Anthony Greene1 Regression Using Correlation To Make Predictions

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Regression

Using Correlation To Make Predictions

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Making a prediction

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To obtain the predicted value of y based on a known value of x and a known correlation.

Note what happens for positive and negative values of r and for high and low values of r and for near-zero values of r.

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Graph of y = 5 – 3 x

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y-Intercept and Slope

For a linear equation y = a + bx, the constant a is the y-intercept and the constant b is the slope.

x and y are related variables

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Straight-line graphs of three linear equations

Y = a + bXa = y-interceptb = slope (rise/run)

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Graphical Interpretation of Slope

The straight-line graph of the linear equation y = a +bx slopes upward if b > 0, slopes downward if b < 0, and is horizontal if b = 0

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Graphical interpretation of slope

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Four data points

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Scatter plot

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Two possible straight-line fits to the data points

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Determining how well the data points in are fit by Line A Vs.Line B

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Least-Squares Criterion

The straight line that best fits a set of data points is the one having the smallest possible sum of squared errors. Recall that the sum of squared errors is error variance.

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Regression Line and Regression Equation

Regression line: The straight line that best fits a set of data points according to the least-squares criterion.Regression equation: The equation of the regression line.

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The best-fit line minimizes the distance between the actual data and the predicted value

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Residual, e, of a data point

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We define SSx, SSP and SSy by

Notation Used in Regression and Correlation

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Regression Equation

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The regression equation for a set of n data points is

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The relationship between b and r

• That is, the regression slope is just the correlation coefficient scaled up to the right size for the variables x and y.

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Criterion for Finding a Regression Line

Before finding a regression line for a set of data points, draw a scatter diagram. If the data points do not appear to be scattered about a straight line, do not determine a regression line.

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Linear regression requires linear data:(a) Data points scattered about a curve (b) Inappropriate straight line fit to the dataHigher order regression equations exist but are outside the range of this course

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Uniform Variance

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Math Proficiency By Grade

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Assumptions for Regression Inferences

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Table for obtaining the three sums of squares for the used car data

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Regression line and data points for used car data

What is a fair asking price for a 2.5 year old car?

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5.226.2047.195ˆ

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So since the price unit is $100s, the best prediction is $17,271

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Extrapolation in the used car example

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Total sum of squares, SST: The variation in the observed values of the response variable:

Regression sum of squares, SSR: The variation in the observed values of the response variable that is explained by the regression:

Error sum of squares, SSE: The variation in the observed values of the response variable that is not explained by the regression:

Sums of Squares in Regression

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Regression Identity

The total sum of squares equals the regression sum of squares plus the error sum of squares. In symbols,

SST = SSR + SSE.

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Graphical portrayal of regression for used cars

y = a + bx

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What sort of things could regression be used for?

Any instance where a known correlation exists, regression can be used to predict a new score. Examples:

1. If you knew that there was a past correlation between the amount of study time and the grade on an exam, you could make a good prediction about the grade before it happened.

2. If you knew that certain features of a stock correlate with its price, you can use regression to predict the price before it happens.

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Regression Example: Low Correlation

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Find the regression equation for predicting height based on knowledge of weight. The existing data is for 10 male stats students?

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287.00 75.00 21,525.00 82,369.00 5,625.00

300.00 71.00 21,300.00 90,000.00 5,041.00

255.00 80.00 20,400.00 65,025.00 6,400.00

180.00 69.00 12,420.00 32,400.00 4,761.00

130.00 70.00 9,100.00 16,900.00 4,900.00

215.00 77.00 16,555.00 46,225.00 5,929.00

165.00 71.00 11,715.00 27,225.00 5,041.00

240.00 71.00 17,040.00 57,600.00 5,041.00

160.00 72.00 11,520.00 25,600.00 5,184.00

150.00 65.00 9,750.00 22,500.00 4,225.002,082.00 721.00 151,325.00 465,844.00 52,147.00

X Y

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287.00 75.00 21,525.00 82,369.00 5,625.00

300.00 71.00 21,300.00 90,000.00 5,041.00

255.00 80.00 20,400.00 65,025.00 6,400.00

180.00 69.00 12,420.00 32,400.00 4,761.00

130.00 70.00 9,100.00 16,900.00 4,900.00

215.00 77.00 16,555.00 46,225.00 5,929.00

165.00 71.00 11,715.00 27,225.00 5,041.00

240.00 71.00 17,040.00 57,600.00 5,041.00

160.00 72.00 11,520.00 25,600.00 5,184.00

150.00 65.00 9,750.00 22,500.00 4,225.002,082.00 721.00 151,325.00 465,844.00 52,147.00

X Y XY X2 Y2

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287.00 75.00 21,525.00 82,369.00 5,625.00

