chapter 9.3 ppt · 2018-04-26 · corresponding predicted y-value. ... constructing a prediction...

M E A S U R E S O F R E G R E S S I O N A N D P R E D I C T I O N I N T E R V A L S

Larson/Farber 4th ed.

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Section 9.3

Section 9.3 Objectives


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� Interpret the three types of variation about a regression line

� Find and interpret the coefficient of determination� Find and interpret the standard error of the estimate

for a regression line� Construct and interpret a prediction interval for y

Variation About a Regression Line


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� Three types of variation about a regression line¡ Total variation¡ Explained variation¡ Unexplained variation

� To find the total variation, you must first calculate¡ The total deviation¡ The explained deviation¡ The unexplained deviation



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iy y-îy y-

î iy y-

(xi, ŷi)

x

y (xi, yi)

(xi, yi)

Unexplained deviation

î iy y-Total

deviationiy y- Explained

deviationîy y-

y

x

Total Deviation = Explained Deviation = Unexplained Deviation =



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Total variation• The sum of the squares of the differences between

the y-value of each ordered pair and the mean of y.

Explained variation� The sum of the squares of the differences between

each predicted y-value and the mean of y.

( )2iy yå -Total variation =

Explained variation = ( )2ˆiy yå -



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Unexplained variation� The sum of the squares of the differences between

the y-value of each ordered pair and each corresponding predicted y-value.

( )2ˆi iy yå -Unexplained variation =

The sum of the explained and unexplained variation is equal to the total variation.

Total variation = Explained variation + Unexplained variation

Coefficient of Determination


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Coefficient of determination� The ratio of the explained variation to the total

variation.� Denoted by r2

2 Explained variationTotal variationr =

Example: Coefficient of Determination


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The correlation coefficient for the advertising expenses and company sales data as calculated in Section 9.1 isr ≈ 0.913. Find the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

22 (0.913)0.834

r »»

About 83.4% of the variation in the company sales can be explained by the variation in the advertising expenditures. About 16.9% of the variation is unexplained.

Solution:

The Standard Error of Estimate


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Standard error of estimate� The standard deviation of the observed yi -values

about the predicted ŷ-value for a given xi -value.� Denoted by se.

� The closer the observed y-values are to the predicted y-values, the smaller the standard error of estimate will be.

2( )ˆ2

i ie

y ys nå -= -

n is the number of ordered pairs in the data set

Larson/Farber 4th ed. 10

2( )ˆ2

i ie

y ys nå -= -

2

, , , ( ), ˆ ˆ( )ˆi i i i i

i i

x y y y yy y

--

The Standard Error of Estimate

1. Make a table that includes the column heading shown.

2. Use the regression equation to calculate the predicted y-values.

3. Calculate the sum of the squares of the differences between each observed y-value and the corresponding predicted y-value.

4. Find the standard error of estimate.

ˆ iy mx b= +

2 ( )ˆi iy yå -

In Words In Symbols

Example: Standard Error of Estimate


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The regression equation for the advertising expenses and company sales data as calculated in section 9.2 is

ŷ = 50.729x + 104.061Find the standard error of estimate. (Calculated in 1000s)

Solution:Use a table to calculate the sum of the squared differences of each observed y-value and the corresponding predicted y-value.

Larson/Farber 4th ed. 12

Solution: Standard Error of Estimatex y ŷ i (yi – ŷ i)2

2.4 225 225.81 (225 – 225.81)2 = 0.6561

1.6 184 185.23 (184 – 185.23)2 = 1.5129

2.0 220 205.52(220 – 205.52)2 =

209.6704

2.6 240 235.96 (240 – 235.96)2 = 16.3216

1.4 180 175.08 (180 – 175.08)2 = 24.2064

1.6 184 185.23 (184 – 185.23)2 = 1.5129

2.0 186 205.52(186 – 205.52)2 =

381.0304

2.2 215 215.66 (215 – 215.66)2 = 0.4356Σ = 635.3463 unexplained

variation

Solution: Standard Error of Estimate


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• n = 8, Σ(yi – ŷ i)2 = 635.3463

2( )ˆ2

i ie

y ys nå -= =-

The standard error of estimate of the company sales for a specific advertising expense is about $10,290.

635.3463 10.2908 2 »-

Prediction Intervals


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� Two variables have a bivariate normal distribution if for any fixed value of x, the corresponding values of y are normally distributed and for any fixed values of y, the corresponding x-values are normally distributed.

Prediction Intervals


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� A prediction interval can be constructed for the true value of y.

� Given a linear regression equation ŷ = mx + b and x0, a specific value of x, a c-prediction interval for y is

ŷ – E < y < ŷ + E where

� The point estimate is ŷ and the margin of error is E. The probability that the prediction interval contains yis c.

202 2( )11

( )c en x xE t s n n x x

-= + +å - å

Constructing a Prediction Interval for y for a Specific Value of x


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1. Identify the number of ordered pairs in the data set n and the degrees of freedom.

2. Use the regression equation and the given x-value to find the point estimate ŷ.

3. Find the critical value tc that corresponds to the given level of confidence c.

ˆi iy mx b= +

Use Table 5 in Appendix B.

In Words In Symbols

d.f. = n – 2

Constructing a Prediction Interval for y for a Specific Value of x


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2( )ˆ2

i ie

y ys nå -= -

4. Find the standard error of estimate se.

5. Find the margin of error E.

6. Find the left and right endpoints and form the prediction interval.

In Words In Symbols

202 2( )11


-= + +å - å

Left endpoint: ŷ – E Right endpoint: ŷ + E Interval: ŷ – E < y < ŷ + E

202 2( )11


-= + +å - å

Example: Constructing a Prediction Interval


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Construct a 95% prediction interval for the company sales when the advertising expenses are $2100. What can you conclude?Recall, n = 8, ŷ = 50.729x + 104.061, se = 10.290 S.E.= 10.975

Solution:Point estimate:ŷ = 50.729(2.1) + 104.061 ≈ 210.592

Critical value:d.f. = n –2 = 8 – 2 = 6 tc = 2.447

215.8, 32.44, 1.975x x xS = S = =

Solution: Constructing a Prediction Interval


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0

2

2

2

2 2

1 8(2.1 1.975)(2.447)(10.290) 1 26.8578 8(32.44) (15

( )

8)

1

.

1( )c e

n x xE t s n n x x-

-= + +

=

»

+ +å - å

-

Left Endpoint: ŷ – E Right Endpoint: ŷ + E

183.735 < y < 237.449

210.592 – 26.857≈ 183.735

210.592 + 26.857≈ 237.449

You can be 95% confident that when advertising expenses are $2100, the company sales will be between $183,735 and $237,449.

Section 9.3 Summary


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� Interpreted the three types of variation about a regression line

� Found and interpreted the coefficient of determination

� Found and interpreted the standard error of the estimate for a regression line

� Constructed and interpreted a prediction interval for y

chapter 9.3 ppt · 2018-04-26 · corresponding predicted y-value. ... constructing a prediction...

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