Chapter 4: More about Relationships Between Two Variables

4.1 – Transforming to Achieve LinearityExponential Growth

Not all data can be expressed with a linear model.

PROBLEM! We cannot use least-squares regression for nonlinear data because least-squares regression depends upon correlation, which only measures the strength of linear relationships.

SOLUTION! Transform the data into a linear set, then use the least-squares regression to determine the best fitting line for the transformed data. Finally, do a reverse transformation equation which will model our original nonlinear data.

Properties of Logarithms

1. log ab = log a + log b

2. log = log a – log bab

3. log xp = p log x

Remember: log has a base of 10 and natural logs (ln) have a base of e. It doesn’t matter which one you use.

Linearizing Exponential Functions:

We want to write an exponential function of the form y = abx as a linear model. (where x, y are variables and a,b are constants)

y = abx

log y = log (abx)

log y = log a + log bx

log y = log a + xlog b

(x, log y)(x, y)

CONCLUSIONS:

1. If the graph of (x, y) is exponential, then the graph of (x, log y) is linear.

2. If the graph of (x, log y) is linear, then the graph of (x, y) is exponential.

Example #1Transform the exponential data to a linear model using logs and then natural logs.

y = 5(2)x

log y = log (5 2x)

log y = log 5 + log 2x

log y = log 5 + xlog 2

log y = 0.69897 + 0.3010x

ln y = ln (5 2x)

ln y = ln 5 + log 2x

ln y = ln 5 + xln 2

ln y = 1.6094 + 0.6931x

Example #2

Convert the equation back to an exponential function.

ln y = 16 + 9x

y = e(16 + 9x)

y = e(16) e(9x)

y = e(16) e(9)x

y = 8,886,110.521 8103.0839x

Example #3

Convert the equation back to an exponential function.

log y = 4 + 2x

y = 10(4 + 2x)

y = 10(4) 10(2x)

y = 10(4) 10(2)x

y = 10,000 100x

Calculator Tip: Exponential Functions

L1: xL2: y

L3: leave blank for now!

L4: log y

LinReg(L1, L4, Y1) - (x, log y, Y1)

To prevent Overload error: convert years to a smaller number

Calculator Tip: Residual Plot

After calculating the line of regression:In Lists!

ExpReg(L1, L2, Y2) - (x, y, Y2)

Calculator Tip: Exponential Equation

Exponential to Linear Change:

1. The ratio of the y’s should be fairly constant

2. Graph x and y and look at the pattern

3. Calculate the transformed linear model

4. Describe the r value and the residual plot

Example#4: Consider the following data representing the population for Asian and Pacific Islander.

Year 1950 1960 1970 1980 1990 2000

Population (in thousands)

1131 1620 2320 3330 4770 6850

1. Make a scatterplot of the data and describe the graph.

D: Positive, as year increases, population increases

F: Nonlinear

S: Strong

2. Describe the pattern of change and find the percent of change for each y (ratio of y’s).

4324.11131

1620 4321.1

2320 4353.1

432.13330

4770 436.1

Year 1950 1960 1970 1980 1990 2000

Population (in thousands)

1131 1620 2320 3330 4770 6850

The ratios of the y’s are fairly consistent, suggesting an exponential model

3. Find r and describe its meaning

r = 0.968

D: Positive

S: Strong

4. Graph and comment on the residual plot for x and y.

Curve, not a good linear model

5. Take the log of the y-values and make a new scatterplot.

D: Positive

F: Linear

S: Strong

D: Positive

F: Nonlinear

S: Strong

6. Find the least squares regression line of the transformed data.

Log(Population) = 2.27095 + 0.0156432(Year)