300.00 71.00 21,300.00 90,000.00 5,041.00

255.00 80.00 20,400.00 65,025.00 6,400.00

180.00 69.00 12,420.00 32,400.00 4,761.00

130.00 70.00 9,100.00 16,900.00 4,900.00

215.00 77.00 16,555.00 46,225.00 5,929.00

165.00 71.00 11,715.00 27,225.00 5,041.00

240.00 71.00 17,040.00 57,600.00 5,041.00

160.00 72.00 11,520.00 25,600.00 5,184.00

150.00 65.00 9,750.00 22,500.00 4,225.002,082.00 721.00 151,325.00 465,844.00 52,147.00

X Y XY X2 Y2

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SSx = x2 - (x)2/n = 465,844-433,472.4 = 32,372

SP = xy - x y/n = 151,325-150, 112.2

b=SP/SSx, so b = 1,213/32,372=0.03

a = (1/n)(y-bx), so a = 0.1(721-60.38) = 66

So, Y=0.03x+66

X Y XY X2 Y2

2,082 721 151,325 465,844 52,147

^

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Y=0.03x+66^

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Regression Example: High Correlation

Find the regression equation for predicting probability of a teenage suicide attempt based on weekly heroine usage.

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X Y XY X2 Y2

1 0.2 0.2 1 0.04

1 0.31 0.31 1 0.0961

1 0.18 0.18 1 0.0324

2 0.27 0.54 4 0.0729

2 0.38 0.76 4 0.1444

2 0.46 0.92 4 0.2116

3 0.9 2.7 9 0.81

3 0.58 1.74 9 0.3364

3 0.45 1.35 9 0.2025

4 0.84 3.36 16 0.7056

4 0.74 2.96 16 0.5476

4 0.68 2.72 16 0.4624

5 0.85 4.25 25 0.7225

5 0.78 3.9 25 0.6084

5 0.73 3.65 25 0.5329

6 0.88 5.28 36 0.7744

6 0.82 4.92 36 0.6724

6 0.78 4.68 36 0.6084

7 0.92 6.44 49 0.8464

7 0.85 5.95 49 0.7225

7 0.91 6.37 49 0.8281

84 13.51 63.18 420 9.9779

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X Y XY X2 Y2

1 0.2 0.2 1 0.04

1 0.31 0.31 1 0.0961

1 0.18 0.18 1 0.0324

2 0.27 0.54 4 0.0729

2 0.38 0.76 4 0.1444

2 0.46 0.92 4 0.2116

3 0.9 2.7 9 0.81

3 0.58 1.74 9 0.3364

3 0.45 1.35 9 0.2025

4 0.84 3.36 16 0.7056

4 0.74 2.96 16 0.5476

4 0.68 2.72 16 0.4624

5 0.85 4.25 25 0.7225

5 0.78 3.9 25 0.6084

5 0.73 3.65 25 0.5329

6 0.88 5.28 36 0.7744

6 0.82 4.92 36 0.6724

6 0.78 4.68 36 0.6084

7 0.92 6.44 49 0.8464

7 0.85 5.95 49 0.7225

7 0.91 6.37 49 0.8281

84 13.51 63.18 420 9.9779

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X Y XY X2 Y2

1 0.2 0.2 1 0.04

1 0.31 0.31 1 0.0961

1 0.18 0.18 1 0.0324

2 0.27 0.54 4 0.0729

2 0.38 0.76 4 0.1444

2 0.46 0.92 4 0.2116

3 0.9 2.7 9 0.81

3 0.58 1.74 9 0.3364

3 0.45 1.35 9 0.2025

4 0.84 3.36 16 0.7056

4 0.74 2.96 16 0.5476

4 0.68 2.72 16 0.4624

5 0.85 4.25 25 0.7225

5 0.78 3.9 25 0.6084

5 0.73 3.65 25 0.5329

6 0.88 5.28 36 0.7744

6 0.82 4.92 36 0.6724

6 0.78 4.68 36 0.6084

7 0.92 6.44 49 0.8464

7 0.85 5.95 49 0.7225

7 0.91 6.37 49 0.8281

84 13.51 63.18 420 9.9779Σ

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n = 21

SSx = x2 - (x)2/n = 420 - 336 = 84

SP = xy - x y/n = 63.18 – 54.04 = 9.14

b=SP/SSx, so b = 9.14/84 = 0.109

a=(1/n)(y-bx), so a = (1/21)(13.51-9.156) = 0.207

So, Y= 0.109x + 0.207

X Y XY X2 Y2

84 13.51 63.18 420 9.9779Σ

^

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Why Is It Called Regression?

• For low correlations, the predicted value is close to the mean

• For zero correlations the prediction is the mean• Only for perfect correlations R2= 1.0 do the

predicted scores show as much variation as the actual scores

• Since perfect correlations are rare, we say that the predicted scores show regression towards the mean