Class 10: Tuesday, Oct. 12

• Hurricane data set, review of confidence intervals and hypothesis tests

• Confidence intervals for mean response• Prediction intervals• Transformations• Upcoming:

– Thursday: Finish transformations, Example Regression Analysis

– Tuesday: Review for midterm– Thursday: Midterm– Fall Break!

Hurricane Data • Is there a trend in the number of

hurricanes in the Atlantic over time (possibly an increase because of global warming)?

• hurricane.JMP contains data on the number of hurricanes in the Atlantic basin from 1950-1997.

Bivariate Fit of Hurricanes By Year

0

2

4

6

8

10

12

14H

urric

ane

s

1950 1960 1970 1980 1990 2000

Year

Summary of Fit RSquare 0.047995 RSquare Adj 0.0273 Root Mean Square Error 2.33302 Mean of Response 5.75 Observations (or Sum Wgts)

48

Parameter Estimates Term Estimate Std Error t Ratio Prob>|t| Lower 95% Upper 95% Intercept 78.80292 47.97217 1.64 0.1073 -17.76005 175.36589 Year -0.037017 0.024308 -1.52 0.1346 -0.085946 0.0119117

-4-202468

Res

idua

l

1950 1960 1970 1980 1990 2000

Year

Distributions Residuals Hurricanes

-4

-3

-2

-1

0

1

2

3

4

5

6

7.01 .05.10 .25 .50 .75 .90.95 .99

-3 -2 -1 0 1 2 3

Normal Quantile Plot

Inferences for Hurricane Data

• Residual plots and normal quantile plots indicate that assumptions of linearity, constant variance and normality in simple linear regression model are reasonable.

• 95% confidence interval for slope (change in mean hurricanes between year t and year t+1): (-0.086,0.012)

• Hypothesis Test of null hypothesis that slope equals zero: test statistic = -1.52, p-value =0.13. We accept since p-value > 0.05. No evidence of a trend in hurricanes from 1950-1997.

0: 10 =βH

• Scale for interpreting p-values:

• A large p-value is not strong evidence in favor of H0, it only shows that there is not strong evidence against H0.

little or no evidence against H0>.1

weak evidence against H0.05-.10

strong evidence against H0.01-.05

very strong evidence against H0<.01

Evidencep-value

Inference in Regression

• Confidence intervals for slope

• Hypothesis test for slope

• Confidence intervals for mean response• Prediction intervals

Car Price Example

• A used-car dealer wants to understand how odometer reading affects the selling price of used cars.

• The dealer randomly selects 100 three-year old Ford Tauruses that were sold at auction during the past month. Each car was in top condition and equipped with automatic transmission, AM/FM cassette tape player and air conditioning.

• carprices.JMP contains the price and number of miles on the odometer of each car.

Bivariate Fit of Price By Odometer

13500

14000

14500

15000

15500

16000P

rice

15000 25000 30000 35000 40000 45000

Odometer

Linear Fit Price = 17066.766 - 0.0623155 Odometer

Summary of Fit RSquare 0.650132 RSquare Adj 0.646562 Root Mean Square Error 303.1375 Mean of Response 14822.82 Observations (or Sum Wgts)

100

Parameter Estimates Term Estimate Std Error t Ratio Prob>|t| Lower 95% Upper 95% Intercept 17066.766 169.0246 100.97 <.0001 16731.342 17402.19 Odometer -0.062315 0.004618 -13.49 <.0001 -0.071479 -0.053152

• The used-car dealer has an opportunity to bid on a lot of cars offered by a rental company. The rental company has 250 Ford Tauruses, all equipped with automatic transmission, air conditioning and AM/FM cassette tape players. All of the cars in this lot have about 40,000 miles on the odometer. The dealer would like an estimate of the average selling price of all cars of this type with 40,000 miles on the odometer, i.e., E(Y|X=40,000).

• The least squares estimate is 575,14$40000*0623.017067)40000|(ˆ =−==XYE

Confidence Interval for Mean Response

• Confidence interval for E(Y|X=40,000): A range of plausible values for E(Y|X=40,000) based on the sample.