7. Find the value of r and describe its meaning.

r = 0.999999

D: Positive

S: Strong

r = 0.968

D: Positive

S: Strong

8. Construct the residual plot and describe its meaning.

No pattern, so good linear model

Curve, not a good linear model

9. Perform the inverse transformation to express y-hat as an exponential equation.

y = 10(2.27095 + 0.0156432x)

y = 10(2.27095) 10(0.0156432x)

y = 10(2.27095) 10(0.0156432)x

y = 186.6162 1.0367x

xy 0156432.027095.2ˆlog

10. Check your work on your calculator using ExpReg.

11. Make a prediction for the population in 2010 using both equations.

log y = 2.27095 + 0.0156432(110)

log y = 3.991697

y = 9810.6342

y = 186.6162 1.0367x

y = 186.6162 1.0367(110)

y = 9,810.6342

Example#5: Consider the following data representing an account balance over time:

x: time (months)

0 48 96 144 192 240

y: account balance ($)

100 161.22 259.93 419.06 675.62 1089.30

D: Positive, as time increases, account balance increases

F: Nonlinear

S: Strong

6122.1100

22.161 6123.1

22.161

93.259 6122.1

93.259

06.419

612.106.419

62.675 6123.1

62.675

30.1089

x: time (months)

0 48 96 144 192 240

y: account balance ($)

100 161.22 259.93 419.06 675.62 1089.30

r = 0.9481

D: Positive

S: Strong

Curved, not good linear model

5. Take the natural log of the y-values and make a new scatterplot.

D: Positive

F: Nonlinear

S: Strong

D: Positive

F: Linear

S: Strong

ln(Account Balance) = 4.60516 + 0.00995047(Months)

7. Find r and describe its meaning.

r = 0.999999

D: Positive

S: Strong

r = 0.9481

D: Positive

S: Strong

No pattern, so good linear modelCurved, not good

linear model

y = e(4.60516 + 0.00995047x)

y = e(4.60516) e (0.00995047x)

y = e(4.60516) e (0.00995047)x

y = 99.9988 1.01x

xy 0.00995047 4.60516ˆln

10. Check your work on your calculator using ExpReg.

11. Make a prediction for the account balance in 60 months using both equations.

ln y = 4.60516 + 0.00995047(60)

ln y = 5.20218656728

y = $181.67

xy 0.00995047 4.60516ˆln y = 99.9988 1.01x

y = 99.9988 1.01(60)

y = $181.67

4.1 – Transforming to Achieve Linearity – Power Model

A power model is in the form y = axp. To transform this equation into a linear model you must apply the log transformation to both variables x and y.

y = axp

log y = log (axp)

log y = log a + log xp

log y = log a + plog x

How is this different than exponential functions?

You have to take the log of both x and y to make a linear model.

Example #6Find the LSRL by taking the logs and then the natural logs.

y = 4x5

log y = log (4x5)

log y = log 4 + log x5

log y = log 4 + 5log x

log y = 0.6021 + 5log x

y = 4x5

ln y = ln (4x5)

ln y = ln 4 + ln x5

ln y = ln 4 + 5ln x

ln y = 1.3863 + 5ln x

Example #7

Convert the equation back to a power equation.

ln y = -5 + 9ln x

y = e(-5 + 9lnx)

y = e(-5) e(9lnx)

y = e(-5) e(lnx)9

y = 0.0067x9

Example #8

Convert the equation back to a power equation.

log y = 0.5 + 2log x

y = 10(0.5 + 2logx)

y = 10(0.5) 10(2logx)

y = 10(0.5) 10(logx)2

y = 3.1623x2

Calculator Tip: Power Functions

L1: xL2: y

L3: log x

L4: log y

LinReg(L3, L4, Y1) - (log x, log y, Y1)

PwrReg(L1, L2, Y2) - (x, y, Y2)

Calculator Tip: Power Equation

Example #9

The distances from our sun and the periods of the 9 planets in the solar system are given below.

