Applied Linear Statistical Models KEY
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. You may unstaple the test if it is convenient for you.
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 / Y / X1 / 199 / 16
2 / 205 / 16
3 / 196 / 16
4 / 200 / 16
5 / 218 / 24
6 / 220 / 24
7 / 215 / 24
8 / 223 / 24
9 / 237 / 32
10 / 234 / 32
11 / 235 / 32
12 / 230 / 32
13 / 250 / 40
14 / 248 / 40
15 / 253 / 40
16 / 246 / 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
is to be fit to the data. Using this model, answer the following problems.
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 a = .05. The alternatives, decision rule, and conclusion is
a) H0: b1 = 0 Ha: b1 > 0; Reject H0 if p-value < .05; p-value @ 1.08X10-12, therefore reject H0.
b) H0: b1 = 0 Ha: b1 > 0; Reject H0 if p-value < .05; p-value @ 2.16X10-12, therefore reject H0.
c) H0: b1 = 0 Ha: b1 ¹ 0; Reject H0 if p-value < .05; p-value @ 2.16X10-12, therefore reject H0.
d) H0: b1 = 0 Ha: b1 > 0; Reject H0 if p-value > .05; p-value @ 2.16X10-12, therefore reject H0.
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 a = .05. The alternatives, decision rule, and conclusion is
a) H0: b1 = 0 Ha: b1 = 2; Reject H0 if p-value < .05; p-value @ 0.709444456, therefore do not reject H0.
b) H0: b1 = 0 Ha: b1 = 2; Reject H0 if p-value < .05; p-value @ 0.354722228, therefore reject H0.
c) H0: b1 = 2 Ha: b1 ¹ 2; Reject H0 if p-value > .05; p-value @ 0.354722228, therefore do reject H0.
d) H0: b1 = 2 Ha: b1 ¹ 2; Reject H0 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 s{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 a = .05. The alternatives, decision rule, and conclusion is
a) H0: E(Y) = b0 + b1X H0: E(Y) ¹ b0 + b1X; Reject H0 if F> F(.95; 2, 14) ; p-value @ 0.462, therefore do not reject H0.
b) H0: E(Y) ¹ b0 + b1X H0: E(Y) = b0 + b1X; Reject H0 if F< 3.89; F = 0.82, therefore reject H0.
c) H0: E(Y) = b0 + b1X H0: E(Y) ¹ b0 + b1X; Reject H0 if F> 3.89; F @ 0.82, therefore do not reject H0.
d) H0: E(Y) = b0 + b1X H0: E(Y) ¹ b0 + b1X; Reject H0 if F> 3.89 ; F @ 493.75, therefore reject H0.
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 / Lower / Upper20 / 206.9610 / 211.6140
30 / 227.8544 / 231.4082
40 / 247.0733 / 252.8767
b) / X / Lower / Upper
20 / 206.3395 / 212.2355
30 / 227.3797 / 231.8828
40 / 246.2982 / 253.6518
c) / X / Lower / Upper
20 / 227.8544 / 231.4082
30 / 206.9610 / 211.6140
40 / 247.0733 / 252.8767
d) / X / Lower / Upper
20 / 205.9587 / 212.6163
30 / 227.0889 / 232.1736
40 / 245.8233 / 254.1267
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 / Lower / Upper20 / 206.3212 / 212.2538
30 / 227.3658 / 231.8967
40 / 246.2755 / 253.6745
b) / X / Lower / Upper
20 / 201.2605 / 217.3145
30 / 223.5006 / 235.7619
40 / 239.9636 / 259.9864
c) / X / Lower / Upper
20 / 227.8544 / 231.4082
30 / 206.9610 / 211.6140
40 / 247.0733 / 252.8767
d) / X / Lower / Upper
20 / 205.9587 / 212.6163
30 / 227.0889 / 232.1736
40 / 245.8233 / 254.1267
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 / Description1 / ID / Identification Number / 1 -- 113
2 / Y / Infection Risk / Average estimated probability acquiring infection in hospital (in percent)
3 / X1 / Routine culturing ratio / Ratio of number of cultures performed to number of patients without signs or symptoms of hospital-acquired infection, times 100
4 / 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
is to be fit to the data. For this model, answer the following problems.
Problem 9
What is the value of ? (Hint: Do not use sweep.)
a) / 0.168187 / 0.000264411 / -0.002003140.000264 / 0.000104023 / -2.3365E-05
-0.002003 / -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
Problem 10
What is the value of b? (Hint: Use SPSS.)
a) / b(Constant) / 0.018235
X1 / 1.940969
X2 / 0.058598
b) / b
(Constant) / 1.940969
X1 / 0.018235
X2 / 0.058598
c) / b
(Constant) / 0.018235
X1 / 0.058598
X2 / 1.940969
d) / b
(Constant) / 1.940969
X1 / 0.058598
X2 / 0.018235
Problem 11
What is the value of SStot?
a) .358
b) .608
c) 74.405
d) 201.380
Problem 12
What is the value of standard error of the estimate?
a) 126.975
b) .369
c) 1.07439
d) 37.203
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