Project #2 Answers
STAT 870
Fall 2012
Complete the following problems below. Within each part, include your R program output with code
inside of it and any additional information needed to explain your answer. Note that you will need to
edit your output and code in order to make it look nice after you copy and paste it into your Word
document.
1) (27 total points) Continue using taste as the response variable and acetic acid as the predictor
variable as in project #1 for the cheese data. Complete the following.
a) (2 points) Give the ANOVA table.
> library(RODBC)
> z<-odbcConnectExcel(xls.file = "C:\\chris\\unl\\Dropbox\\NEW\\
STAT870\\projects\\Fall2012\\cheese.xls")
> cheese<-sqlFetch(channel = z, sqtable = "Sheet1")
> close(z)
> head(cheese)
Case taste Acetic
H2S Lactic
1
1 12.3 4.543 3.135
0.86
2
2 20.9 5.159 5.043
1.53
3
3 39.0 5.366 5.438
1.57
4
4 47.9 5.759 7.496
1.81
5
5
5.6 4.663 3.807
0.99
6
6 25.9 5.697 7.601
1.09
> mod.fit1<-lm(formula = taste ~ Acetic, data = cheese)
> summary(mod.fit1)
Call:
lm(formula = taste ~ Acetic, data = cheese)
Residuals:
Min
1Q
-29.642 -7.443
Median
2.082
3Q
6.597
Max
26.581
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -61.499
24.846 -2.475 0.01964 *
Acetic
15.648
4.496
3.481 0.00166 **
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.82 on 28 degrees of freedom
Multiple R-squared: 0.302,
Adjusted R-squared: 0.2771
F-statistic: 12.11 on 1 and 28 DF, p-value: 0.001658
> anova(mod.fit1)
Analysis of Variance Table
Response: taste
Df Sum Sq Mean Sq F value
Pr(>F)
Acetic
1 2314.1 2314.14 12.114 0.001658 **
Residuals 28 5348.7 191.03
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
1
> var(cheese$taste)*(nrow(cheese) - 1)
[1] 7662.887
> save.anova<-anova(mod.fit1)
> names(save.anova)
[1] "Df"
"Sum Sq" "Mean Sq" "F value" "Pr(>F)"
> save.anova$"Sum Sq"
[1] 2314.142 5348.745
> sum(save.anova$"Sum Sq")
[1] 7662.887
Source of variation
Regression
Error
Total
df
1
28
29
SS
MS
F
2314.1 2314.14 12.114
5348.7 191.03
7662.8
Note that there 2314.1 + 5348.7 = 7662.8 in the ANOVA table although SSTO is actually
7662.9 when rounded to one decimal place. The difference is due to rounding error.
b) (3 points) Using the relevant information from the ANOVA table, perform an F-test for 1 = 0
vs. 1  0. Use  = 0.05.
i) H0: 1 = 0 vs. Ha: 1  0
ii) F = 12.114, p-value = 0.0017
iii)  = 0.05
iv) Because 0.0017 < 0.05, reject H0
v) There is sufficient evidence of a linear relationship between taste and acetic acid.
c) (3 points) What is R2 for the sample regression model? Fully interpret its value.
R2 = 0.30 as given in the summary(mod.fit1) output. Approximately 30% of the variation in
taste is accounted for by the acetic acid value.
d) (8 points) Using my examine.mod.simple() function, comment on the following items with
regards to the model:
i) Linearity of the regression function
ii) Constant error variance
iii) Normality of i
iv) Outliers
Make sure to specifically refer to plots and numerical values in your comments.
