Chapter 5

Simple Linear regression (solutions to

exercises)

Chapter 5 CONTENTS

Contents

5 Simple Linear regression (solutions to exercises)

5.1 Plastic ﬁlm folding machine . . . . . . . . . . . . . . . . . . . . . .

5.2 Linear regression life time model

. . . . . . . . . . . . . . . . . . .

5.3 Yield of chemical process . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . .

5.4 Plastic material

5.5 Water polution .

. . . . . . . . . . . . . . . . . . . . . . . . . .

5.6 Membrane pressure drop . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . .

5.7 Membrane pressure drop (matrix form)

Independence and correlation . . . . . . . . . . . . . . . . . . . . .

5.8

Chapter 5

5.1 PLASTIC FILM FOLDING MACHINE

5.1 Plastic ﬁlm folding machine

Exercise 5.1

Plastic ﬁlm folding machine

On a machine that folds plastic ﬁlm the temperature may be varied in the range

of 130-185 ◦C. For obtaining, if possible, a model for the inﬂuence of tempera-

ture on the folding thickness, n = 12 related set of values of temperature and

the fold thickness were measured that is illustrated in the following ﬁgure:

a) Determine by looking at the ﬁgure, which of the following sets of esti-

mates for the parameters in the usual regression model is correct:

ˆβ0 = 0, ˆβ1 = −0.9, ˆσ = 36

ˆβ0 = 0, ˆβ1 = 0.9, ˆσ = 3.6

ˆβ0 = 252, ˆβ1 = −0.9, ˆσ = 3.6

ˆβ0 = −252, ˆβ1 = −0.9, ˆσ = 36

ˆβ0 = 252, ˆβ1 = −0.9, ˆσ = 36

Solution

First of all, the only possible intercept ( ˆβ0) among the ones given in the answers is

252. And then the slope estimate of -0.9 in these two options looks reasonable. We

13014015016017018090100110120130TemperatureThicknessChapter 5

5.1 PLASTIC FILM FOLDING MACHINE

just need to decide on whether the estimated standard deviation of the error se = ˆσ

is 3.6 or 36. From the ﬁgure it is clear that the points are NOT having an average

vertical distance to the line in the size of 36, so 3.6 must be the correct number and

hence the correct answer is:

3 )

ˆβ0 = 252, ˆβ1 = −0.9, ˆσ = 3.6

b) What is the only possible correct answer:

1) The proportion of explained variation is 50% and the correlation is

0.98

2) The proportion of explained variation is 0% and the correlation is

−0.98

3) The proportion of explained variation is 96% and the correlation is

−1

4) The proportion of explained variation is 96% and the correlation is

0.98

5) The proportion of explained variation is 96% and the correlation is

−0.98

Solution

The proportion of variation explained must be pretty high, so 0 can be ruled out.

Answer 1 and 4 is also ruled out since the correlation clearly is negative. This also

narrows the possibilities down to answer 3 and 5. And since the correlation is NOT

exactly -1 (in which case the observations would be exactly on the line), the correct

answer is:

5) The proportion of explained variation is 96% and the correlation is −0.98

Chapter 5

5.2 LINEAR REGRESSION LIFE TIME MODEL

5.2 Linear regression life time model

Exercise 5.2

Linear regression life time model

A company manufactures an electronic device to be used in a very wide tem-

perature range. The company knows that increased temperature shortens the

life time of the device, and a study is therefore performed in which the life time

is determined as a function of temperature. The following data is found:

Temperature in Celcius (t)

Life time in hours (y)

420

365

285

220

176

117 69 34

a) Calculate the 95% conﬁdence interval for the slope in the usual linear re-

gression model, which expresses the life time as a linear function of the

temperature.

