Applied Linear Statistical Models
Midterm Test – November 2, 2003
Dr. Douglas H. Jones
Instructions: Please mark your answers in the answer sheet (attached). Mark in the
upper right hand side the header of answer sheet with your name and student number.
You are not to consult with any person. You may use only the notes, textbook, SPSS, and
Excel.
Problem scenario: Sixteen batches of plastic were made, and from each batch one test
item was molded. Each test item was randomly assigned to one of the four predetermined
times levels, and the hardness was measured after the assigned elapsed time. The results
are shown below; X is the elapsed time in hours, and Y is hardness in Brinell units.
OBS
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Y
199
205
196
200
218
220
215
223
237
234
235
230
250
248
253
246
X
16
16
16
16
24
24
24
24
32
32
32
32
40
40
40
40
The actual data are to be found at the website
http://www.rci.rutgers.edu/~dhjones/APPLIED_LINEAR_STATISTICAL_MODELS(PHD)/
TEST1/ under the file name plastic.xls.
The first-order regression model
Yi   0  1 X i   i
is to be fit to the data. Using this model, answer the following problems.
1
Problem 1
The 95 percent confidence interval for the change in the mean hardness when the
elapsed time increases by one hour is
a) Lower = 1.830; Upper = 2.248
b) Lower = 162.901; Upper = 174.299
c) Lower = 1.840; Upper = 2.228
d) Lower = 162.941; Upper = 174.259
Problem 2
The percent of variation in Y accounted for by the model is
a) 97.1 percent
b) 98.6 percent
c) 97.3 percent
d) 97.3 percent
Problem 3
The plastic should get harder as time increases. Conduct a test to decide whether this is
being true; use  = .05. The alternatives, decision rule, and conclusion is
a) H0: 1 = 0 H0: 1 > 0; Reject Ha if p-value < .05; p-value  1.08X10-12, therefore reject
H0.
b) H0: 1 = 0 H0: 1 > 0; Reject Ha if p-value < .05; p-value  2.16X10-12, therefore reject
H0.
c) H0: 1 = 0 H0: 1  0; Reject Ha if p-value < .05; p-value  2.16X10-12, therefore reject
H0.
d) H0: 1 = 0 H0: 1 > 0; Reject Ha if p-value > .05; p-value  2.16X10-12, therefore reject
H0.
2
Problem 4
The plastic manufacturer has stated that the mean hardness should increase by 2 Brinell
units per hour. Conduct a two-sided test to decide whether this standard is being satisfied;
use  = .05. The alternatives, decision rule, and conclusion is
a) H0: 1 = 0 H0: 1 = 2; Reject Ha if p-value < .05; p-value  0.709444456, therefore do
not reject H0.
b) H0: 1 = 0 H0: 1 = 2; Reject Ha if p-value < .05; p-value  0.354722228, therefore
reject H0.
c) H0: 1 = 2 H0: 1  2; Reject Ha if p-value > .05; p-value  0.354722228, therefore do
reject H0.
d) H0: 1 = 2 H0: 1  2; Reject Ha if p-value < .05; p-value  0.709444456, therefore do
not reject H0.
Problem 5
Obtain the power of your test in Problem 4 if the standard actually is being exceeded by .3
Brinell units per hour. Assume {b1} = .1. Select the closest value from the following
choices.
a) .05
b) .80
c) .53
d) .95
Problem 6
Perform the F test determine whether or not there is lack of fit of a linear regression
function; use  = .05. The alternatives, decision rule, and conclusion is
a) H0: E(Y) = 0 + 1X Ha: E(Y)  0 + 1X; Reject H0 if F> F(.95; 2, 14) ; p-value  0.462,
therefore do not reject H0.
b) H0: E(Y)  0 + 1X Ha: E(Y) = 0 + 1X; Reject H0 if F< 3.89; F = 0.82, therefore reject
H0.
c) H0: E(Y) = 0 + 1X Ha: E(Y)  0 + 1X; Reject H0 if F> 3.89; F  0.82, therefore do not
reject H0.
d) H0: E(Y) = 0 + 1X Ha: E(Y)  0 + 1X; Reject H0 if F> 3.89 ; F  493.75, therefore
reject H0.
