exercise 10 - california state university fullertonmathfaculty.fullerton.edu/sbehseta/hw4 (without...

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Exercise 10.1

(a) Plot wages versus LOS. Describe the relationship. There is one woman with relatively high

wages for her length of service. Circle this point and do not use it in the rest of this exercise.

(b) Find the least-squares line. Summarize the significance test for the slope. What do you

conclude?

The least-squares line is

( )

The standard error for the estimate of the slope is 0.02560. The corresponding t- and p-values

are 2.85 and 0.00605, respectively. With a p-value so small, we conclude that there is significant

evidence against the mull hypothesis that the slope is zero.

(c) State carefully what the slope tells you about the relationship between wages and length of

service.

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The slope tells us that for every unit increase in LOS, the average wage will increase by 0.07295

units.

(d) Give a 95% confidence interval for the slope.

A 95% confidence interval for the slope is [0.02169093, 0.1242139].

Code and Output: > confint( m1)

2.5 % 97.5 %

(Intercept) 38.95561398 47.9195425

LOS 0.02169093 0.1242139

Exercise 10.2

Refer to the previous exercise. Analyze the data with the outlier included. How does this change the

estimates of the parameters , , and ? What effect does the outlier have on the results of the

significance test for the slope?

The estimates of the parameters from the previous exercise (without the outlier) were:

With the outlier included, the estimates become:

The outlier has the effect of lowering the significance of the test for the slope—increasing the p-value

from 0.00605 to 0.0186.

Exercise 10.5

In Example 10.8 we examined the yield in bushels per acre of corn for the years 1966, 1976, 1986, and

1996. Data for all years between 1957 and 1996 appear in Table 10.2

(a) Plot the yield versus year. Describe the relationship. Are there any outliers or unusual years?

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Comment:

The relationship looks roughly linear. There don’t appear to be any outliers. As for unusual years,

there are a few years in the 70s and 80s that appear to have a lower yield than what might be

expected.

(b) Perform the regression analysis and summarize the results. How rapidly has yield increased

over time?


( )

1960 1970 1980 1990

60

80

10

01

20

14

0Exercise 10.5

Year

Yie

ld

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The significance tests for intercept and slope are both highly significant—the p-value for the test

of the intercept being equal to zero was p = 2.77e-15 and for the test of the slope being equal

to zero was p = 1.26e-15.

The average yield has increases by approximately 1.84 bushels per acre per year.

Code and Output: > m5= lm( Yield ~ Year)

> summary( m5)

Call:

lm(formula = Yield ~ Year)

Residuals:

Min 1Q Median 3Q Max

-27.890 -6.102 1.240 7.270 15.072

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -3544.6463 278.3787 -12.73 2.77e-15 ***

Year 1.8396 0.1408 13.06 1.26e-15 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.28 on 38 degrees of freedom

Multiple R-squared: 0.8178, Adjusted R-squared: 0.813

F-statistic: 170.6 on 1 and 38 DF, p-value: 1.257e-15

Exercise 10.6

(a) Find the equation of the least-squares line.

The equation is:

( )

Code and Output: > m6= lm( Y ~ X)

> summary( m6)

Call:

lm(formula = Y ~ X)

Residuals:


-0.55162 -0.17595 -0.09349 0.17381 0.81315

Coefficients:


(Intercept) 1.23235 0.28604 4.308 0.00353 **

X 0.20221 0.01145 17.663 4.6e-07 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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F-statistic: 312 on 1 and 7 DF, p-value: 4.596e-07

(b) Test the null hypothesis that the slope is zero and describe your conclusion.

The p-value for this test is p = 4.6e-07, so we reject at the 0.05 level. The conclusion is that it is

extremely unlikely that the slope is zero.

(c) Give a 95% confidence interval for the slope.

A 95% confidence interval for the slope is

[ ]

Code and Output: > confint( m6)

2.5 % 97.5 %

(Intercept) 0.5559843 1.9087240

X 0.1751401 0.2292829

(d) The parameter corresponds to natural gas consumption for cooking, hot water, and other

uses when there is no demand for heating. Give a 95% confidence interval for this parameter.

A 95% confidence interval for the intercept is

[ ]

Exercise 10.8

(a) Plot the data. Does the trend in lean over time appear to be linear?