> save.it1<-examine.mod.simple(mod.fit.obj = mod.fit1, const.var.test = TRUE,
boxcox.find = TRUE)
> save.it1$levene
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 4.5572 0.04167 *
28
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> save.it1$bp
2
Breusch-Pagan test
data: Y ~ X
BP = 5.4974, df = 1, p-value = 0.01905
> save.it1$lambda.hat
[1] 0.6
Box plot
Residuals vs. predictor
Residuals
-20 -10
-30
6.5
6.0
5.5
5.0
Predictor variable
Predictor variable
Residuals vs. estimated mean response
ei vs. estimated mean response
*
25
35
30
40
Density
15
20
25
1
0
20
25
30
35
40
0.0 0.1 0.2 0.3 0.4 0.5 0.6
20
10
0
Residuals
-20 -10
10
15
Histogram of semistud. residuals
-30
5
10
Estimated mean response
Estimated mean response
0
-1
-3
20
15
10
Residuals vs. observation number
-2
Semistud. residuals
2
20
Residuals
-20 -10
-30
0
10
50
40
30
20
Response variable
10
0
40
30
20
10
0
30
-2
Observation number
-1
0
1
2
Semistud. residuals
Normal Q-Q Plot
Box-Cox transformation plot
-100
-180
-140
log-Likelihood
1
0
-1
Semistud. residuals
-60
2
95%
-2
Response variable
50
3
Dot plot
4.5
6.5
6.0
5.5
5.0
4.5
Box plot
0
10
50
40
30
20
Response variable
0
4.5
10
6.0
5.5
Predictor variable
20
6.5
Response vs. predictor
5.0
6.0
5.5
5.0
4.5
Predictor variable
6.5
Dot plot
-2
-1
0
1
2
-2
-1
0
1
2
Theoretical Quantiles
i) Linearity of the regression function – The plot of the residuals vs. acetic acid shows no
pattern among plotting points. This indicates the linearity assumption is reasonable.
ii) Constant error variance – The plot of the residuals vs. the estimated mean response show
less variability at the smaller values of estimated mean response than at larger values.
3
However, there are much fewer observations at the smaller values so concluding nonconstant error variance is difficult. The BP test has a p-value of 0.0191 and Levene’s test
has a p-value of 0.0417, so there is marginal evidence of non-constant variance from these
hypothesis tests. The Box-Cox transformation plot shows the upper bound for the 95%
confidence interval value for  does not contain 1, but it is quite close. Again, there is
marginal evidence then of non-constant variance.
iii) Normality of i – The normal QQ-plot has some deviation from the straight line toward the
tails of the distribution. The histogram of the semi-studentized residuals also has possible
deviation from normality. However, the sample size is only 30 so it may be difficult to
assess normality with this size of a sample.
iv) Outliers – A plot of the semistudentized residuals vs. the estimated potency shows no
points outside of the 3 bounds. Therefore, there are no outliers.
e) (6 points) You should detect one or more potential problems with the model through the work
in part d). Transform the response variable to be taste . Why do you think this transformation
was chosen? Determine if the transformation helps to solve a problem with the model.
This transformation was chosen because ̂ was estimate to be 0.6. Rather than choosing 0.6,
I chose 0.5 because it results in a more meaningful transformation (square root).
