Session 23 Notes, Section 10-5

Coefficient of Determination - Standard Error of the Estimate

In this unit, we have learned that, if the correlation coefficient is significant, we can determine the equation of the linear regression line.  Using this regression line and given values for the independent variable x, we can "predict" values for the dependent variable "y".  In this session we will learn other measures associated with the correlation coefficient: the coefficient of determination, the standard error of the estimate, and the prediction interval.

At the end of this sesson you will be able to:

• Understand the terms and vocabulary of variation
• Compute the coefficient of determination and the coefficient of nondetermination.
  • Summary: Coefficient of determination
  • Example: determination
  • Example: nondetermination
• Compute the standard error of estimate.
  • Example  (y-y') method
  • Example a,b method
• Find a prediction interval.
  • Example
  • Example
• Your Turn:  Summary

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Vocabulary

• Total Variation
  • The total variation, , is the sum of the squares of the vertical distance each point is from the mean.
  • The total variation can be divided into two parts:
    • that which is attributed to the relationship of x and y,
      • (i.e., from the predicted y' values) is and is called the explained variation.
    • that which is due to chance, found by , is called the unexplained variation. This variation cannot be attributed to the relationships.
  • Hence, the total variation is equal to the sum of the explained variation and the unexplained variation.
    • 
    • For a single point, the differences are called deviations.

Figure 10-17 is a nice illustration of the relationship between these variations.  You do not have go memorize each kind of variation.

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Compute the coefficient of determination.

• The coefficient of determination is the ratio of the explained variation to the total variation.  The symbol for the coefficient of determination is r² and is given by

• The coefficient of determination is a measure of the variation of the dependent variable that is explained by the regression line and the independent variable
  • Also he coefficient of determination is found by squaring correlation coefficient  r and converting it to a percent.
• The coefficient of nondetermination is a measure of the unexplained variation.
  • The formula for the coefficient of nondetermination is:
    • 1.00 - r²

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Summary: Coefficient of Determination

If the correlation relationship between two sets of data is significant, then we can make a regression  line.

A regression line is the data line of best fit for values of x and y and can be used to predict results for y given certain values of x.  To predict values of y means mathematically that y depends on x.  If you studied functions in your algebra classes, this means y is a function of x, y is the dependent variable, and x is the independent variable.

Keep in mind that the regression line is a data line of best fit and, therefore, is only a predictor of y-values.  Therefore, the is a variation between the y-value on the regression line and the y-value for a point.

The "explained" variation is the difference between the y-value of the regression line and the average y-value.  See all of this discussion implies y is the dependent variable depending on the x-values.

Coefficient of determination is a measure of explained variation.  Therefore, the coefficient of determination is a measure of the variation of the dependent variable y (from the regression line) accounted for by the variation of the independent variable x.

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Example of Cofficient of Determination.

Find the coefficient of determination given r = 0.15.

• The coefficient of determination is found by squaring r and converting it to a percent.
  • If r = 0.15, then r² = 0.0225.
    • This result means that 2.25% of the variation in the dependent variable is accounted for by the variations in the independent variable.
    • Note:  it is not surprising that if the correlation coefficient is only .15, the variation that is due to the independent variable is not very great.

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Example of Coefficient of Nondetermination

• Find the coefficient of nondetermination given r = 0.24.
  1. The coefficient of nondetermination is found by subtracting the square of the coefficient of determination from 1.
  2. The symbol for the coefficient of determination is r².
     • Therefore if r = 0.24,
       • then r² = 0.0576.
     • Subtracting from 1 we obtain 1 - r² = 1 - 0.0576 = 0.9424.
  3. Converting this to a percent we have 94.24%.
     • This result means that 94.24% of the variation in the dependent variable is not explained by the variations in the independent variable.
     • Note:  it is not surprising that if the correlation coefficient is only .24, the majority of the variation is not due to the independent variable.

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Compute the standard error of estimate.

The standard error of estimate, denoted by S_est is the standard deviation of the observed y values about the predicted y' values. The formula for the standard error of estimate is:

• 
• A second formula that can be used for the calculation of the standard error of estimate is:
  • 
    • Where a and b are from the equation of the regression line, written as y'=a + bx, where b is the slope of the line and a is the y' intercept.

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Example of Standard of Estimate

A researcher collects the following data and determines that there is a significant relationship between the age of a copy machine and its monthly maintenance cost. The regression equation is y = 54.57 + 8.63x. Find the standard error of the estimate.

Machine       Age x (years)    Monthly cost y A 1 63 B 2 75 C 3 72 D 4 91 E 4 94 F 6 105

Since the formula for the standard error of estimate is:

x             y             y             y - y       (y - y)2 1 63       2 75       3 72       4 91       4 94       6 105

Using the regression line equation y = 54.57 + 8.63x, compute the predicted values y for each y and place the results in the column labeled y
x = 1, y' = 54.57 + (8.63)(1) = 63.2
x = 2, y' = 54.57 + (8.63)(2) = 71.83
x = 3, y' = 54.57 + (8.63)(3) = 80.46
x = 4, y' = 54.57 + (8.63)(4) = 89.09
x = 6, y' = 54.57 + (8.63)(6) = 106.35

For each y, subtract y and place the answer in the column labeled y - y.
63 - 63.2 = -0.20
75 - 71.83 = 3.17
72 - 80.46 = -8.46
 91 - 89.09 = 1.91
94 - 89.09 = 4.91
105 - 106.35 = -1.35

Square the numbers found in the last step and place the squares in the column labeled (y - y)2 and find the sum of the numbers in the last column. The completed table should look like this:

x	y	y	y - y	(y - y)2
1	63	63.2	-0.20	0.04
2	75	71.83	3.17	10.0489
3	72	80.46	-8.46	71.5716
4	91	89.09	1.91	3.6481
4	94	89.09	4.91	24.1081
6	105	106.35	-1.35	1.8225
				111.2392

Substitute in the formula and find S_est.

sest =

(y - y)2 n - 2

=

111.2392 6 - 2

= 5.27

The standard error of the estimate is 5.27.