Distance (astronomical units)

.39 .72 1 1.5 5.2 9.5 19 30 40

Period (earth years)

.24 .62 1 1.9 12 29 84 160 250

D: Positive, as distance increases, period increases

F: Nonlinear

S: Strong

Distance (astronomical units)

.39 .72 1 1.5 5.2 9.5 19 30 40

Period (earth years)

.24 .62 1 1.9 12 29 84 160 250

583.224.0

62.0 61.1

9.1 316.6

42.212

29 89.2

84 90.1

160 56.1

Ratio of y’s are not similar, perhaps not exponential

r = 0.9779

D: Positive

S: Strong

4. Graph an exponential model and discuss if it is appropriate to use this model.

Curved, not good linear model

5. Transform the data to a linear model by taking the log of the x’s and the y’s. Make a sketch of the new scatterplot.

D: Positive

F: Nonlinear

S: Strong

D: Positive

F: Linear

S: Strong

log(Period) = 0.002916 + 1.49627log(Distance)

7. Find the value of r and describe its meaning.

r = 0.9999765

D: Positive

S: Strong

r = 0.9779

D: Positive

S: Strong

No pattern, so good linear model

y = 10(0.002916+ 1.49627logx)

y = 10(0.002916) 10(1.49627logx)

y = 10(0.002916) 10(logx)1.49627

y = 1.0067x1.49627

) log1.49627( 0.002916 ˆlog xy

10. Check your work on your calculator using PwrReg.

11. If a planet were discovered 35 astronomical units from our sun, predict its period using both equations.

) log1.49627( 0.002916 ˆlog xy

)35 log1.49627( 0.002916 ˆlog y

3723.52 ˆlog y10 10

709.205 ˆ y

y = 1.0067x1.49627

y = 1.0067(35)1.49627

y = 205.709

How do you determine if the model is exponential or power?

1. Graph the original data. Do you see a curve?

2. Look for the ratio of the y values to see if maybe exponential

3. Take the logs of both x and y. Then graph (x, log y) and (log x, log y). Which graph looks more linear?

4. Use the r value and the residual plot to determine the strength of the linear relationship.

Example #10An experiment was conducted to determine the effect of practice time (in seconds) on the percent of unfamiliar words recalled. Here is a Fathom scatterplot of the results with a least-squares regression line superimposed.

(a) Sketch a residual plot below.

(b) Does a linear model fit the data well? Justify your answer.

No, the residual plot has a curve in it, so it isn’t linear

We used Fathom to transform the original data in hopes of achieving linearity. The screen shots below show the results of two different transformations.

(c) Would an exponential model or a power model fit the original data better? Justify your answer.

Power, Stronger r value and residual plot is not as curved

(d) Use the model you chose in (c) to predict word recall for 25 seconds of practice. Show your method.

) ln0.293( 3.48 ˆln xy

4.42313 ˆln ye e

3568.83 ˆ y

)25 ln0.293( 3.48 ˆln y

Example #11

Foresters are interested in predicting the amount of usable lumber they can harvest from various tree species. The following data have been collected on the diameter of Ponderosa pine trees, measured at chest height, and the yield in board feet. Note that a board foot is defined as a piece of lumber 12 inches by 12 inches by 1 inch. Determine if an exponential or power model would make a better model. Support your reasoning. Using the model you have chosen, predict the yield in board feet from a diameter of 40.

Diameter Bd Feet

36 192

28 113

41 294

32 123

38 252

31 141

39 231

33 187

37 205

39 265

1. Graph the original data. Do you see a curve?

2. Look for the ratio of the y values to see if maybe exponential

Diameter Bd Feet

36 192

28 113

41 294

32 123

38 252

31 141

39 231

33 187

37 205

39 265

3. Take the logs of both x and y. Then graph (x, log y) and (log x, log y). Which graph looks more linear?

Power is more linear

4. Use the r value and the residual plot to determine the strength of the linear relationship.

r = 0.9751 r = 0.9880

Power has a stronger r value and doesn’t have a curve in the residual plot, therefore, it is a power model.

Using the model you have chosen, predict the yield in board feet from a diameter of 40.