> mod.fit2<-lm(formula = sqrt(taste) ~ Acetic, data = cheese)
> summary(mod.fit2)
Call:
lm(formula = sqrt(taste) ~ Acetic, data = cheese)
Residuals:
Min
1Q
-3.5072 -0.9050
Median
0.4291
3Q
0.8931
Max
2.3050
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -4.5639
2.7805 -1.641 0.11191
Acetic
1.6719
0.5031
3.323 0.00249 **
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1.547 on 28 degrees of freedom
Multiple R-squared: 0.2828,
Adjusted R-squared: 0.2572
F-statistic: 11.04 on 1 and 28 DF, p-value: 0.002489
> save.it2<-examine.mod.simple(mod.fit.obj = mod.fit2, const.var.test = TRUE,
boxcox.find = TRUE, Y = sqrt(cheese$taste))
> save.it2$levene
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.5849 0.4508
28
> save.it2$bp
Breusch-Pagan test
data: Y ~ X
BP = 1.0556, df = 1, p-value = 0.3042
4
> save.it2$lambda.hat
[1] 1.19
Response vs. predictor
Residuals vs. predictor
1
0
-3
1
4.5
4.5
2
-2
-1
Residuals
6
5
4
3
Response variable
6.0
5.5
Predictor variable
5.0
6.0
5.5
5.0
Predictor variable
7
2
6.5
Dot plot
6.5
Box plot
4.5
Dot plot
5.5
6.0
6.5
4.5
5.0
5.5
6.0
6.5
Predictor variable
Predictor variable
Residuals vs. estimated mean response
ei vs. estimated mean response
*
2
1
0
-1
Semistud. residuals
-3
-3
-2
1
0
-2
-1
Residuals
6
5
4
3
2
1
Response variable
6
5
4
3
2
1
3.0
3.5
4.0
4.5
5.0
5.5
6.0
3.0
Estimated mean response
Density
5
10
15
20
25
30
-2
Observation number
1
2
-80 -70 -60 -50 -40 -30 -20
log-Likelihood
1
0
-1
0
5.0
5.5
6.0
-1
0
1
Box-Cox transformation plot
-2
-1
4.5
Semistud. residuals
Normal Q-Q Plot
-2
4.0
Estimated mean response
0.0 0.1 0.2 0.3 0.4 0.5 0.6
1
0
-1
-3
-2
Residuals
0
3.5
Histogram of semistud. residuals
2
Residuals vs. observation number
Semistud. residuals
Response variable
2
7
7
3
Box plot
5.0
95%
-2
-1
0
1
2
Theoretical Quantiles
The transformation appears to help solve the non-constant error variance problem, although
there is still a little less variability for smaller values of Ŷ than for larger values. The 95%
confidence interval for  now contains 1. Also, both the BP and Levene’s tests have large pvalues.
f) (3 points) Using the new model that was estimated for part e), find the 95% confidence
intervals for taste (not taste ) when acetic acid has a value of 4.5 and 6.4. Compare the
5
intervals to those found in project #1. Which intervals (project #1 or #2) are more likely to have
95% confidence? Explain.
The purpose of this problem is to make sure you understand how to find the response variable
in its original form. Also, I wanted you to think about what happens if model assumptions are
not satisfied.
> pred1<-predict(object = mod.fit1, newdata = data.frame(Acetic = c(4.5, 6.4)),
level = 0.95, interval = "confidence")
> pred2<-predict(object = mod.fit2, newdata = data.frame(Acetic = c(4.5, 6.4)),
level = 0.95, interval = "confidence")
> pred1
fit
lwr
upr
1 8.91634 -1.628502 19.46118
2 38.64710 28.863754 48.43044
> pred2^2
fit
lwr
upr
1 8.759054 3.166644 17.13656
2 37.652245 25.414686 52.28717
The 95% confidence intervals for taste (using the transformation-based model) are 3.17 <
E(taste) < 17.14 when acetic acid is 4.5 and 25.41 < E(taste) < 52.29 when acetic acid is 6.4.
In project #1, the corresponding intervals were -1.63 < E(taste) < 19.46 and 28.86 < E(taste) <
48.43, so we do see some differences. The new intervals for this project are more likely to
have 95% confidence because the model’s assumptions are closer to being satisfied.
g) (2 points) Are there any other problems with the model after what was done in part e)? Justify
your answer. Note that you do not need to actually implement any changes to the model.
There still may be problems with the normality of taste . The normal QQ-plot has some
deviation from the straight line toward the tails of the distribution. The histogram of the semistudentized residuals also has possible deviation from normality. However, the sample size is
only 30 so it may be difficult to assess normality with this size of a sample.
2) (12 total points) The extra credit of project #1 asked you to simulate data from a sample
regression model using taste as the response variable (not transformed) and acetic acid as the
predictor variable. Complete the following using my simulated data from the answer key.
a) (1 point) Run my code to simulate the data. Give the simulated Y values to show that you
simulated the data correctly.