Comment:

Yes, the trend appears to be linear.

Plot:

(b) What is the equation of the least-squares line? What percentage of the variation in lean is

explained by this line?

76 78 80 82 84 86

640

660

680

700

720

740

760

Exercise 10.8

Year

Lean

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The equation of the least-squares line is

( )

From the R-squared value, approximately 98.7% of the variation is explained by this line.

Code and Output: > m8= lm( Lean ~ Year)

> summary( m8)

Call:

lm(formula = Lean ~ Year)

Residuals:


-5.9670 -3.0989 0.6703 2.3077 7.3956

Coefficients:


(Intercept) -61.1209 25.1298 -2.432 0.0333 *

Year 9.3187 0.3099 30.069 6.5e-12 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1




(c) Give a 95% confidence interval for the average rate of change (tenths of a millimeter per year)

of the lean.

A 95% confidence interval for the average rate of change (tenths of a millimeter per year) of the

lean is

[ ]

Exercise 10.9

(a) In 1918 the lean was 2.9071. (The coded value is 71.) Using the least-squares equation for the

years 1975 to 1987, calculate a predicted value for the lean in 1918.

The predicted value for the lean in 1918 is 106.6154 tenths of a millimeter.

Code and Output: > predict( m8, newdata=data.frame( Year=c( 18)), interval="prediction",

level=0.95)

fit lwr upr

1 106.6154 62.59376 150.637

(b) Although the least-squares line gives an excellent fit to the data for 1975 to 1987, this pattern

did not extend back to 1918. Write a short statement explaining why this conclusion follows

from the information available. Use numerical and graphical summaries to support your

explanation.

The reason why the conclusion (that the linear pattern does not extend back to 1918) follows

from the information available is because the predicted value for lean in 1918 (2.9107

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meters/coded value 106.6) does not match the value of the actual lean in 1918 (2.9071 meters/

coded value 71). As for the part of the question asking for numerical and graphical summaries,

I’m not sure which numerical/graphical summaries the author has in mind; also, I don’t see how

any numerical/graphical summary would explain why the pattern does not extend back to 1918.

Exercise 10.10

(a) The engineers working on the Leaning Tower of Pisa are most interested in how much the

tower will lean if no corrective action is taken. Use the least-squares equation to predict the

tower’s lean in the year 1997.

The predicted value for the lean in 1997 is 842.7912 tenths of a millimeter.

Code and Output: > predict( m8, newdata=data.frame( Year=c( 97)), interval="prediction",

level=0.95)

fit lwr upr

1 842.7912 828.2892 857.2932

(b) To give a margin of error for the lean in 1997, would you use a confidence interval for a mean

response or a prediction interval? Explain your choice.

We would use a prediction interval because this was a prediction.

Exercise 10.13

Exercise 10.6 gives information about the regression of natural gas consumptionon degree-days for a

particular household.

(a) What is the t statistic for testing ?

The t statistic was 17.663.

(b) For the alternative , what critical value would you use for a test at the

significance level? Do you reject at this level?

We would use the critical value .

Yes, we would reject at this level.

(c) How would you report the P-value for this test?

We would report p= 2.297770e-07.

Code and Output: > pt( q=17.663, df=7, lower.tail=F)

[1] 2.297770e-07

Exercise 10.15

(a) Find and ∑( ) from the data.

We have:

and from the equation:

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( )

we have:

( )

( )

Thus

∑( )

( )

Code and Output: > mean( Spheres)

[1] 13.07

> m15= lm( Vein ~ Spheres)

> summary( m15)

Call:

lm(formula = Vein ~ Spheres)

Residuals:


-2.2136 -1.2829 -0.3227 0.9887 3.0295

Coefficients:


(Intercept) 1.03119 1.22391 0.843 0.424

Spheres 0.90197 0.08344 10.810 4.73e-06 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1




(b) We expect x and y to be positively associated. State hypotheses in terms of the slope of the

population regression line that expresses this expectation, and carry out a significance test.

What conclusions do you draw?