• Approximate 95% Confidence interval:

• Notes about formula for SE: Standard error becomes smaller as sample size n increases, standard error is smaller the closer is to

• In JMP, after Fit Line, click red triangle next to Linear Fit and click Confid Curves Fit. Use the crosshair tool by clicking Tools, Crosshair to find the exact values of the confidence interval endpoints for a given X0.

)}|(ˆ{*2)|(ˆ00 XXYESEXXYE =±=

∑ =−

−+== n

i i XX

XX

nRMSEXXYESE

1

2

20

0)(

)(1)}|(ˆ{

0XX

Bivariate Fit of Price By Odometer

13500

14000

14500

15000

15500

16000P

rice

15000 25000 30000 35000 40000 45000

Odometer

Approximate 95% confidence interval for == )000,40|(ˆ XYE ($14,514, $14,653)

A Prediction Problem

• The used-car dealer is offered a particular 3-year old Ford Taurus equipped with automatic transmission, air conditioner and AM/FM cassette tape player and with 40,000 miles on the odometer. The dealer would like to predict the selling price of this particular car.

• Best prediction based on least squares estimate:

575,14$40000*0623.017067)40000|(ˆ =−==XYE

Range of Selling Prices for Particular Car

• The dealer is interested in the range of selling prices that this particular car with 40,000 miles on it is likely to have.

• Under simple linear regression model, Y|X follows a normal distribution with mean and standard deviation . A car with 40,000 miles on it will be in interval about 95% of the time.

• Class 5: We substituted the least squares estimates for for and said car with 40,000 miles on it will be in interval about 95% of the time. This is a good approximation but it ignores potential error in least square estimates.

X*10 ββ + σ

σββ *240000*10 ±+

RMSE,ˆ,ˆ10 ββ σββ ,, 10

RMSE*240000*ˆˆ10 ±+ ββ

Prediction Interval

• 95% Prediction Interval: An interval that has approximately a 95% chance of containing the value of Y for a particular unit with X=X0 ,where the particular unit is not in the original sample.

• Approximate 95% prediction interval:

• In JMP, after Fit Line, click red triangle next to Linear Fit and click Confid Curves Indiv. Use the crosshair tool by clicking Tools, Crosshair to find the exact values of the prediction interval endpoints for a given X0.

∑ =−++±= n

i i XXn

RMSEXXYE1

20 )(

11*2)|(ˆ

Bivariate Fit of Price By Odometer

13500

14000

14500

15000

15500

16000P

rice

15000 30000 40000Odometer

95% Confidence Interval for )14653 ,14514( }40000|{ ≈=XYµ 95% Prediction Interval for X=40000 15194) ,13972( ≈

A Violation of LinearityBivariate Fit of Life Expectancy By Per Capita GDP

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Life

Exp

ecta

ncy

0 5000 10000 15000 20000 25000 30000

Per Capita GDP

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-15

-5

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0 5000 10000 15000 20000 25000 30000

Per Capita GDP

Y=Life Expectancy in 1999X=Per Capita GDP (in US Dollars) in 1999Data in gdplife.JMP

Linearity assumption of simplelinear regression is clearly violated.The increase in mean life expectancy for each additional dollarof GDP is less for large GDPs thanSmall GDPs. Decreasing returns toincreases in GDP.

Transformations

• Violation of linearity: E(Y|X) is not a straight line.

• Transformations: Perhaps E(f(Y)|g(X)) is a straight line, where f(Y) and g(X) are transformations of Y and X, and a simple linear regression model holds for the response variable f(Y) and explanatory variable g(X).

Bivariate Fit of Life Expectancy By log Per Capita GDP

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80

Life

Exp

ecta

ncy

6 7 8 9 10

log Per Capita GDP

Linear Fit Life Expectancy = -7.97718 + 8.729051 log Per Capita GDP

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-15

-5

5

15

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6 7 8 9 10

log Per Capita GDP

The mean of Life Expectancy | Log Per Capita appears to be approximatelya straight line.