> sum.fit1<-summary(mod.fit1)
> set.seed(7172)
> Y.star<-mod.fit1$coefficients[1] + mod.fit1$coefficients[2]*cheese$Acetic +
rnorm(n = 30, mean = 0, sd = sum.fit1$sigma)
> Y.star
[1] -10.310369 19.442977 46.226673 22.374284
1.770760 39.405165 28.965504
37.815467 14.692547
4.394592
[11] 47.148690 49.616239 21.646524 12.658015 27.213700 15.769941 12.332883
14.699950 27.294941 14.316974
[21]
5.417620
2.693531 24.563771 27.522578 26.537968 52.735022 42.552976
19.317254 30.032733 36.492530
6
b) (8 points) Estimate the appropriate regression model for the simulated data. Using my
examine.mod.simple() function, comment on the following items with regards to the model:
i) Linearity of the regression function
ii) Constant error variance (do not examine the Box-Cox transformation value)
iii) Normality of i
iv) Outliers
Make sure to specifically refer to plots and numerical values in your comments.
> mod.fit.sim<-lm(formula = Y.star ~ cheese$Acetic)
> summary(mod.fit.sim)
Call:
lm(formula = Y.star ~ cheese$Acetic)
Residuals:
Min
1Q
-24.523 -6.374
Median
0.103
3Q
4.641
Max
24.887
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-80.467
20.509 -3.924 0.000516 ***
cheese$Acetic
18.973
3.711
5.113 2.04e-05 ***
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 11.41 on 28 degrees of freedom
Multiple R-squared: 0.4828,
Adjusted R-squared: 0.4643
F-statistic: 26.14 on 1 and 28 DF, p-value: 2.038e-05
> save.it.sim<-examine.mod.simple(mod.fit.obj = mod.fit.sim, const.var.test = TRUE,
boxcox.find = TRUE)
> save.it.sim$levene
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.1333 0.7178
28
> save.it.sim$bp
Breusch-Pagan test
data: Y ~ X
BP = 0.0608, df = 1, p-value = 0.8052
7
-20
5.0
5.5
6.0
6.5
5.0
5.5
6.0
6.5
Predictor variable
Residuals vs. estimated mean response
ei vs. estimated mean response
3
*
0
-1
Semistud. residuals
-3
-2
10
0
Residuals
-10
-20
1
2
20
50
20
10
0
-10
4.5
Predictor variable
40
30
Response variable
40
30
20
10
0
-10
10
20
30
40
10
Estimated mean response
30
40
Estimated mean response
Histogram of semistud. residuals
0.1
0.2
Density
0
-10
0.0
-20
Residuals
10
0.3
20
Residuals vs. observation number
20
0
5
10
15
20
25
30
Observation number
-3
-2
-1
0
1
2
3
Semistud. residuals
1
0
-1
Semistud. residuals
2
Normal Q-Q Plot
-2
Response variable
50
Dot plot
0
Residuals
20
0
-10
4.5
4.5
4.5
Box plot
-10
30
10
40
20
50
Residuals vs. predictor
10
Response variable
6.0
5.0
5.5
Predictor variable
6.0
5.5
5.0
Predictor variable
Response vs. predictor
6.5
Dot plot
6.5
Box plot
-2
-1
0
1
2
Theoretical Quantiles
i) Linearity of the regression function – The plot of the residuals vs. acetic acid shows no
pattern among plotting points. This indicates the linearity assumption is reasonable.
ii) Constant error variance – The plot of the residuals vs. the estimated mean response show
similar levels of variability in the residuals. This indicates that there is not sufficient
evidence against the constant variance assumption. The BP test has a p-value of 0.81 and
Levene’s test has a p-value of 0.72, so there again is not sufficient evidence against the
constant variance assumption.
iii) Normality of i – The normal QQ-plot has very little deviation from the straight line in the
tails. The histogram of the semi-studentized residuals roughly follows a normal distribution,
especially for a sample of size 30 only.
