128
48 120
56 135
61 143
67 141
70 152
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Notice…… Notice the following:
A. Strong Positive – as x increases, y
also increases.
B. Linear - it is a graph of a line.
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Example 2……By HandIndependent Dependent
Variable Variable
# of Absences Final Grade
6 82
2 86
15 43
9 74
12 58
5 90
8 78
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Example 2……With TIIndependent Dependent
Variable Variable
# of Absences Final Grade
6 82
2 86
15 43
9 74
12 58
5 90
8 78
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Notice…… Notice the following:
A. Strong Negative – As x increases, y decreases
B. Linear – it’s the graph of a line.
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Example 3……By HandIndependent Dependent
Variable Variable
Hrs. of Exercise Amt of Milk
3 48
0 8
2 32
5 64
8 10
5 32
10 56
2 72
1 48
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Example 3……With TIIndependent Dependent
Variable Variable
Hrs. of Exercise Amt of Milk
3 48
0 8
2 32
5 64
8 10
5 32
10 56
2 72
1 48
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Notice…… Notice:
There seems to be no correlation between the hours or exercise a person performs and the amount of milk they drink.
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Steps to see on Calculator…… Put x’s in L1 and y’s in L2
Click on “2nd y=“
Set scatter plot to look like the screen to the right.
Press zoom 9 or set your own window and then press graph.
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Linear CorrelationSection 3.2
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Correlation…… Definition – a
statistical method used to determine whether a relationship exists between variables.
3 Types of Correlation:
A. Positive
B. Negative
C. No Correlation
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Positive Correlation: as x increases, y increases or as x decreases, y decreases.
Negative Correlation: as x increases, y decreases.
No Correlation: there is no relationship between the variables.
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Linear Correlation Analysis …… Primary Purpose: to measure the strength of
the relationship between the variables.
*This is a test question!!!!
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Coefficient of Linear Correlation The numerical measure
of the strength and the direction between 2 variables.
This number is called the correlation coefficient.
The symbol used to represent the correlation coefficient is “r.”
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The range of “r” values…… The range of the correlation coefficient is -1
to +1.
The closer to 0 you get, the weaker the correlation.
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Range……
Strong
Negative No Linear Relationship Strong
Positive
____________________________________ -1 0
+1
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Computational Formula using z-scores of x and y……
1
n
zzr yx
deviationst
meanvaluez
.
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Example 1…… Find the correlation
coefficient (r) of the following example.
Use the lists in the calculator.
x y
2 80
5 80
1 70
4 90
2 60
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Find mean and st. dev first…… Since you will be using a
formula that uses z-scores, you will need to know the mean and standard deviation of the x and y values.
Put x’s in L1 Put y’s in L2 Run stat calc one var
stats L1 – Write down mean & st. dev.
Run stat calc one var stats L2 – Write down mean & st. dev.
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X values: Y values:
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Write down on your paper……You’ll use them later. X Values:
Mean = 2.8
St. Dev = 1.643167673
Y Values:
Mean = 76
St. Dev = 11.40175425
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Calculator Lists……
Set Formula Set Formula Set Formula
L1 L2 L3 = (L1-2.8)/1.643167673 L4 = (L2-76)/11.40175425 L5 = L3 x L4
x y z(of x) z (of y) z (of x) times z(of y)
2 80 -0.4869 0.35082 -0.1708
5 80 1.3389 0.35082 0.46971
1 70 -1.095 -0.5262 0.57646
4 90 0.7303 1.2279 0.89672
2 60 -0.4869 -1.403 0.68321
2.455298358
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Calculate “r”…… From the lists….. n = 5
455298395.2 yx zz61.0
4
455298395.2
1
n
zzr yx
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What does that mean? Since r = 0.61, the
correlation is a moderate correlation.
Do we want to make predictions from this?
It depends on how precise the answer needs to be.
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Example 2…… Find the correlation
coefficient (r) for the following data.
Do you remember what we found from the scatter plot?
Age Blood Pressure
43 128
48 120
56 135
61 143
67 141
70 152
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Let’s do this one together…… Remember to use your lists in the calculator. Don’t round numbers until your final answer. Find the mean and st. dev. for x and y. Explain what you found.
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X Values: Y Values:
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List values you should have……
L1 L2 L3 L4 L5
43 128 -1.368 -0.7458 1.0205
48 120 -0.8965 -1.448 1.2978
n=6 56 135 -0.1415 -0.1316 0.01863
61 143 0.33028 0.57031 0.18836
67 141 0.89647 0.39483 0.35395
70 152 1.1796 1.36 1.6042
4.483364073
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Compute “r”……
897.05
483364073.4
1
n
zzr yx
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Describe it…… Since r = 0.897
Strong Positive Correlation
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Example 3…… Find the correlation
coefficient for the following data.
Do you remember what we found from the scatter plot?
# of Absences Final Grade
6 82
2 86
15 43
9 74
12 58
5 90
8 78
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X Values: Y Values:
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List Values you should have……
L1 L2 L3 L4 L5
6 82 -0.4898 0.53626 -0.2626
2 86 -1.404 0.7746 -1.088
15 43 1.5673 -1.788 -2.802
n=7 9 74 0.19591 0.05958 0.01167
12 58 0.88158 -0.8938 -0.7879
5 90 -0.7183 1.0129 -0.7276
8 78 -0.0327 0.29792 -0.0097
-5.66529102
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Compute “r”……
944.06
66529102.5
1
n
zzr yx
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Describe it…… Since r = -0.944
Strong Negative Correlation
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Example 4…… Find the correlation
coefficient of the following data.
Do you remember what we found from the scatter plot?
Hrs of Exercise Amt of Milk
3 48
0 8
2 32
5 64
8 10
5 32
10 56
2 72
1 48
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X Values: Y Values:
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List Values you should have……Hrs of Exercise Amt of Milk L3 L4 L5
3 48 -0.3015 0.30713 -0.0926
0 8 -1.206 -1.476 1.7804
2 32 -0.603 -0.4062 0.24495
5 64 0.30151 1.0205 0.30768
n=9 8 10 1.206 -1.387 -1.673
5 32 0.30151 -0.4062 -0.1225
10 56 1.8091 0.66379 1.2008
2 72 -0.603 1.3771 -0.8304
1 48 -0.9045 0.30713 -0.2778
0.537689672
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Compute “r”……
067.8
5376896717.
1
n
zzr yx
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Describe It…… Since r = .067
No Correlation…..No correlation exists
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What is It is the coefficient of determination.
It is the percentage of the total variation in y which can be explained by the relationship between x and y.
A way to think of it: The value tells you how much your ability to predict is improved by using the regression line compared with NOT using the regression line.
?2r
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For Example…… If it means that 89% of the variation
in y can be explained by the relationship between x and y.
It is a good fit.
89.2 r
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Assignment…… Worksheet