Similar presentations:

# Correlation and Regression

## 1. Chapter 9: Correlation and Regression

• 9.1 Correlation• 9.2 Linear Regression

• 9.3 Measures of Regression and Prediction Interval

Larson/Farber

1

## 2. Correlation

Correlation• A relationship between two variables.

• The data can be represented by ordered pairs (x, y)

x is the independent (or explanatory) variable

y is the dependent (or response) variable

A scatter plot can be used to determine whether a linear (straight line)

correlation exists between two variables.

y

x

y

1

2

3

–4 –2 –1

4

5

0

2

2

x

2

4

6

–2

–4

Larson/Farber

2

## 3. Types of Correlation

yy

As x increases, y

tends to decrease.

As x increases, y

tends to increase.

x

Negative Linear Correlation

y

Larson/Farber

Positive Linear Correlation

y

x

No Correlation

x

x

Nonlinear Correlation

3

## 4. Example: Constructing a Scatter Plot

A marketing manager conducted a study todetermine whether there is a linear relationship

between money spent on advertising and

company sales. The data are shown in the table.

Display the data in a scatter plot and determine

whether there appears to be a positive or negative

linear correlation or no linear correlation.

Company sales

(in thousands of dollars)

y

Larson/Farber

Advertising expenses

(in thousands of dollars)

Advertising Company

expenses,

sales

($1000), x ($1000), y

2.4

225

1.6

184

2.0

220

2.6

240

1.4

180

1.6

184

2.0

186

2.2

215

Positive linear correlation. As

x the advertising expenses increase,

the sales tend to increase.

4

## 5. Constructing a Scatter Plot Using Technology

• Enter the x-values into list L1 and the y-values into list L2.• Use Stat Plot to construct the scatter plot.

STAT > Edit…

Graph

STATPLOT

100

50

Larson/Farber.

1

5

5

## 6. Correlation Coefficient

Correlation coefficient• A measure of the strength and the direction of a linear relationship between

two variables.

• r represents the sample correlation coefficient.

• ρ (rho) represents the population correlation coefficient

r

n xy x y

n x 2 x

2

n y 2 y

2

n is the number of

data pairs

• The range of the correlation coefficient is -1 to 1.

If r = -1 there is a perfect

negative correlation

Larson/Farber

If r is close to 0 there is

no linear correlation

If r = 1 there is a perfect

positive correlation

6

## 7. Linear Correlation

yy

r = 0.91

r = 0.88

x

Strong negative correlation

y

Strong positive correlation

y

r = 0.42

x

Weak positive correlation

Larson/Farber

x

r = 0.07

x

Nonlinear Correlation

7

## 8. Calculating a Correlation Coefficient

In WordsIn Symbols

1.

Find the sum of the x-values.

x

2.

Find the sum of the y-values.

y

3.

Multiply each x-value by its

corresponding y-value and find the

sum.

xy

4.

Square each x-value and find the

sum.

x2

5.

Square each y-value and find the

sum.

y2

6.

Use these five sums to calculate

the correlation coefficient.

Larson/Farber 4th ed.

r

n xy x y

n x 2 x

2

n y 2 y

2

8

## 9. Example: Finding the Correlation Coefficient

Example: Finding the CorrelationAdvertising Company

Coefficient

expenses,

sales

($1000), x ($1000), y

Calculate the correlation coefficient for the advertising

expenditures and company sales data. What can you conclude?