The hypotheses are

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The p-value of the significance test, carried out on a t-value of 10.810 from a distributions

is 2.365e-06 (get this either by dividing the p-value from the two-sided test from the output

from part (a) above (i.e. p= 4.73e-06) by 2 or by using the pt() function in R).

The conclusion is that the slope is positive.

Code and Output:

Either > 4.73e-06/2

[1] 2.365e-06

or > pt( q=10.810 , df=8, lower.tail=F)

[1] 2.365891e-06

(c) Find a 99% confidence interval for the slope.

A 99% confidence interval is

[ ]

Code and Output: > confint( m15, level=.99)

0.5 % 99.5 %

(Intercept) -3.0754859 5.137872

Spheres 0.6219957 1.181953

(d) Suppose that we observe a value of Spheres equal to 15.0 for one dog. Give a 90% interval for

predicting the variable Vein for that dog.

A 90% prediction interval for Vein corresponding to a value of 15.0 for Spheres is

[ ]

Code and Output: > predict( m15, newdata=data.frame( Spheres=c( 15.0)),

interval="prediction", level=0.90)

fit lwr upr

1 14.56081 11.12175 17.99987

Exercise 10.17

(a) Plot the data. Are there any outliers or unusual points?

There are not outliers or unusual points.

Plot:

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(b) Find the least-squares fit to the data, and estimate 1/R for this wire. Then give a 95%

confidence interval for 1/R.

The least-squares fit is

( )

the estimate for 1/R is

and a 95% confidence interval for 1/R is

[ ]

Code and Output: > m17= lm( I ~ V)

> summary( m17)

Call:

lm(formula = I ~ V)

0.5 1.0 1.5 2.0

0.5

1.0

1.5

2.0

Exercise 10.17

V

I

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Residuals:

1 2 3 4 5

-0.007373 0.070402 -0.091823 -0.067158 0.095952

Coefficients:


(Intercept) -0.06485 0.11416 -0.568 0.609729

V 1.18445 0.07790 15.206 0.000618 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



F-statistic: 231.2 on 1 and 3 DF, p-value: 0.0006176

> confint( m17)

2.5 % 97.5 %

(Intercept) -0.4281749 0.2984698

V 0.9365513 1.4323495

(c) If estimates 1/R, then 1/ estimates R. Estimate the resistance R. Similarly, if L and U

represent the lower and upper confidence limits for 1/R, then the corresponding limits for R

are given by 1/U and 1/L, as long as L and U are positive. Use this face and your answer to (b)

to find a 95% confidence interval for R.

A 95% confidence interval for R is

[

] [ ]

(d) Ohm’s law states that in the model is 0. Calculate the test statistic for this hypothesis and

give an approximate P-value.

The test statistic for this hypothesis (see Code and Output for part (b)) is:

and the corresponding p-value from a two-sided test is approximately:

Exercise 10.19

(a) Plot the data. Are there any outliers or unusual points?

There don’t appear to be any outliers. There are a few different VO2 values corresponding to

the same HR value, which might be unusual.

Plot:

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(b) Compute the least-squares regression line for predicting oxygen uptake from hear rate for this

individual.

The least-squares regression line is

( )

Code and Output: > m19= lm( V02 ~ HR)

> m19

Call:

lm(formula = V02 ~ HR)

Coefficients:

(Intercept) HR

-2.80435 0.03865

(c) Test the null hypothesis that the slope of the regression line is 0. Explain in words the

meaning of your conclusions from this test.

100 110 120 130

0.5

1.0

1.5

2.0

Exercise 10.19

HR

V0

2

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The p-value for a two-sided test of the slope being 0 is 1.00e-11.

The meaning is that there is a statistically significant linear relationship between the two

variables.

Code and Output: > summary( m19)

Call:

lm(formula = V02 ~ HR)

Residuals:


-0.355944 -0.035899 0.003056 0.056513 0.220056

Coefficients:


(Intercept) -2.80435 0.25833 -10.86 4.59e-09 ***

HR 0.03865 0.00240 16.10 1.00e-11 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1




(d) Calculate a 95% confidence interval for the oxygen uptake of this individual on a future

occasion when his heart rate is 95. Repeat the calculation for heart rate 110.

95% confidence intervals for the oxygen uptake of this individual when his heart rate is 95 and

110 are

[ ]

and

[ ]

respectively.

