Stat 401B Final Exam
(slightly revised version)
Fall 2015
I have neither given nor received unauthorized assistance on this exam.
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ATTENTION!
Incorrect numerical answers unaccompanied by supporting reasoning will receive NO
partial credit.
Correct numerical answers to difficult questions unaccompanied by supporting
reasoning may not receive full credit.
SHOW YOUR WORK/EXPLAIN YOURSELF!
1
1. Some
4 pts
6 pts
1
2
inch finished hex nuts have weights that are normally distributed with mean 17gm and
standard deviation .6gm .
a) What fraction of these nuts have weights above 17.4gm ?
b) These nuts are packaged by weight. A package (intended to hold at least 100 of these hex nuts)
will be filled with a weight of nuts that is at least 1710gm . Approximate the probability that 99 nuts
have a total weight of at least 1710gm (so that the actual count of nuts is less than desired number).
(Hint: What would the average weight of these 99 nuts have to be for this to happen?)
2. A data base is segmented as below in terms of "type of record" and "completeness of record."
Suppose that one will select a single record at random from this data base.
Type A
Type B
Type C
200 records 350 records 450 records
Complete
Incomplete 50 records 50 records 100 records
5 pts
a) Evaluate P  Type B | Incomplete .
5 pts
b) Are the events "Type B" and "Incomplete" independent? Say why or why not.
2
3. Some experimental data on the page "Using Central Composite Design for Process Optimization"
on the weibull.com web site concern the tensile strength of welds made on steel. We will use
these data in various ways in this problem.
5 pts
First, n  7 welds made at a standard set of process conditions produced y  6611 kgf and
s  145 kgf .
a) Give 95% confidence limits for the mean strength of steel welds made under standard process
conditions. (Plug in completely, but you need not do arithmetic.)
5 pts
b) Interpret your interval from a). (Say carefully what is meant by the "95%" figure.)
5 pts
c) A weld is made at a non-standard set of process conditions and its strength tested. The value
y  5830 kgf is observed. Give 95% confidence limits for the difference in mean strengths for the
two sets of welding process conditions. (Plug in completely, but you need not do arithmetic.)
5 pts
d) The non-standard set of process conditions referred to in part c) actually differs from the standard
set only in the electrical current applied. Coded values of the current are x  1 for the non-standard
conditions and x  0 for the standard conditions. A plot of the data is below. What model
assumptions (be complete in stating them) would you make in order to support a prediction interval
for the strength of a weld made with coded current x  .5 (and all other process conditions standard)?
3
5 pts
e) In fact, for the situation of d) sLF  145 kgf , x  .125 and
8
 x  x 
i 1
i
2
.875 . Give 95%
prediction limits for the next y at x  .5 under your assumptions of d). (The least squares line
should be obvious to you from the plot in d) and the information given at the beginning of the
problem.) (Plug in completely, but you need not do arithmetic.)
5 pts
f) There was also a weld made at coded electrical current x  1 that had strength y  6210 . A plot of
this data point, the 7 data points mentioned at the start of the problem, and the one mentioned in part c)
is below. If strength is a linear function of current,  1  0    0   1   0 . Give 95% confidence
limits for
 1  0    0  1  .
Is it plausible that strength is linear over this range of x ?
The entire data set was used to fit a model for strength, y , as a linear function of coded current, x1 ,
coded voltage, x2 , coded "stick out", x3 , and coded angle, x4 . (See the R output beginning on page 7.)
5 pts
g) What fraction of the raw variability in y is accounted for using x1 , x2 , x3 , and x4 as predictor
variables?
5 pts h) What is the meaning of b3  305.56 ? (Interpret this fitted coefficient.)
4
5 pts
i) There is R output for the fit of a full quadratic model for y in the predictor variables
x1 , x2 , x3 , and x4 beginning on page 8. Give the value of an F statistic and degrees of freedom for
testing whether the quadratic model is a statistically significant improvement over the linear model in
x1 , x2 , x3 , and x4 for explaining y .
F  ___________
d . f .  _______ , _______
4. Lab #10 used balanced 23 factorial data of Example 4 of Chapter 8 of Vardeman and Jobe. We will
here continue use of that scenario.
5 pts
a) Below is the ANOVA table produced in Lab #10. Suppose that one runs the regsubsets()
function from the leaps package using the 7 dummy variables created in the lab. Which model with
4 predictors will be identified as best, and what value of R 2 is associated with it?
5 pts
b) Why would it be impossible to answer part a) based on only the table above if the data were not
balanced?
5 pts
c) CVSSE values for the best (in terms of R 2 ) models with k  1, 2, , 7 factorial effects were
computed using 8-fold cross validation. A plot of these is below. In light of this plot and the
ANOVA table above, what "few effects" model for Power appears best? How does its CVSSE
compare to the SSE that would be obtained fitting it?
5
5 pts 5. Below is a toy data set consisting of N  5 pairs  x, y  . Find the LOO cross-validation SSE for 1-
nn prediction. (You don't need to do arithmetic, but write out a complete numerical expression. )
y
2.5
4.0
4.5
7.5
8.0
x
1.0
2.0
2.5
3.5
4.0
6. Below are fake regression trees that you may assume come from B  3 bootstrap samples of a large
number, N , of  x1i , x2i , yi  data points.
5 pts a) What is the random forest prediction at  x1 , x2   .5,.5  ? (Assume that the forest includes only the
B  3 trees represented above.)