225

xy

540

x2

5.76

1.6

184

294.4

2.56

33,856

2.0

220

440

4

48,400

2.6

240

624

6.76

57,600

1.4

180

252

1.96

32,400

1.6

184

294.4

2.56

33,856

2.0

186

372

4

34,596

2.2

215

473

4.84

46,225

x

y

2.4

Σx

= 15.8 4th ed. Σy = 1634

Larson/Farber

Σxy = 3289.8

y2

50,625

Σx2 = 32.44 Σy2 = 337,558

2.4

1.6

2.0

2.6

1.4

1.6

2.0

2.2

225

184

220

240

180

184

186

215

9

## 10. Finding the Correlation Coefficient Example Continued…

Σx = 15.8r

Σy = 1634 Σxy = 3289.8 Σx2 = 32.44 Σy2 = 337,558

n xy x y

n x x

2

2

n y y

2

2

8(3289.8) 15.8 1634

8(32.44) 15.82 8(337, 558) 1634 2

501.2

0.9129

9.88 30, 508

r ≈ 0.913 suggests a strong positive linear correlation. As the amount spent on

advertising increases, the company sales also increase.

Larson/Farber

Ti83/84

Catalog – Diagnostic ON

Stat-Calc-4:LinReg(ax+b) L1, L2

10

## 11. Using a Table to Test a Population Correlation Coefficient ρ

• Once the sample correlation coefficient r has been calculated, we need todetermine whether there is enough evidence to decide that the population

correlation coefficient ρ is significant at a specified level of significance.

• Use Table 11 in Appendix B.

• If |r| is greater than the critical value, there is enough evidence to

decide that the correlation coefficient ρ is significant.

For Example: To determine whether ρ is significant for five pairs of data

(n = 5) at a level of significance of α = 0.01

If |r| > 0.959, the correlation is significant.

Otherwise, there is not enough evidence to

conclude that the correlation is significant.

Larson/Farber

11

## 12. Hypothesis Testing for a Population Correlation Coefficient ρ

• A hypothesis test (one or two tailed) can also be used to determine whetherthe sample correlation coefficient r provides enough evidence to conclude

that the population correlation coefficient ρ is significant at a specified

level of significance.

• Left-tailed test

H0: ρ 0 (no significant negative correlation)

Ha: ρ < 0 (significant negative correlation)

• Right-tailed test

H0: ρ 0 (no significant positive correlation)

Ha: ρ > 0 (significant positive correlation)

• Two-tailed test

H0: ρ = 0 (no significant correlation)

Ha: ρ 0 (significant correlation)

Larson/Farber

12

## 13. Using the t-Test for ρ

In WordsIn Symbols

1.

State the null and alternative hypothesis.

2.

Specify the level of significance.

Identify .

3.

Identify the degrees of freedom.

d.f. = n – 2.

4.

Determine the critical value(s) and

rejection region(s).

Use Table 5 in Appendix B.

5.

Find the standardized test statistic.

6.

Make a decision to reject or fail to reject

the null hypothesis and interpret the

decision in terms of the original claim.

Larson/Farber

State H0 and Ha.

t

r

1 r2

n 2

If t is in the rejection region,

reject H0. Otherwise fail to

reject H0.

13

## 14. Example: t-Test for a Correlation Coefficient

For the advertising data, we previouslycalculated r ≈ 0.9129. Test the significance of

this correlation coefficient. Use α = 0.05.

Advertising Company

expenses,

sales

($1000), x ($1000), y

ρ=0

H0

ρ≠0

2.4

225

Ha

0.025

0.025

α

0.05

1.6

184

d.f. 8 – 2 = 6

t

2.0

220

-2.447 0 2.447

0.9129

2.6

240

Test Statistic: t

5.478

2

1

(0.9129)

1.4

180

Stat-Tests

LinRegTTest

8 2

1.6

184

Decision: Reject H0

2.0

186

At the 5% level of significance, there is enough

2.2

215

evidence to conclude that there is a significant

linear correlation between advertising expenses

Larson/Farber 4th ed.

and

company sales.

14

## 15. Correlation and Causation

• The fact that two variables are strongly correlated does not in itself imply acause-and-effect relationship between the variables.

• If there is a significant correlation between two variables, you should

consider the following possibilities:

1. Is there a direct cause-and-effect relationship between the variables?

• Does x cause y?

2.

Is there a reverse cause-and-effect relationship between the variables?

• Does y cause x?

3.