How do we use the transformation?

•

• Testing for association between Y and X: If the simple linear regression model holds for f(Y) and g(X), then Y and X are associated if and only if the slope in the regression of f(Y) and g(X) does not equal zero. P-value for test that slope is zero is <.0001: Strong evidence that per capita GDP and life expectancy are associated.

• Prediction and mean response: What would you predict the life expectancy to be for a country with a per capita GDP of $20,000?

Linear Fit Life Expectancy = -7.97718 + 8.729051 log Per Capita GDP Parameter Estimates Term Estimate Std Error t Ratio Prob>|t| Intercept -7.97718 3.943378 -2.02 0.0454 log Per Capita GDP

8.729051 0.474257 18.41 <.0001

47.789035.9*7291.89772.7)9035.9log|(ˆ

)000,20loglog|(ˆ)000,20|(ˆ

=+−==

====

XYE

XYEXYE

How do we choose a transformation?

• Tukey’s Bulging Rule.

• See Handout.

• Match curvature in data to the shape of one of the curves drawn in the four quadrants of the figure in the handout. Then use the associated transformations, selecting one for either X, Y or both.

Transformations in JMP2. Use Tukey’s Bulging rule (see handout) to determine

transformations which might help. 3. After Fit Y by X, click red triangle next to Bivariate Fit

and click Fit Special. Experiment with transformations suggested by Tukey’s Bulging rule.

4. Make residual plots of the residuals for transformed model vs. the original X by clicking red triangle next to Transformed Fit to … and clicking plot residuals. Choose transformations which make the residual plot have no pattern in the mean of the residuals vs. X.

5. Compare different transformations by looking for transformation with smallest root mean square error on original y-scale. If using a transformation that involves transforming y, look at root mean square error for fit measured on original scale.

Bivariate Fit of Life Expectancy By Per Capita GDP

40

50

60

70

80Li

fe E

xpec

tanc

y

0 5000 10000 15000 20000 25000 30000

Per Capita GDP

Linear Fit

Transformed Fit to Log

Transformed Fit to Sqrt

Transformed Fit Square

`

•

Linear Fit Life Expectancy = 56.176479 + 0.0010699 Per Capita GDP Summary of Fit RSquare 0.515026 RSquare Adj 0.510734 Root Mean Square Error 8.353485 Mean of Response 63.86957 Observations (or Sum Wgts) 115 Transformed Fit to Log Life Expectancy = -7.97718 + 8.729051 Log(Per Capita GDP) Summary of Fit RSquare 0.749874 RSquare Adj 0.74766 Root Mean Square Error 5.999128 Mean of Response 63.86957 Observations (or Sum Wgts) 115

Transformed Fit to Sqrt Life Expectancy = 47.925383 + 0.2187935 Sqrt(Per Capita GDP) Summary of Fit RSquare 0.636551 RSquare Adj 0.633335 Root Mean Square Error 7.231524 Mean of Response 63.86957 Observations (or Sum Wgts) 115 Transformed Fit Square Square(Life Expectancy) = 3232.1292 + 0.1374831 Per Capita GDP Fit Measured on Original Scale Sum of Squared Error 7597.7156 Root Mean Square Error 8.1997818 RSquare 0.5327083 Sum of Residuals -70.29942

By looking at the root mean square error on the original y-scale, we see thatall of the transformations improve upon the untransformed model and that the transformation to log x is by far the best.

Linear Fit

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0 5000 10000 15000 20000 25000 30000

Per Capita GDP

Transformation to Log X

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0 5000 10000 15000 20000 25000 30000

Per Capita GDP

Transformation to X

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0 5000 10000 15000 20000 25000 30000

Per Capita GDP

Transformation to 2Y

-25

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0 5000 10000 15000 20000 25000 30000

Per Capita GDP

The transformation to Log X appears to have mostly removed a trend in the meanof the residuals. This means that . There is still a problem of nonconstant variance.

XXYE log)|( 10 ββ +≈