5 pts b) Suppose that in fact  x1 , x2 , y   .5,.5,3 was part of the bootstrap samples that were used to
produce trees #1 and #2, but not #3. What is this data point's contribution to an OOB error sum of
squares?
5 pts
7. Consider the ordinary least squares predictor ŷ OLS  x  (from standard multiple linear regression)
and a lasso predictor ŷ Lasso  x  (for some  ) computed for the same data. Is it true that the SSE for
ŷ Lasso  x  is always at least as big as that for ŷ OLS  x  ? Say why or why not.
6
R Code and Output
> Welds
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
current voltage stickout angle strength
-1
-1
-1
-1
4730
1
-1
-1
-1
4990
-1
1
-1
-1
4240
1
1
-1
-1
7320
-1
-1
1
-1
7130
1
-1
1
-1
4920
-1
1
1
-1
4110
1
1
1
-1
5020
-1
-1
-1
1
5560
1
-1
-1
1
4910
-1
1
-1
1
5330
1
1
-1
1
7490
-1
-1
1
1
6820
1
-1
1
1
4030
-1
1
1
1
3690
1
1
1
1
4210
-1
0
0
0
5830
1
0
0
0
6210
0
-1
0
0
6230
0
1
0
0
6530
0
0
-1
0
6370
0
0
1
0
5510
0
0
0
-1
6390
0
0
0
1
6110
0
0
0
0
6550
0
0
0
0
6650
0
0
0
0
6750
0
0
0
0
6610
0
0
0
0
6340
0
0
0
0
6600
0
0
0
0
6780
> summary(lm(strength~.,Welds))
Call:
lm(formula = strength ~ ., data = Welds)
Residuals:
Min
1Q
-1740.7 -1023.5
Median
312.6
3Q
803.2
Max
1607.1
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 5805.16
199.69 29.070
<2e-16 ***
current
92.22
262.06
0.352
0.728
voltage
-76.67
262.06 -0.293
0.772
stickout
-305.56
262.06 -1.166
0.254
angle
-38.89
262.06 -0.148
0.883
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 1112 on 26 degrees of freedom
Multiple R-squared: 0.05766,
Adjusted R-squared: -0.08731
F-statistic: 0.3977 on 4 and 26 DF, p-value: 0.8084
7
> anova(lm(strength~.,Welds))
Analysis of Variance Table
Response: strength
Df
Sum Sq Mean Sq F value
current
1
153089 153089 0.1238
voltage
1
105800 105800 0.0856
stickout
1 1680556 1680556 1.3595
angle
1
27222
27222 0.0220
Residuals 26 32141108 1236196
Pr(>F)
0.7277
0.7722
0.2542
0.8832
> summary(lm(strength~.,data.frame(Welds2)))
Call:
lm(formula = strength ~ ., data = data.frame(Welds2))
Residuals:
Min
1Q
-262.13 -142.13
Median
5.84
3Q
104.91
Max
350.64
Coefficients:
Estimate Std. Error t value
(Intercept)
6544.16
69.59 94.045
current
190.00
165.87
1.145
voltage
150.00
165.87
0.904
stickout
-305.56
55.29 -5.526
angle
-38.89
55.29 -0.703
current2
-445.68
145.61 -3.061
voltage2
-85.68
145.61 -0.588
stickout2
-525.68
145.61 -3.610
angle2
-215.68
145.61 -1.481
currentvoltage
753.75
58.64 12.853
currentstickout -526.25
58.64 -8.974
currentangle
-110.00
175.93 -0.625
voltagestickout -628.75
58.64 -10.722
voltageangle
-255.00
175.93 -1.449
stickoutangle
-277.50
58.64 -4.732
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’
Pr(>|t|)
< 2e-16
0.268853
0.379236
4.60e-05
0.491936
0.007469
0.564470
0.002348
0.157979
7.56e-10
1.21e-07
0.540624
1.03e-08
0.166533
0.000226
***
***
**
**
***
***
***
***
0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 234.6 on 16 degrees of freedom
Multiple R-squared: 0.9742,
Adjusted R-squared: 0.9516
F-statistic: 43.13 on 14 and 16 DF, p-value: 5.149e-10
> anova(lm(strength~.,data.frame(Welds2)))
Analysis of Variance Table
Response: strength
Df
current
1
voltage
1
stickout
1
angle
1
current2
1
voltage2
1
stickout2
1
angle2
1
currentvoltage
1
currentstickout 1
currentangle
1
Sum Sq
153089
105800
1680556
27222
8379097
554530
990677
120721
9090225
4431025
21511
Mean Sq F value
Pr(>F)
153089
2.7822 0.1147672
105800
1.9228 0.1845697
1680556 30.5418 4.601e-05 ***
27222
0.4947 0.4919356
8379097 152.2787 1.371e-09 ***
554530 10.0778 0.0058839 **
990677 18.0042 0.0006200 ***
120721
2.1939 0.1579786
9090225 165.2025 7.560e-10 ***
4431025 80.5279 1.212e-07 ***
21511
0.3909 0.5406241
8
voltagestickout 1 6325225 6325225 114.9524 1.033e-08 ***
voltageangle
1 115600 115600
2.1009 0.1665326
stickoutangle
1 1232100 1232100 22.3917 0.0002256 ***
Residuals
16 880396
55025
--Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
9