Solution

Either one could do all the regression computations to ﬁnd the ˆβ1 = −5.3133 and

then subsequently use the formula for the conﬁdence interval for β1 in Method 5.15

ˆβ1 ± t1−α/2 · ˆσβ1 = ˆβ1 ± tα/2 · ˆσ

(cid:115)

i=1(xi − ¯x)2 ,

∑n

or just run lm in R to ﬁnd:

Chapter 5

5.2 LINEAR REGRESSION LIFE TIME MODEL

D <- data.frame(t=c(10,20,30,40,50,60,70,80,90),

y=c(420,365,285,220,176,117,69,34,5))

fit <- lm(y ~ t, data=D)

summary(fit)

Call:

lm(formula = y ~ t, data = D)

Residuals:

Min

-21.02 -12.62

Coefficients:

1Q Median

Max

-9.16 17.71 29.64

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

(Intercept) 453.556

-5.313

---

Signif. codes:

14.394

0.256

31.5

8.4e-09 *

-20.8 1.5e-07 *

0 ’’ 0.001 ’’ 0.01 ’’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 19.8 on 7 degrees of freedom

Multiple R-squared:

F-statistic: 432 on 1 and 7 DF,

0.982

p-value: 0.000000151

0.984, Adjusted R-squared:

and use the knowledge of the information in the R-output that wht is know as the

"standard error for the slope” can be directly read off as

(cid:115)

ˆσβ1 = ˆσ

i=1(xi − ¯x)2

∑n

= 0.2558,

and t0.025(7) = 2.364 - in R:

qt(.975,7)

[1] 2.365

to get −5.31 ± 2.365 · 0.2558, or in R:

-5.31+c(-1,1)qt(.975,7)0.2558

[1] -5.915 -4.705

b) Can a relation between temperature and life time be documented on level

Chapter 5

5.2 LINEAR REGRESSION LIFE TIME MODEL

5%?

Solution

Since the conﬁdence interval does not include 0, it can be documented that there

is a relationship between life time and temperature, also the p-value is 1.5 · 10−7 <

0.05 = α, which also give strong evidence against the null-hypothesis.

Chapter 5

5.3 YIELD OF CHEMICAL PROCESS

5.3 Yield of chemical process

Exercise 5.3

Yield of chemical process

The yield y of a chemical process is a random variable whose value is considered

to be a linear function of the temperature x. The following data of correspond-

ing values of x and y is found:

Temperature in ◦C (x)

Yield in grams (y)

100

The average and standard deviation of temperature and yield are

¯x = 50, sx = 39.52847, ¯y = 55.4, sy = 31.66702,

In the exercise the usual linear regression model is used

Yi = β0 + β1xi + εi,

εi ∼ N(0, σ2

ε ),

i = 1, . . . , 5

a) Can a signiﬁcant relationship between yield and temperature be docu-

mented on the usual signiﬁcance level α = 0.05?

Chapter 5

5.3 YIELD OF CHEMICAL PROCESS

Solution

It could most easily be solved by running the regression in R as:

D <- data.frame(x=c(0,25,50,75,100),

y=c(14,38,54,76,95))

fit <- lm(y ~ x, data=D)

summary(fit)

Call:

lm(formula = y ~ x, data = D)

Residuals:

-1.4 2.6 -1.4 0.6 -0.4

Coefficients:

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

(Intercept) 15.4000

0.8000

---

Signif. codes:

1.4967

0.0244

10.3

32.7 0.000063 *

0.002 **

0 ’’ 0.001 ’’ 0.01 ’’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 1.93 on 3 degrees of freedom

Multiple R-squared:

F-statistic: 1.07e+03 on 1 and 3 DF,

0.997, Adjusted R-squared:

p-value: 0.0000627

0.996

Alternatively one could use hand calculations and use the formula in Theorem 5.12

for the t-test of the null hypothesis: H0 : β1 = 0.

The relevant test statistic and p-value can be read off in the R output as 32.7 and

0.000063. So the answer is:

Yes, as the relevant test statistic and p-value are resp. 32.7 and 0.00006 < 0.05 = α.

b) Give the 95% conﬁdence interval of the expected yield at a temperature of

xnew = 80 ◦C.