8
iv) Outliers – A plot of the semistudentized residuals vs. the estimated potency shows no
points outside of the 3 bounds. Therefore, there are no outliers.
c) (3 points) Answer one of the following:
i) If you found potential problems, what could be a reason for them?
Because we simulated the data, we know the model is correct and that there should not be
any problems. Why did we perhaps have some problems detected then with the QQ-plot?
One potential reason could be due to a sample size of only 30.
ii) If you did not find potential problems, why is this expected?
Based on what I saw in the plots and the summary measures, I would have been satisfied
with the model. This is expected because we simulated the data with all of the correct
assumptions!
Out of curiosity, I increased the sample size to 900 using the following code.
> set.seed(7172)
> Y.star2<-mod.fit1$coefficients[1] + mod.fit1$coefficients[2]*rep(cheese$Acetic,
times = 30) + rnorm(n = 900, mean = 0, sd = sum.fit1$sigma)
> mod.fit.sim2<-lm(formula = Y.star2 ~ rep(cheese$Acetic, times = 30))
> summary(mod.fit.sim2)
Call:
lm(formula = Y.star2 ~ rep(cheese$Acetic, times = 30))
Residuals:
Min
1Q
-41.952 -9.361
Median
-0.046
3Q
8.797
Max
43.113
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)
-67.6876
4.4980 -15.05
<2e-16 ***
rep(cheese$Acetic, times = 30) 16.7935
0.8139
20.63
<2e-16 ***
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 13.7 on 898 degrees of freedom
Multiple R-squared: 0.3216,
Adjusted R-squared: 0.3209
F-statistic: 425.8 on 1 and 898 DF, p-value: < 2.2e-16
> save.it.sim<-examine.mod.simple(mod.fit.obj = mod.fit.sim2, const.var.test =
TRUE)
9
Dot plot
Dot plot
40
Residuals
-20
-40
5.0
5.5
6.0
6.5
4.5
5.5
6.0
6.5
Predictor variable
Residuals vs. estimated mean response
ei vs. estimated mean response
*
1
0
-1
Semistud. residuals
-3
-2
20
0
Residuals
-20
-40
15
20
25
30
35
40
10
Estimated mean response
15
20
25
30
35
40
Estimated mean response
Histogram of semistud. residuals
0.2
0.0
-40
-20
0.1
0
Density
20
0.3
40
0.4
Residuals vs. observation number
Residuals
2
3
40
60
40
20
0
Response variable
-20
10
0
200
400
600
800
-3
Observation number
-2
-1
0
1
2
3
Semistud. residuals
2
1
0
-1
-3
Semistud. residuals
3
Normal Q-Q Plot
-2
60
40
20
0
-20
Response variable
5.0
Predictor variable
80
Box plot
0
20
60
40
20
-20
0
Response variable
6.0
5.5
Predictor variable
4.5
4.5
80
Residuals vs. predictor
80
6.5
Response vs. predictor
5.0
6.0
5.5
5.0
4.5
Predictor variable
6.5
Box plot
-3
-2
-1
0
1
2
3
Theoretical Quantiles
Notice the histogram has a much closer shape to a normal distribution. Also, the QQ-plot has
almost all of its points very close to the straight line. Thus, normality looks to be satisfied. This
occurs because the larger sample size makes it easier to differentiate non-normal from normal.
Also, notice there are 4 semistudentized residuals just outside of the -3 and 3. For a normal
distribution, we would expect about
> 900 - 900*(pnorm(q = 3) - pnorm(q = -3))
[1] 2.429816
to be outside of the boundaries. Thus, even though we have some values just outside of these
boundary lines, this is to be expected. In actual application, one should be worried about a
model if there were a considerable number more than 2.4 outside and if some were much
farther outside than what we observe here.
10