) ln3.13667( 5.9157- ˆln xy

5.6551 ˆln ye e

7443.285 ˆ y

)40 ln3.13667( 5.9157- ˆln y

4.2 – Relationship between Categorical Variables

http://www.ruf.rice.edu/~lane/stat_sim/transformations/index.html

Because we cannot perform direct calculation on categorical data, we use the counts or percents of individuals by category.

Two-Way Table: Classifies categorical data according to two variables.

Marginal Distribution: The total of each margin, column and row.

Conditional Distribution: Distribution of one variable for given categories of another variable.

Segmented bar graph:

The following segmented bar graph represents the conditional distributions of living arrangements for each race category:

Adds up conditional probabilities to 100% based on categories

Example #12In a national survey of adult Americans in 1998, people were asked to indicate their age and to classify their interest in politics as very much, somewhat, or not much. The ages were grouped in ranges.

18-35 36-55 56-94

Not Much 146 146 89

Somewhat 192 260 154

Very Much 47 125 106

a. Calculate the row and column totals.

385 531 349

381606

b. What proportion of the survey respondents were between ages 18 and 35?

18-35 36-55 56-94

Not Much 146 146 89

Very Much 47 125 106385 531 349

381606

2781265

= 0.3043

18-35 36-55 56-94

Not Much 146 146 89

Very Much 47 125 106385 531 349

381606

2781265

c. What proportion of the survey respondents were between 36 and 55?

531= 0.41976

18-35 36-55 56-94

Not Much 146 146 89

Very Much 47 125 106385 531 349

381606

2781265

d. What proportion of the survey respondents were between 56 and 94?

349= 0.27589

18-35 36-55 56-94

Not Much 146 146 89

Very Much 47 125 106385 531 349

381606

2781265

e. Restrict your attention (for the moment) to just the respondents under 35 years of age. What proportion of these young respondents classify themselves as having not much interest in politics?

146= 0.3792

18-35 36-55 56-94

Not Much 146 146 89

Very Much 47 125 106385 531 349

381606

2781265

f. What proportion of the young respondents classify themselves as somewhat interested in politics?

192= 0.4987

18-35 36-55 56-94

Not Much 146 146 89

Very Much 47 125 106385 531 349

381606

2781265

g. What proportion of the young respondents classify themselves as having very much interest in politics?

47= 0.1221

h. Record the conditional distribution that you have just calculated in the table below.

18-35 36-55 56-94

Not Much .2749 .2550

Somewhat

Very Much

Total 1.000 1.000 1.000

0.3792

0.4987

0.1221

0.4896

0.2354

0.4413

0.3037

i. Construct a segmented bar graph

18-35 36-55 56-94

Age Category

Very Much

Somewhat

Not Much

Example #13The University of CA at Berkeley was charged with having discriminated against women in their graduate admissions process for the fall quarter of 1973. The table below identifies the number of acceptances and denials for both men and women applicants in each of the six largest graduate programs at the institution at that time.

Men Accepted

Men Denied

Women Accepted

Women denied

Program A 511 314 89 19

Program B 352 208 17 8

Program C 120 205 202 391

Program D 137 270 132 243

Program E 53 138 95 298

Program F 22 351 24 317

total 1195 1486 559 1276

a. Start by ignoring the program distinction, collapsing the data into a two-way table of gender by admissions status. To do this, find the total number of men accepted and denied and the total number of women accepted and denied. Fill in the table below:

Accepted Denied Total

1195 1486

559 1276

1754 2762 4516

b. Consider for the moment just the men applicants. Of the men who applied to one of these programs, what proportion were accepted? Now consider the women applicants; what proportion of them were accepted? Do these proportions seem to support the claim that men were given preferential treatment in admissions decisions?