Chapter 5

5.3 YIELD OF CHEMICAL PROCESS

Solution

We use the formula in Equation (5-59) for the conﬁdence limit of the line (the ex-

pected value of Yi for a value xnew):

ˆβ0 + ˆβ1xnew ± t1−α/2 ˆσ

(cid:115)

(xnew − ¯x)2

Sxx

and we have to compute ˆβ0, ˆβ1 and se either by hand OR in R as above:

ˆβ0 = 15.4, ˆβ1 = 0.8, ˆσ = 1.932.

So the conﬁdence interval becomes

(15.4 + 0.8 · 80) ± 3.182 · 1.932

(cid:114)

(80 − 50)2

6250

since

sx =

n − 1

∑

i=1

(xi − ¯x)2 =

n − 1

Sxx ⇔

Sxx = (n − 1)sx = 4 · 39.5282 = 6250.

79.40 ± 3.61 = [75.79, 83.01].

Thus the answer is

In R this could be by:

predict(fit, newdata=data.frame(x=80), interval="confidence",

level=0.95)

fit

upr

1 79.4 75.79 83.01

lwr

c) What is the upper quartile of the residuals?

Solution

The ﬁve residuals become: -1.4, 2.6, -1.4, 0.6 og -0.4.

We use the basic deﬁnition of ﬁnding a quantile (from Deﬁnition 1.7) and the upper

quartile is q0.75 (see Deﬁnition 1.8). We set n = 5, p = 0.75, so

np = 3.75

Chapter 5

5.3 YIELD OF CHEMICAL PROCESS

So the upper quartile is the 4th observation in the ordered sequence:

−1.4, −1.4, −0.4, 0.6, 2.6.

This is also found in the summary() output above under

Residuals:

-1.4 2.6 -1.4 0.6 -0.4

So the answer is: 0.6.

Chapter 5

5.4 PLASTIC MATERIAL

5.4 Plastic material

Exercise 5.4

Plastic material

In the manufacturing of a plastic material, it is believed that the cooling time has

an inﬂuence on the impact strength. Therefore a study is carried out in which

plastic material impact strength is determined for 4 different cooling times. The

results of this experiment are shown in the following table:

Cooling times in seconds (x)

Impact strength in kJ/m2 (y)

42.1

36.0

31.8

28.7

The following statistics may be used:

¯x = 28.75, ¯y = 34.65, Sxx = 368.75.

a) What is the 95% conﬁdence interval for the slope of the regression model,

expressing the impact strength as a linear function of the cooling time?

Chapter 5

5.4 PLASTIC MATERIAL

Solution

The easiest way to get to the conﬁdence interval is to use the standard error for the

slope ( ˆσβ1 or denoted with SEβ1) given in the R output:

x <- c(15,25,35,40)

y <- c(42.1,36.0,31.8,28.7)

summary(lm(y ~ x))

Call:

lm(formula = y ~ x)

Residuals:

0.2814 -0.6051

0.4085 -0.0847

Coefficients:

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

(Intercept)

---

Signif. codes:

49.639

-0.521

0.878

0.029

56.5 0.00031 *

-18.0 0.00308 **

0 ’’ 0.001 ’’ 0.01 ’’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 0.556 on 2 degrees of freedom

0.991

Multiple R-squared:

F-statistic: 324 on 1 and 2 DF,

0.994, Adjusted R-squared:

p-value: 0.00308

the standard error for the slope is ˆσβ1 = 0.029 (also known as the sampling distribu-

tion standard deviation for ˆβ1). Finding the relevant t-quantile (with ν = 2 degrees

of freedom (either of):

c(qt(0.025, df=2), qt(0.975, df=2))

[1] -4.303

4.303

|t0.025| = 4.303, which using Theorem 5.15 gives

giving

−0.521 ± 4.303 · 0.029,

−0.521 ± 0.125,

or, that we say with high conﬁdence that the true parameter value is in the interval,

i.e.