Code and Output: > predict( m19, newdata=data.frame( HR=c( 95, 110)),

interval="confidence", level=0.95)

fit lwr upr

1 0.8675958 0.7833762 0.9518154

2 1.4473774 1.3871265 1.5076283

(e) From what you have learned in (a), (b), (c), and (d) of this exercise, do you think that the

researchers should use predicted VO2 in place of measured VO2 for this individual under

similar experimental conditions? Explain your answer.

Yes, researchers may use predicted VO2 in place of measured VO2 for this individual.

I don’t know how to explain it using parts (a), (b), (c), and (d), but I would say, using the Adjusted

R-squared value of 0.9348 that a straight-line model does a good job of explaining the

relationship between VO2 and heart rate.

Exercise 10.20

Calculate the t statistic for testing Specify an appropriate alternative hypothesis for this

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problem and give an approximate p-value for the test. Then explain your conclusion in words a

physician can understand.

The t statistic is

, an appropriate alternative hypothesis is and an

approximate p-value is

The conclusion is that there is a statistically significant straight line relationship between the traditional

procedure and the new procedure; and every unit increase in the new method results in 0.83 units

increase in the old method?

Code and Output:

> 2* pt( 0.83/0.065, df=81-2, lower.tail=F)

[1] 6.71074e-21

Exercise 10.21

(a) It is reasonable to suppose that greater airflow will cause more evaporation. State

hypotheses to test this belief and calculate the test statistic. Find an approximate P-value for

the significance test and report your conclusion.

The hypotheses are

The test statistic is

and an approximate P-value is

The conclusion is that greater airflow will cause more evaporation.

(b) Construct a 95% confidence interval for the additional evaporation experience when airflow

increases by 1 unit.

A 95% confidence interval or the slope is

( )( )

[ ]

Exercise 10.26

Return to the data on current versus voltage given in the Ohm’s law experiment in Exercise 10.17.

(a) Compute all values for the ANOVA table.

Code and Output: > anova( m17)

Analysis of Variance Table

Response: I

Df Sum Sq Mean Sq F value Pr(>F)

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V 1 2.09316 2.09316 231.21 0.0006176 ***

Residuals 3 0.02716 0.00905

(b) State the null hypothesis tested by the ANOVA F statistic, and explain in plain language what

this hypothesis says.

The null hypothesis test by the ANOVA F statistic is (Moore and McCabe 2006, p 655):

In plain language, this hypothesis says that y is not linearly related to x.

(c) What is the distribution of this F statistic when is true? Find an approximate P-value for

the test of

The distribution of this F statistic when is true is an ( ) distribution.

An approximate P-value is .

Exercise 10.28

(a) The correlation between monthly income and birth weight was r=0.39. Calculate the t statistic

for testing the null hypothesis that the correlation is 0 in the entire population of infants.

The t statistic for testing the null hypothesis that the correlation is 0 is given by (Moore and

McCabe, p 664)

√

√

√

√

(b) The researchers expected that higher birth weights would be associated with higher incomes.

Express this expectation as an alternative hypothesis for the population correlation.

The alternative hypothesis expressing this expectation is:

(c) Determine a P-value for versus the alternative that you specified in (b). What conclusion

does your test suggest?

The P-value is 0.006427687. This suggests that monthly income and birth weight are related—

specifically, it suggests that there is a positive correlation between the two.

Code and output: > pt( 2.61086302, df=38, lower.tail=F)

[1] 0.006427687

Exercise 10.29

(a) The correlation between parental control and self-esteem was r = -0.19. Calculate the t

statistic for testing the null hypothesis that the population correlation is 0.

The t statistic is:

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√

√ ( )

(b) Find an approximate P-value for testing versus the two-sided alternative and report your

conclusion.

An approximate p-value for testing the null hypothesis against the two-sided alternative is p =

3.202573e-07. The conclusion is that the correlation coefficient is significant and that since the

value of the correlation coefficient is only -0.19, we can conclude that there is no strong linear

relationship?

Exercise 10.34

(a) Plot the data and describe the pattern. Is it reasonable to summarize this kind of relationship

with a correlation?