Is it possible that the relationship between the variables can be caused

by a third variable or by a combination of several other variables?

4.

Is it possible that the relationship between two variables may be a

coincidence?

Larson/Farber

15

## 16. 9.2 Objectives

• Find the equation of a regression line• Predict y-values using a regression equation

After verifying that the linear correlation between two variables is significant,

we determine the equation of the line that best models the data (regression

line) - used to predict the value of y for a given value of x.

y

x

Larson/Farber

16

## 17. Residuals & Equation of Line of Regression

Residuals & Equation of Line ofRegression

Residual

• The difference between the observed y-value and the predicted y-value for

a given x-value on the line.

Regression line

For a given x-value,

Line of best fit

di = (observed y-value) – (predicted y-value)

• The line for which the sum y Observed y-value

of the squares of the

d6{

residuals is a minimum.

d4

}d5

d

{

3

• Equation of Regression

Predicted y-value

}

d

2

ŷ = mx + b

}d1

x

ŷ - predicted y-value

n xy x y

y

x

m

2

b

y

mx

m

m – slope

2

n x x

n

n

b – y-intercept

y - mean of y-values in the data

- mean of x-values in the data

x

Larson/Farber 4th ed.

The regression line always passes through x , y17

{

## 18. Finding Equation for Line of Regression

Recall the data from section 9.1225

xy

540

x2

5.76

1.6

184

294.4

2.56

33,856

2.0

220

440

4

48,400

2.6

240

624

6.76

57,600

1.4

180

252

1.96

32,400

1.6

184

294.4

2.56

33,856

2.0

186

372

4

34,596

2.2

215

473

4.84

46,225

x

y

2.4

Σx

= 15.8 4th ed. Σy = 1634

Larson/Farber

m

Σxy = 3289.8

501.2

50.72874

9.88

Equation of Line of Regression :

50,625

Advertising Company

expenses,

sales

($1000), x ($1000), y

2.4

1.6

2.0

2.6

1.4

1.6

2.0

2.2

225

184

220

240

180

184

186

215

Σx2 = 32.44 Σy2 = 337,558

n xy x y 8(3289.8) (15.8)(1634)

8(32.44) 15.82

2

n x 2 x

y2

b y mx 1634 (50.72874) 15.8

8

8

204.25 (50.72874)(1.975) 104.0607

yˆ 50.729 x 104.061

18

## 19. Solution: Finding the Equation of a Regression Line

Company sales(in thousands of

dollars)

• To sketch the regression line, use any two x-values within the range of the

data and calculate the corresponding y-values from the regression line.

260

y

240

yˆ 50.729 x 104.061

220

200

180

160

1,2

1,4

1,6

1,8

2

2,2

2,4

2,6

x

2,8

Advertising expenses

(in thousands of dollars)

StatPlot and Graph

Ax + b

Ti83/84

Catalog – Diagnostic ON

Stat-Calc-4:LinReg(ax+b) L1, L2

Larson/Farber 4th ed.

50.729

104.061

19

## 20. Example: Predicting y-Values Using Regression Equations

The regression equation for the advertising expenses (in thousands of dollars) andcompany sales (in thousands of dollars) data is ŷ = 50.729x + 104.061. Use this

equation to predict the expected company sales for the advertising expenses below:

1.1.5 thousand dollars :

ŷ =50.729(1.5) + 104.061 ≈ 180.155

When advertising expenses are $1500, company sales are about $180,155.

2.1.8 thousand dollars

ŷ =50.729(1.8) + 104.061 ≈ 195.373

When advertising expenses are $1800, company sales are about $195,373.

3. 2.5 thousand dollars ŷ =50.729(2.5) + 104.061 ≈ 230.884

When advertising expenses are $2500, company sales are about $230,884.

Prediction values are meaningful only for x-values in (or close to) the range of the data. Xvalues in the original data set range from 1.4 to 2.6. It is not appropriate to use the regression

line to predict company sales for advertising expenditures such as 0.5 ($500) or 5.0 ($5000).