−0.646 ≤ β1 ≤ −0.396.

Chapter 5

5.4 PLASTIC MATERIAL

b) Can you conclude that there is a relation between the impact strength and

the cooling time at signiﬁcance level α = 5%?

Solution

The relevant p-value can be read off directly from the summary output: 0.00308, and

we can conclude: Yes, as the relevant p-value is 0.00308, which is smaller than 0.05.

c) For a similar plastic material the tabulated value for the linear relation

between temperature and impact strength (i.e the slope) is −0.30. If the

following hypothesis is tested (at level α = 0.05)

H0 : β1 = −0.30

H1 : β1 (cid:54)= −0.30

with the usual t-test statistic for such a test, what is the range (for t) within

which the hypothesis is accepted?

Solution

The so-called critical values for the t-statistic with ν = 2 degrees of freedom is found

as (or at least the negative one of the two): t0.025 = −4.303 - in R: qt(0.975,2)). So

the answer becomes:

[−4.303, 4.303].

Chapter 5

5.5 WATER POLUTION

5.5 Water polution

Exercise 5.5

Water polution

In a study of pollution in a water stream, the concentration of pollution is mea-

sured at 5 different locations. The locations are at different distances to the

pollution source. In the table below, these distances and the average pollution

are given:

Distance to the pollution source (in km)

Average concentration

11.5

10.2

10.3

9.68

9.32

a) What are the parameter estimates for the three unknown parameters in

the usual linear regression model: 1) The intercept (β0), 2) the slope (β1)

and 3) error standard deviation (σ)?

Chapter 5

5.5 WATER POLUTION

Solution

The question is solved by considering the following R-output:

D <- data.frame(concentration=c(11.5, 10.2, 10.3, 9.68, 9.32),

distance=c(2, 4, 6, 8, 10))

fit <- lm(concentration ~ distance, data=D)

summary(fit)

Call:

lm(formula = concentration ~ distance, data = D)

Residuals:

0.324 -0.488

Coefficients:

0.100 -0.032

0.096

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

(Intercept)

distance

---

Signif. codes:

11.664

-0.244

0.365

0.055

31.96 0.000067 *

-4.43

0.021 *

0 ’’ 0.001 ’’ 0.01 ’’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 0.348 on 3 degrees of freedom

0.823

Multiple R-squared:

F-statistic: 19.7 on 1 and 3 DF, p-value: 0.0213

0.868, Adjusted R-squared:

Given the knowledge of the R-output structure, the three values can be read off

directly from the output.

So the correct answer is: ˆβ0 = 11.7, ˆβ1 = −0.244 and SE ˆσ = ˆσ = 0.348.

b) How large a part of the variation in concentration can be explained by the

distance?

Solution

The amount of variation in the model output (Y) explained by the variable input

(x) can be found from the squared correlation, that can be read off directly from the

Chapter 5

5.5 WATER POLUTION

output as "Multiple R-squared". So the correct answer is: R2 = 86.8% (it is actually

an estimate of the variation in concentration which can be explained by distance,

since it is what we found with the particular data at hand. If the sample was taken

again, then this value would vary. We should actually calculate a conﬁdence interval

for R2 to understand how accurate this estimate is!).

c) What is a 95%-conﬁdence interval for the expected pollution concentra-

tion 7 km from the pollution source?