The pattern looks somewhat linear. It seems reasonable to summarize this kind of relationship

with a correlation because the correlation between the variables is

(b) Find the correlation and perform the significance test. Summarize the results and report your

conclusion.

The correlation is , the t statistic is ⏞

, and the

corresponding p-value for testing the null hypothesis against the two-sided alternative is

⏞

40 45 50 55 60 65 70 75

4050

6070

80

Exercise 10.34

Femur

Hum

erus

40 50 60 70 80

4050

6070

Humerus

Fem

ur

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Code and Output: > cor.test( Humerus, Femur)

Pearson's product-moment correlation

data: Humerus and Femur

t = 15.9405, df = 3, p-value = 0.0005368

alternative hypothesis: true correlation is not equal to 0

95 percent confidence interval:

0.910380 0.999633

sample estimates:

cor

0.9941486

Exercise 10.35

(a) Plot the data and describe the relationship between the two scores.

There appears to be a somewhat linear relationship between the two scores.

(b) Find the correlation between the two scores and test the null hypothesis that the population

correlation is 0. Summarize your results.

The correlation is , the t statistic is ⏞

, and the corresponding p-

value for testing the null hypothesis against the two-sided alternative is ⏞

In

summary, there is evidence of a linear relationship.

Code and Output:

80 85 90 95 100 105

7580

8590

9510

010

5

Exercise 10.35

Round1

Rou

nd2

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> cor.test( Round1, Round2)


data: Round1 and Round2

t = 2.9903, df = 10, p-value = 0.01357



0.1868394 0.9043687

sample estimates:

cor

0.6870684

(c) The plot shows one outlier. Recompute the correlation and redo the significance test without

this observation. Write a short summary explaining the effect of the outlier on the correlation

and significance test in (b).

The correlation becomes , the t statistic ⏞

, and the

corresponding p-value for testing the null hypothesis against the two-sided alternative

⏞

In summary, the outlier in part (b) had the effect of reducing the correlation

and increasing the p-value.

Code and Output: > detach( data)

> data=data[-8,]

> attach( data)

> cor.test( Round1, Round2)


data: Round1 and Round2

t = 4.6806, df = 9, p-value = 0.001151



0.4890076 0.9579716

sample estimates:

cor

0.841913

Exercise 10.36

(a) Find the equation of the least-squares line for predicting GHP from FVC.

The equation is

( )

where the slope and intercept were found using the equations (Moore and McCabe, p 157)

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and

(b) Give the results of the significance test for the null hypothesis that the slope is 0. (Hint: What

is the relation between this test and the test for a zero correlation?)

Testing for the null hypothesis that the slope is 0 is the same as testing for the null hypothesis

that the correlation is zero? Recall that the t statistic for testing zero correlation is

√

√

√

√

and hence the p-value for a test of zero correlation against the two-sided alternative is

⏞

Exercise 10.39

(a) Plot the data with SAT on the x axis and ACT on the y axis. Describe the overall pattern and

any unusual observations.

The overall relationship looks linear. There’s a potential outlier (observation 42).

Plot:

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(b) Find the least-squares regression line and draw it on your plot. Give the results of the

significance test for the slope.


( )

The significance test for the slope yields a p-value of Thus we strongly reject

the null hypothesis that the slope is zero.

Code and Outputs: > m39=lm( ACT ~ SAT)#-> a=3.47576, b=0.01894

> summary(m39)

Call:

lm(formula = ACT ~ SAT)

Residuals:


-6.29776 -1.78047 0.06872 1.29578 10.39677

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Coefficients:


(Intercept) 1.626282 1.844230 0.882 0.382

SAT 0.021374 0.001983 10.778 1.80e-15 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1




(c) What is the correlation between the two tests?

The correlation is .

Exercise 10.40

(a) What is the mean of these predicted values? Compare it with the mean of the ACT scores.

The mean of predicted values is 21.13333. The mean of the ACT scores is also 20.75833.

Code and Outputs: > ACT.predicted=predict( m39, newdata=data.frame( HR=SAT))

> mean( ACT.predicted)

[1] 20.75833

(b) Compare the standard deviation of the predicted values with the standard deviation of the

actual ACT scores. If least-squares regression is used to predict ACT scores for a large number

of students such as these, the average predicted value will be accurate but the variability of

the predicted scores will be too small.