1195= 0.4457

559= 0.3046

MEN WOMEN

c. To try to isolate the program responsible for the alleged mistreatment of women applicants, calculate the proportion of men and the proportion of women within each program who were accepted. Record your results in the table below:

Proportion of men Accepted

Proportion of women Accepted

Program A

Program B

Program C

Program D

Program E

Program F

511/1195 = 0.4276

352/1195 = 0.2946

120/1195 = 0.1004

137/1195 = 0.1146

53/1195 = 0.0444

22/1195 = 0.0184

89/559 = 0.1592

17/559 = 0.0304

202/559 = 0.3614

132/559 = 0.2361

95/559 = 0.1699

24/559 = 0.0429

d. Does it seem as if any program is responsible for the large discrepancy between women in the overall proportions admitted?

Yes, program A and program B accepted less women than men.

Example #14:The following two-way table classifies hypothetical hospital patients according to the hospital that treated them and whether they survived or died:

Survived Died Total

Hospital A 800 200 1000

Hospital B 900 100 1000

a. Calculate the proportion of hospital A’s patients who survived and the proportion of hospital B’s patients who survived. Which hospital saved the higher percentage of its patients?

800= 0.80

900= 0.90

Hospital A Hospital B

Suppose that when we further categorize each patient according to whether they were in fair condition or poor condition prior to treatment we obtain the following two-way table:

FAIR CONDITION Survived Died Total

Hospital A 590 10 600

Hospital B 870 30 900

POOR CONDITION Survived Died Total

b. Among those who were in fair condition, compare the recovery rates for the two hospitals. Which hospital saved the greater percentage of its patients who had been in fair condition?

590= 0.9833

870= 0.9667

c. Among those who were in poor condition, compare the recovery rates for the two hospitals. Which hospital saved the greater percentage of its patients who had been in poor condition?

210= 0.525

30= 0.3

Simpson’s Paradox:

When you combine data sometimes it reverses the direction of the relationship in the individual pieces.

d. Write a few sentences explaining (arguing from the given data given) how it happens that hospital B has the higher recovery rate overall, yet hospital A has the higher recovery rate for each type of patient.

e. Which hospital would you rather go to if you were ill? Explain.

4.3 – Establishing Causation

The only time you can determine causation is when you conduct an experiment.

What if you can’t do an experiment?

• Look for a strong, consistent association

• The increase in the explanatory variable leads to a stronger increase in response

• The cause is plausible

x causes y

Seems x causes y, but z has an effect on x and y, making it look like x causes y

“Z is common to both”

Seems x causes y, but z also has an effect on y, making it look like x causes y

Example #15A soccer coach wanted to improve the team's playing ability, so he had them run two miles a day. At the same time the players decided to take vitamins. In two weeks the team was playing noticeably better, but the coach and players did not know whether it was from the running or the vitamins. What type of variable is this?

Confounding.

Running Improve teams ability

vitamins

Example #16An article that appeared in the San Luis Obispo Tribune (November 11, 1999) was titled “Study Points Out Dangerous Side to SUV Popularity: Half of All 1996 Ejection Deaths Occur in SUVs.” This article states that SUV’s have a much higher rate of passengers being thrown from a window during an accident than do automobiles. The article also states that more than half of all deaths caused by ejection involved SUVs – the basis for the conclusion that SUVs are more dangerous than cars. Later in the article, there is a comment that about 98% of those injured or killed in ejection accidents were not wearing seat belts. Comment on the conclusion that SUVs are more dangerous than cars.

Confounding variable

SUV Roll over

Seat belts

Example #17A study showed that households with more TV sets tend to have longer life expectancies. Describe a possible common response relationship.

More TVs Longer life expectancy

More $

COMMON RESPONSE!

Example #18Based on a survey conducted on the DietSmart.com website, investigators concluded that women who regularly watched Oprah were only one-seventh as likely to crave fattening foods as those who watched other daytime talk shows. Is it reasonable to conclude that watching Oprah causes a decrease in cravings for fattening foods? Explain.

NO, Not an experiment!

Chapter 4: More about Relationships Between Two Variables

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