Solution

The wanted number is estimated by the point on the line (using xnew = 7)

and the conﬁdence interval is given by

−0.244 · 7 + 11.664 = 9.96,

9.96 ± t0.025(3) · ˆσ

(cid:115)

(7 − 6)2

Sxx

where Sxx = 42 + 22 + 02 + 22 + 42 = 40 and t0.025(3) = 3.182 (in R: qt(0.975,3))

we have that

(cid:114)

= 0.525,

3.182 · 0.348

where sx is:

sd(D$distance)

[1] 3.162

and thus

This could also have been found by

Sxx = (n − 1) · s2

x = 4 · 3.1622 = 40.

predict(fit, newdata=data.frame(distance=7), interval="confidence",

level=0.95)

fit

upr

1 9.956 9.431 10.48

lwr

So the correct answer is:

9.96 ± 0.525 = [9.43, 10.5].

Chapter 5

5.6 MEMBRANE PRESSURE DROP

5.6 Membrane pressure drop

Exercise 5.6

Membrane pressure drop

When purifying drinking water you can use a so-called membrane ﬁltration.

In an experiment one wishes to examine the relationship between the pressure

drop across a membrane and the ﬂux (ﬂow per area) through the membrane.

We observe the following 10 related values of pressure (x) and ﬂux (y):

Pressure (x)

Flux (y)

1.02

1.15

2.08

0.85

2.89

1.56

4.01

1.72

5.32 5.83

4.32 5.07

7.26

5.00

7.96

5.31

9.11

6.17

9.99

7.04

Copy this into R to avoid typing in the data:

D <- data.frame(

pressure=c(1.02,2.08,2.89,4.01,5.32,5.83,7.26,7.96,9.11,9.99),

flux=c(1.15,0.85,1.56,1.72,4.32,5.07,5.00,5.31,6.17,7.04)

)

a) What is the empirical correlation between pressure and ﬂux estimated to?

Give also an interpretation of the correlation.

Chapter 5

5.6 MEMBRANE PRESSURE DROP

Solution

The questions are most easily solved by using lm in R:

D <- data.frame(

pressure=c(1.02,2.08,2.89,4.01,5.32,5.83,7.26,7.96,9.11,9.99),

flux=c(1.15,0.85,1.56,1.72,4.32,5.07,5.00,5.31,6.17,7.04)

)

fit <- lm(flux ~ pressure, data=D)

summary(fit)

Call:

lm(formula = flux ~ pressure, data = D)

Residuals:

Min

-0.989 -0.318 -0.140 0.454

1Q Median

Max

1.046

Coefficients:

Estimate Std. Error t value

-0.43

10.23 0.0000072 *

Pr(>|t|)

0.68

0.4417

0.0706

(Intercept) -0.1886

0.7225

pressure

---

Signif. codes:

0 ’’ 0.001 ’’ 0.01 ’’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 0.645 on 8 degrees of freedom

Multiple R-squared:

F-statistic: 105 on 1 and 8 DF,

0.92

p-value: 0.00000718

0.929, Adjusted R-squared:

The found coefﬁcient of determination (see Theorem 5.25) can be read off the R out-

put to be 0.929. The sign of the correlation is the same as the sign of the slope, which

can be read off to be positive ( ˆβ1 = 0.7225), so the correlation is

√

ˆρ = r =

0.929 = 0.964.

So the empirical correlation is 0.964, and thus ﬂux is found to increase with increas-

ing pressure.

b) What is a 90% conﬁdence interval for the slope β1 in the usual regression

model?

Chapter 5

5.6 MEMBRANE PRESSURE DROP

Solution

We use the formula for the slope (β1, see Method 5.15) conﬁdence interval, and can

actually just realize that the correct t-quantile to use is the t1−0.05(8) = 1.860 (in R:

qt(0.95,8)), and the other values we read of the summary output.

So the conﬁdence interval is: 0.7225 ± 1.860 · 0.0706.

c) How large a part of the ﬂux-variation (∑10

i=1(yi − ¯y)2) is not explained by

pressure differences?

Solution

The squared correlation, r2 = 0.929 express the explained variation, this means that

1 − 0.929 = 0.071 express the unexplained variation by the model.

d) Can you at signiﬁcance level α = 0.05 reject the hypothesis that the line

passes through (0, 0)?