The standard deviation of predicted values is 3.410658, while the standard deviation of actual

ACT scores is 5.263797.

Code and Output: > s.ACT.predicted=sd( ACT.predicted)#== 3.410658

> s.ACT=sd( ACT)#== 5.263797

(c) Find the SAT score for a student who is one standard deviation above the mean ( (

) ). Find the predicted ACT score and standardize this score. (Use the means and

standard deviations from this set of data for these calculations.)

Student #6 scored a 1440 on the SAT, which is 2.92781269 standard deviations above the mean.

The predicted ACT score for this student is 32.40439, which when standardized also becomes

2.927812—in other words it is 2.927812 standard deviations above the mean predicted ACT

score.

Code and Output: > predict( m39, newdata=data.frame( SAT=c(1440)))#== 32.40439

> (32.40439-mean( ACT.predicted))/s.ACT.predicted#== 2.927812

(d) Repeat part (c) for a student whose SAT score is one standard deviation below the mean (z = -

1).

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Student #7 scored a 490 on the SAT, which is 2.34669184 standard deviations below the mean.

The predicted ACT score for this student is 12.09939, which is also 2.34669184 standard

deviations below the mean predicted ACT score.

(e) What do you conclude from parts (c) and (d)? Perform additional calculation for different z’s

if needed.

We conclude that when using this least-squares line to predict values, the prediction will be the

same number of standard deviations above/below the mean of predicted values as the

explanatory variable is above/below the mean of explanatory variables.

Exercise 10.41

(a) Using the data in Table 10.4, find the values of and .

Using the formula

We get the values and , which seems wrong.

However, if we use the formula

we get the values and

(b) Plot the data with the least-squares line and the new prediction line.

(c) Use the new line to find predicted ACT scores. Find the mean and the standard deviation of

these scores. How do they compare with the mean and standard deviation of the ACT scores?

400 600 800 1000 1200 1400

1020

30

Exercise 10.41: Using a.1= s.x/s.y formula

SAT

ACT

least-squares line

new prediction line

400 600 800 1000 1200 1400

1020

30

Exercise 10.41: Using a.1= s.y/s.x formula

SAT

ACT

least-squares line

new prediction line

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Using the formula

instead of the formula

to determine the slope of the least-squares equation yields new predicted ACT scores with a

mean of 21.13333 and a standard deviation of 4.713725. Compared with the mean and

standard deviation of the ACT scores, they are the same.

Cod and Outpue: > ACT.predicted.new <- (function(SATscore){ -2.752178 +

0.02617112*SATscore})( SAT)

> mean( ACT.predicted.new)#== 21.13333

[1] 21.13333

> mean( ACT)#== 21.13333

[1] 21.13333

> sd( ACT.predicted.new)#== 4.713725

[1] 4.713725

> sd( ACT)#== 4.713726

[1] 4.713726

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Exercise 3.4.1

(a) Based on the output for model (3.7) a business analyst concluded:

[…]

Provide a detailed critique of this conclusion.

The discernable pattern indicates that an improper model has been fit. Also, the outlier

(observation 13) in the plot of studentized residuals versus Distance warrants some concern.

(b) Does the ordinary straight line regression model (3.7) seem to fit the data well? If not,

carefully describe how the model can be improved.

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The ordinary straight line regression does sort of fit the data well, but the model can be

improved by removing the outlier. Even after the model is refit with this outlier removed, there

is still a discernable pattern in the plot of studentized residuals versus Distance:

The pattern indicated that a quadratic term should be added. Addition of a quadratic term

yields:

500 1000 1500

-2-1

01

Distance

Sta

nd

ard

ize

d R

esid

ua

ls

Exercise 3.4.1:Standardized Residuals versus Distance

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The problem of non-random residuals appears to have been fixed, however the improvement in

adjusted R-squared when going from the

Fare ~ Distance

model to the

Fare ~ Distance + DistanceSquared

model is small—the change is only from 99.63% to 99.87%. It might be possible to improve the

model further by looking for more outliers and then refitting the model with them removed.