Larson/Farber

20

## 21. 9.3 Measures of Regression and Prediction Intervals (Objectives)

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

Three types of variation about a regression line

● Total variation

● Explained variation

y

First calculate

The total deviation yi y

ˆi y

The explained deviation y

The unexplained deviation yi y

ˆi

(xi, yi)

Unexplained

deviation i

y yˆi

Total

deviation

y

Larson/Farber 4th ed.

● Unexplained variation

yi y

(xi, ŷi)

Explained deviation

yˆi y

(xi, yi)

x

x

21

## 22. Variation About a Regression Line

Total variation = y y 2i

• The sum of the squares of the differences between the y-value of each

ordered pair and the mean of y.

Total variation = Explained variation + Unexplained variation

Explained variation yˆi y 2

• The sum of the squares of the differences between each predicted y-value

and the mean of y.

Unexplained variation y yˆ 2

i

i

• The sum of the squares of the differences between the y-value of each

ordered pair and each corresponding predicted y-value.

Coefficient of determination (r2)

Explained variation

r2

Total variation

• Ratio of the explained variation to the total variation.

For the advertising data, correlation coefficient r ≈ 0.913 => r2 = (.913)2 = .834

About 83.4% of the variation in company sales can be explained by variation in

Larson/Farber

4th ed.

22

advertising

expenditures. About 16.9% of the variation is unexplained.

## 23. The Standard Error of Estimate

Standard error of estimate• The standard deviation (se )of the observed yi -values about the predicted ŷ-value for

a given xi -value.

( yi yˆi) 2 n = number of ordered data pairs.

se

n 2

• The closer the observed y-values are to the predicted y-values, the smaller the

standard error of estimate will be.

The regression equation for the advertising expenses and company sales data

as calculated in section 9.2 is : ŷ = 50.729x + 104.061

x

2.4

1.6

2.0

2.6

1.4

1.6

2.0

2.2

Larson/Farber

y

225

184

220

240

180

184

186

215

ŷi

225.81

185.23

205.52

235.96

175.08

185.23

205.52

215.66

(yi – ŷ i)2

(225 – 225.81)2 = 0.6561

(184 – 185.23)2 = 1.5129

(220 – 205.52)2 = 209.6704

(240 – 235.96)2 = 16.3216

(180 – 175.08)2 = 24.2064

(184 – 185.23)2 = 1.5129

(186 – 205.52)2 = 381.0304

(215 – 215.66)2 = 0.4356

Σ = 635.3463

Unexplained variation

635.3463

10.290

8 2

The standard error of

estimate of the company

sales for a specific

advertising expense is

about $10.29.

Stat-Tests

LinRegTTest 23

## 24. Prediction Intervals

• Two variables have a bivariate normal distribution if for any fixed valueof x, the corresponding values of y are normally distributed and for any

fixed values of y, the corresponding x-values are normally distributed.

Point estimate:

Given a linear regression equation ŷi = mxi + b and

x0(a specific value of x), d.f. = n-2, a c-prediction interval ŷ = 50.729(2.1) +

104.061 ≈ 210.592

for y is:

n(x0 x ) 2

1

ŷ – E < y < ŷ + E , where, E tcse 1 n

n x 2 ( x) 2

Critical value:

d.f. = n –2 = 8 – 2 = 6

The point estimate is ŷ and the margin of error is E. The

tc = 2.447

probability that the prediction interval contains y is c.

Example: 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 x 15.8, x 2 32.44,

n(x0 x )2

1

E tcse 1

n n x 2 ( x)2

Prediction Interval:

210.592 – 26.857 to 210.592 + 26.857

183.735 < y < 237.449

x 1.975

1 8(2.1 1.975)2

Larson/Farber

(2.447)(10.290)

1

26.857 You can be 95% confident that when advertising expenses

4th ed. 8 8(32.44) (15.8) 2

24

are $2100, sales will be between $183,735 and $237,449.