Solution

The hypothesis is the same as:

H0 : β1 = 0

which is the hypothesis results provided in the output in the "intercept" row of

summary, so: No, since the relevant p-value is 0.68, which is larger than α

e) A conﬁdence interval for the line at three different pressure levels: xA

new =

3.5, xB

new = 5.0 and xC

new = 9.5 will look as follows:

ˆβ0 + ˆβ1 · xU

new ± CU

where U then is either A, B or C. Write the constants CU in increasing

order.

Chapter 5

5.6 MEMBRANE PRESSURE DROP

Solution

The formula for the Conﬁdence limits of α + βxnew includes the following term:

(xnew − ¯x)2

Sxx

and this is the ONLY term in CU that makes CU different between the three Us. And

since ¯x = 5.547 it is clear that

(5.0 − 5.547)2 < (3.5 − 5.547)2 < (9.5 − 5.547)2

and hence

(xB

new − 5.547)2 < (xA

new − 5.547)2 < (xC

new − 5.547)2

So CB < CA < CC

Chapter 5

5.7 MEMBRANE PRESSURE DROP (MATRIX FORM)

5.7 Membrane pressure drop (matrix form)

Exercise 5.7

Membrane pressure drop (matrix form)

This exercise uses the data presented in Exercise 6 above.

a) Find parameters values, standard errors, t-test statistics, and p-values for

the standard hypotheses tests.

Copy this into R to avoid typing in the data:

D <- data.frame(

pressure=c(1.02,2.08,2.89,4.01,5.32,5.83,7.26,7.96,9.11,9.99),

flux=c(1.15,0.85,1.56,1.72,4.32,5.07,5.00,5.31,6.17,7.04)

)

Solution

Chapter 5

5.7 MEMBRANE PRESSURE DROP (MATRIX FORM)

D <- data.frame(

pressure=c(1.02,2.08,2.89,4.01,5.32,5.83,7.26,7.96,9.11,9.99),

flux=c(1.15,0.85,1.56,1.72,4.32,5.07,5.00,5.31,6.17,7.04)

)

fit <- lm(flux ~ pressure, data=D)

summary(fit)

Call:

lm(formula = flux ~ pressure, data = D)

Residuals:

Min

-0.989 -0.318 -0.140 0.454

1Q Median

Max

1.046

Coefficients:

Estimate Std. Error t value

-0.43

10.23 0.0000072 *

Pr(>|t|)

0.68

0.4417

0.0706

(Intercept) -0.1886

0.7225

pressure

---

Signif. codes:

0 ’’ 0.001 ’’ 0.01 ’’ 0.05 ’.’ 0.1 ’ ’ 1

Residual standard error: 0.645 on 8 degrees of freedom

Multiple R-squared:

F-statistic: 105 on 1 and 8 DF,

0.92

p-value: 0.00000718

0.929, Adjusted R-squared:

The parameter estimates are given in the ﬁrst column, the standard errors in the

second column, the t-test statistics are given in the third column and the p-values of

the standard hypothesis are given in the last column.

b) Reproduce the above numbers by matrix vector calculations. You will

need some matrix notation in R:

– Matrix multiplication (XY): X%*%Y

– Matrix transpose (XT): t(X)

– Matrix inverse (X−1): solve(X)

– Make a matrix from vectors (X = [xT

1 ; xT

2 ]): cbind(x1,x2)

Collection in a table

analasis.table <- cbind(beta, se.beta, t.obs, p.value)

analasis.table

se.beta

[1,] -0.1886 0.44171 -0.4269 0.680696710

[2,] 0.7225 0.07064 10.2269 0.000007177

Put some names on our table

colnames(analasis.table) <- c("Estimates","Std.Error","t.obs","p.value")

rownames(analasis.table) <- c("beta1","beta2")

analasis.table

Estimates Std.Error

t.obs

p.value

0.44171 -0.4269 0.680696710

0.07064 10.2269 0.000007177

beta1

beta2

-0.1886

0.7225

Done!!