500 1000 1500

-3-2

-10

1

Distance

Sta

nd

ard

ize

d R

esid

ua

ls

Exercise 3.4.1:Standardized Residuals versus Distance

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Exercise 3.4.2

Is the following statement true or false? If you believe that the statement is false, provide a brief

explanation.

Suppose that a straight line regression model has been fit to bivariate data set of the form

( ) ( ) ( ) Furthermore, suppose that the distribution of X appears to be normal

while the Y variable is highly skewed. A plot of standardized residuals from the least squares regression

line produce a quadratic pattern with increasing variance when plotted against ( ). In this

case, one should consider adding a quadratic term in X to the regression model and thus consider a

model of the form .

Response:

I agree. Regarding the plot of standardized residuals, Sheather writes that if a plot of residuals against

X that produces a discernible pattern, then the shape of the pattern provides information on the

function of x that is missing from the model (p 49). He goes on to write that if the residuals from the

straight-line fit of Y and X have a quadratic pattern, then we can conclude that there is need for a

quadratic term to be added to the original straight-line regression model (p 50).

Regarding the issue of increasing variance, Sheather writes that there are two methods to deal with

it—transformations and weighted least squares.

Exercise 3.4.3

Part A

(a) Develop a simple linear regression model based on least squares that predicts advertising

revenue per page from circulation (i.e. feel free to transform either the predictor or the

response variable or both variables). Ensure that you provide justification for you choice of

model.

A simple linear regression model is

( ) ( ) ( )

Justification:

The plot of AdRevenue ~ Circulation has x values that are spread too far—a log transformation

on the Xs will help “bring” them closer together. The plot of AdRevenue ~ log( Circulation) has y

values that are too care apart—a log transformation on the Ys will, again, help bring them closer

together. The final model

( ) ( )

looks good visiually. A plot of all three models is below:

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(b) Find a 95% prediction interval for the advertising revenue per page for magazines with the

following circulations:

1) 0.5 million

A 95% prediction interval for advertising revenue is [ ].

2) 20 million

A 95 prediction interval for advertising revenue is [ ].

Code and Output:

> logCirculation= log( Circulation)

> logAdRevenue= log( AdRevenue)

> m343= lm( logAdRevenue ~ logCirculation)

0 5 10 15 20 25 30

200

400

600

800

Circulation

AdR

evenue

Exercise 3.4.3:AdRevenue ~ Circulation

-1 0 1 2 3

200

400

600

800

log(Circulation)

AdR

evenue

Exercise 3.4.3:AdRevenue ~ log( Circulation)

-1 0 1 2 3

4.5

5.0

5.5

6.0

6.5

log(Circulation)

log(A

dR

evenue)

Exercise 3.4.3:log( AdRevenue) ~ log( Circulation)

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> logAdRevenue.predicted= predict( m343,

newdata=data.frame( logCirculation= log( c( 0.5, 20))),

interval="prediction", level=.95)

> AdRevenue.predicted= exp( logAdRevenue.predicted)

> AdRevenue.predicted

fit lwr upr

1 74.30864 51.82406 106.5485

2 522.56626 359.89585 758.7626

(c) Describe any weaknesses in your model.

Interpretation of the least squares coefficients becomes difficult with the log transformation

applied to both explanatory and response variables.

Part B

(a) Develop a polynomial regression model based on least-squares that directly predicts the

effect on advertising revenue per page of an increase in circulation of 1 million people (i.e. do

not transform either the predictor nor the response variable). Ensure that you provide

detailed justification for your choice of model. [Hint: Consider polynomial model of order up

to 3.]

A polynomial regression model based on least-squares that directly predicts the effect on

advertising revenue per page of an increase in circulation of 1 million people is:

( ) ( )

( )

Detailed Justification:

This is how we arrived at the above model:

1) we first fit three models:

i.

ii.

iii.

2) we identified the leverage points

3) For each model, we identified which of the leverage points were bad (using the rule that

identifies points outside the interval -2 to 2 as bad)—the bad leverage points were 2, 20,

and 49 for the first model; 2, 4, 20, 49 for the second; and 2, 8, 20, 49 for the third.

4) we removed the bad leverage points for each model and then refit each model to the

new data set (the data set with the bad leverage points removed)

5) Finally, we compared the R squared vales for all three models.