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Example of Standard Error of Estimate Using the Regression line

Find the standard error of the estimate for the given data using the formula

The regression equation is y = 59.37 + 6.24x.

Machine       Age x (years)    Monthly cost y A 1 67 B 2 70 C 3 76 D 4 82 E 4 90 F 6 96

Round your answer to two decimal places.

1. Make a table
2. Find the product of x and y values, and place the results in the third column.
3. Square the y, and place the results in the fourth column.
4. Find the sums of the second, third, and fourth columns.

The completed table should look like this:

x		y		xy		y2
1		67		67		4489
2		70		140		4900
3		76		228		5776
4		82		328		6724
4		90		360		8100
6		96		576		9216
	y =	481	xy =	1699	y2 =	39205

From the regression equation y = 59.37 + 6.24x,

• a = 59.37 and
• b = 6.24.

Substitute in the formula and find S_est.

sest

=

39205 - (59.37)(481) - (6.24)(1699) 6 - 2  = 3.4

The standard error of the estimate is 3.4

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Find a prediction interval.

• The standard error of estimate can be used for constructing a prediction interval about a y' value.
• The formula for the prediction interval is:
  • 
    • The d.f. = n - 2.

You can see this is similar to a confidence interval that we studied earlier in the course.

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Example of Prediction Interval

For the given data and regression equation y = 58.39 + 5.28x, find the 95% prediction interval for the monthly maintenance cost of a machine that is 4 years old.  Round your answer to two decimal places.

Machine       Age x (years)    Monthly cost y A 1 64 B 2 71 C 3 69 D 4 80 E 4 82 F 6 90

Find x, x2, and  X . x = 20  ,  x2 = 82   , X = 20 6 = 3.3
Find y for x = 4.
y	=	58.39 + 5.28x
y	=	58.39 + 5.28(4) = 79.51
Find y=456, y2=35122, and  xy=1601 Find sest: sest = 3.27
Substitute in the formula and solve: t/2 = 2.776, d.f. = 6 - 2 = 4 for 95%.
y - t/2sest   1 +   1 n + n(x -  X )2  n x2 - (x)2 < y < y + t/2sest   1 +   1 n + n(x -  X )2  n x2 - (x)2
79.51 - (2.776)(3.27)   1 + 1 6 + 6 (4 - 3.3)2  6 (82) - (20)2 < y < 79.51 + (2.776)(3.27)   1 + 1 6 + 6 (4 - 3.3)2  6 (82) - (20)2
79.51 - (2.776)(3.27)(1.09)	< y <	79.51 + (2.776)(3.27)(1.09)
79.51 - 9.92	< y <	79.51 + 9.92
69.97	< y <	89.45
Hence, one can be 95% confident that the interval 69.97 < y < 89.45 contains the actual value of y.

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Example of Prediction Interval.

For the given data and regression equation y = 58.67 + 5.85x, find the 95% prediction interval for the monthly maintenance cost of a machine that is 4 years old. Machine       Age x (years)    Monthly cost y A 1 66 B 2 74 C 3 71 D 4 77 E 4 84 F 6 97 Round your answer to two decimal places.

Find x, x2, and  X . x = 20    x2 = 82    X = 20 6 = 3.3
Find y for x = 4.
y	=	58.67 + 5.85x
y	=	58.67 + 5.85(4) = 82.07
Find y=469, y2=37267, and  xy=1653   Find sest: sest = 4.49
Substitute in the formula and solve: t/2 = 2.776, d.f. = 6 - 2 = 4 for 95%. (page 731)
y - t/2sest   1 +   1 n + n(x -  X )2  n x2 - (x)2 < y < y + t/2sest   1 +   1 n + n(x -  X )2  n x2 - (x)2
82.07 - (2.776)(4.49)   1 + 1 6 + 6 (4 - 3.3)2  6 (82) - (20)2 < y < 82.07 + (2.776)(4.49)   1 + 1 6 + 6 (4 - 3.3)2  6 (82) - (20)2
82.07 - (2.776)(4.49)(1.09)	< y <	82.07 + (2.776)(4.49)(1.09)
82.07 - 13.62	< y <	82.07 + 13.62
68.43	< y <	95.71
Hence, one can be 95% confident that the interval 68.43 < y < 95.71 contains the actual value of y

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Summary

• The coefficient of determination is a better indicator of the strength of a linear relationship than the correlation coefficient.
  • It is better because it identifies the percentage of variation of the dependent variable that is directly attributable to the variation of the independent variable.
  • The coefficient of determination is obtained by squaring the correlation coefficient and converting the result to a percentage.
• Another statistic used in correlation and regression is the standard error of estimate, which is an estimate of the standard deviation of the y values about the predicted y' values.
• The standard error of estimate can be used to construct a prediction interval about a specific value point estimate y' of the mean or the y values for a given x.

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