Chapter 5

5.8 INDEPENDENCE AND CORRELATION

5.8 Independence and correlation

Exercise 5.8

Independence and correlation

Consider the layout of independent variable in Example 5.11,

a) Show that Sxx = n·(n+1)

12·(n−1) .

Hint: you can use the following relations

∑

i=1

∑

i=1

i =

n(n + 1)

i2 =

n(n + 1)(2n + 1)

Solution

¯x becomes

¯x =

i − 1

n − 1

∑

i=1

n(n − 1)

(cid:18) n(n + 1)

− n

∑

i=1

(cid:19)

(i − 1)

and Sxx becomes

Sxx =

(cid:18) i − 1

n − 1

(cid:19)2

−

∑

i=1

= −

(n − 1)2

(i2 + 1 − 2i)

∑

i=1

(cid:18) n(n + 1)(2n + 1) − 6n2

(cid:19)

(n − 1)2

(cid:18) 4n2 + 6n + 2 − 12n − 3(n − 1)2

n(n + 1)

12(n − 1)

(cid:18) n2 − 1

(cid:19)

Chapter 5

5.8 INDEPENDENCE AND CORRELATION

b) Show that the asymptotic correlation between ˆβ0 and ˆβ1 is

n→∞ ρn( ˆβ0, ˆβ1) = −

lim

√

Solution

The correlation between ˆβ0 and ˆβ0 is

ρn( ˆβ0, ˆβ1) =

Cov( ˆβ0, ˆβ1)

(cid:113)

V( ˆβ0) V( ˆβ1)

σ2 ¯x/Sxx

(cid:16) 1

n + ¯x2

Sxx

¯x/Sxx

(cid:114)(cid:16) Sxx

σ4

(cid:114)

Sxx

= −

(cid:17)

n + ¯x2

(cid:17) 1

Sxx

= −

¯x

(cid:113) Sxx

n + ¯x2

Notice that the correlation is not a function of the variance (σ2), but only a function

of the independent variables. Now insert the values of ¯x and Sxx

ρn( ˆβ0, ˆβ1) = −

= −

(cid:113) n+1

12(n−1) + 1

(cid:113) 2n−1

6(n−1)

= −

(cid:115)

6(n − 1)

2(n − 1/2)

= −

(cid:113) n+1+3(n−1)

12(n−1)

(cid:112)

6(n − 1)

√

2n − 1

√

(cid:114)

= −

n − 1

n − 1/2

which converges to −

√

2 for n → ∞.

Consider a layout of the independent variable where n = 2k and xi = 0 for i ≤ k

and xi = 1 for k < i ≤ n.

c) Find Sxx for the new layout of x.

Chapter 5

5.8 INDEPENDENCE AND CORRELATION

Solution

and

Snew

xx =

¯x =

(cid:18)

0 −

(cid:19)2

∑

i=k+1

(cid:18)

1 −

(cid:19)2

∑

i=1

d) Compare Sxx for the two layouts of x.

Solution

Sxx

Snew

n(n + 1)

12(n − 1)

(n + 1)

3(n − 1)

< 1;

f or n > 2

which imply that Snew

xx > Sxx for all n > 2.

e) What is the consequence for the parameter variance in the two layouts?

Solution

The larger Sxx for the new layout imply that the parameter variance is smaller for

the new layout (given that data comes from the same model).

f) Discuss pro’s and cons for the two layouts.

Chapter 5

5.8 INDEPENDENCE AND CORRELATION

Solution

The smaller parameter variance for the new layout would suggest that we should

use this layout. However, we would not be able to check that data is in fact generated

by a linear model. Consider e.g. data generated by the model

yi = β0 + β1x2

i + εi,

εi ∼ N(0, σ2),

if we only look at xi = 0 or xi = 1 we will not be able to detect that the relationship

is in fact non-linear.