Model 3 resulted in the highest adjusted R- squared value, so that is why it was chosen.

Code and Outputs: #IDENTIFY LEVERAGE POINTS:

leverage.vals= lm.influence(m343b.2)$hat

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leverage.vals[ leverage.vals > 4/ length( Circulation)]# these points are

levereage points: 2, 4, 6, 8, 20, 46, 49.

#CALCULATE STANDARDIZED RESIDUALS FOR ABOVE LEVERAGE PTS AND DETERMINE

WHICH ONES ARE OUTLIERS/BAD:

rstandard.vals1= rstandard( m343b.1)[ c( 2, 4, 6, 8, 20, 46, 49)]

rstandard.vals1[ rstandard.vals1 > 2 | rstandard.vals < -2]# these are

bad leverage points: 2, 20, 49



bad leverage points: 2, 4, 20, 49



bad leverage points: 2, 8, 20, 49

#REMOVE BAD LEVERAGE POINTS (i.e., OUTLIERS) AND REFIT EACH MODEL:

ad.new1= ad[ -c(2, 20, 49),]

detach( ad)

attach( ad.new1)

m343b.1= lm( AdRevenue ~ Circulation)

ad.new2= ad[ -c(2, 4, 20, 49),]

detach( ad.new1)

attach( ad.new2)

m343b.2= lm( AdRevenue ~ Circulation + CirculationSquared)

ad.new3= ad[ -c(2, 8, 20, 49),]

detach( ad.new2)

attach( ad.new3)

m343b.3= lm( AdRevenue ~ Circulation + CirculationSquared +

CirculationCubed)

#COMPARE THE THREE MODELS AGAIN WHEN OUTLIER FOR EACH MODEL ARE REMOVED:

> summary( m343b.1)

Call:

lm(formula = AdRevenue ~ Circulation)

Residuals:


-51.266 -22.737 -7.741 12.815 129.115

Coefficients:


(Intercept) 95.8273 4.8479 19.77 <2e-16 ***

Circulation 24.7248 0.9011 27.44 <2e-16 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1



F-statistic: 752.8 on 1 and 65 DF, p-value: < 2.2e-16

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> summary( m343b.2)

Call:

lm(formula = AdRevenue ~ Circulation + CirculationSquared)

Residuals:


-74.773 -13.330 -2.778 11.169 107.490

Coefficients:


(Intercept) 66.2968 7.1849 9.227 2.62e-13 ***

Circulation 43.2744 3.7622 11.502 < 2e-16 ***

CirculationSquared -0.8077 0.1681 -4.806 9.92e-06 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1




> summary( m343b.3)

Call:

lm(formula = AdRevenue ~ Circulation + CirculationSquared +

CirculationCubed)

Residuals:


-81.992 -12.923 -1.105 11.428 99.918

Coefficients:


(Intercept) 47.95376 9.77958 4.903 7.12e-06 ***

Circulation 62.65352 8.13960 7.697 1.33e-10 ***

CirculationSquared -4.67795 1.44208 -3.244 0.00190 **

CirculationCubed 0.10817 0.03726 2.903 0.00511 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1




(b) Find a 95% prediction interval for the advertising revenue per page for magazines with the

following circulations:

(i) 0.5 million

A 95% prediction interval for the first model is [ ].

A 95% prediction interval for the second model is [ ].

A 95% prediction interval for the third model is [ ].

(ii) 20 million

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A 95% prediction interval for the first model is [ ].

A 95% prediction interval for the second model is [ ].

A 95% prediction interval for the third model is [ ].

(c) Describe any weaknesses in your model.

The weakness in our model (i.e. model 3) is how much greater the lengths of its predication

intervals are compared to the other two models.

Part C

(a) Compare the model in Part A with that in Part B. Decide which provides a better model. Give

reasons to justify your choice.

Not sure. Part A? Because the models in Part B give such different prediction intervals?

(b) Compare the prediction intervals in Part A with those in Part B. In each case, decide which

interval you would recommend. Give reasons to justify each choice.

Not sure—the intervals are all so different and it isn’t clear why any particular one would be

better than the others.

exercise 10 - california state university fullertonmathfaculty.fullerton.edu/sbehseta/hw4 (without...

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