3
Problem 7
Management wishes to obtain interval estimates of the mean hardness when the elapsed
time is 20, 30, and 40 hours, respectively. Calculate the desired confidence intervals,
using the Bonferroni procedures and a 95 percent family confidence coefficient.
a)
X
20
30
40
Lower
206.9610
227.8544
247.0733
Upper
211.6140
231.4082
252.8767
b)
X
20
30
40
Lower
206.3395
227.3797
246.2982
Upper
212.2355
231.8828
253.6518
c) X
20
30
40
Lower
227.8544
206.9610
247.0733
Upper
231.4082
211.6140
252.8767
d)
Lower
205.9587
227.0889
245.8233
Upper
212.6163
232.1736
254.1267
X
20
30
40
4
Problem 8
Management wishes to obtain interval estimates of the mean hardness when the elapsed
time is 20, 30, and 40 hours, respectively. Calculate the desired confidence intervals,
using the Working-Hotelling procedures and a 95 percent family confidence coefficient.
a)
X
20
30
40
Lower
206.3212
227.3658
246.2755
Upper
212.2538
231.8967
253.6745
b)
X
20
30
40
Lower
201.2605
223.5006
239.9636
Upper
217.3145
235.7619
259.9864
c) X
20
30
40
Lower
227.8544
206.9610
247.0733
Upper
231.4082
211.6140
252.8767
d)
Lower
205.9587
227.0889
245.8233
Upper
212.6163
232.1736
254.1267
X
20
30
40
5
Problem scenario: The primary objective of the Study on the Efficacy of Nosocomial
Infection Control (SENIC Project) was to determine whether infection surveillance and
control programs have reduced the rates of nosocomial (hospital-acquired) infection in the
United States hospitals. This data set consists of a random sample of 113 hospitals
selected from the original 338 hospitals surveyed.
Variable Number
Variable
Name
Variable Label
Description
1
ID
Identification Number
1 -- 113
4
Y
Infection Risk
Average estimated probability acquiring infection
in hospital (in percent)
5
X1
Routine culturing ratio
Ratio of number of cultures performed to
number of patients without signs or symptoms of
hospital-acquired infection, times 100
6
X2
Routine chest X-ray ratio
Ratio of number of X-rays performed to number
of patients without signs or symptoms of
pneumonia, times 100
The actual data are to be found at the website
http://www.rci.rutgers.edu/~dhjones/APPLIED_LINEAR_STATISTICAL_MODELS(PHD)/
TEST1/ under the file name senic.sav. The first-order regression model in two predictors
Yi  0  1 X i1  1 X i 2   i
is to be fit to the data. For this model, answer the following problems.
Problem 9
What is the value of
( XX)1 ?
a) 0.168187 0.000264411
0.000264 0.000104023
-0.002003 -2.3365E-05
-0.00200314
-2.3365E-05
2.90603E-05
b) -1.275929 -0.01658002 0.018947172
0.01852 0.000216022 -0.000268678
0.014508 0.000270169 -0.000230009
c) 0.159428 -0.00318123 -0.001229205
0.004512 0.001775218 -0.000398737
-0.002085 -5.5736E-05 3.63311E-05
d) 0.381408 0.002751463 -0.005096418
-0.109853 -0.00128135 0.001593677
0.006702 0.000179107 -0.000116751
6
Problem 10
What is the value of b?
a)
(Constant)
X1
X2
b
0.018235
1.940969
0.058598
(Constant)
X1
X2
b
1.940969
0.018235
0.058598
(Constant)
X1
X2
b
0.018235
0.058598
1.940969
(Constant)
X1
X2
b
1.940969
0.058598
0.018235
b)
c)
d)
Problem 11
What is the value of SStot?
a) .358
b) .608
c) 74.405
d) 201.380
7
Problem 12
What is the value of standard error of the estimate?
a) 126.975
b) .369
c) 1.07439
d) 37.203
8
Name:
SN:
Answer Sheet (Circle your choice)
Problem
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