Stat 231 Final Exam
December 16, 1997
Professor Vardeman
1. The first page of printout attached to this exam summarizes some data (collected by a student
group) on the diameters of holes bored in certain steel housing covers. A member of the group
measured the diameters of < œ * parts 7 œ 3 times each. Data values were recorded in units of
"!$ mm above the nominal (or ideal) diameter.
Consider first only the measurements made on housing number 1.
(a) Give a 90% two-sided confidence interval for the long run mean measured diameter for
housing 1. (No need to simplify.)
(b) Give a 90% two-sided prediction interval for the next measured diameter for housing
number 1. (No need to simplify.)
(c) Give a 90% upper confidence bound for the long run standard deviation of measured
diameters for housing number 1.
Now consider the measurements made on both housings 1 and 2.
(d) Test the hypothesis that there is no difference between the long run mean measured
diameters for housings 1 and 2 using ! œ Þ!". (Show all 7 steps.)
Finally, consider the measurements made on all 9 housings.
-1-
(e) What model assumptions must be made in order to do inference for comparing these 9 mean
long run measured diameters?
(f) Under the assumptions of part (e), what :-value is associated with a test of
H! À ." œ .# œ â œ .* ? With ! œ Þ!& should this hypothesis be rejected? Explain.
:-value œ __________
Reject? yes no (circle one)
Explanation:
(g) Based on the model assumptions of part (e) give a 90% two-sided confidence interval for
the difference in long run mean measured diameters for housings 1 and 2. (No need to simplify.)
(h) Suppose that the parts measured by these students were in fact a random sample from a large
lot of such housings and there is interest in estimating how much the holes bored in these
housings actually vary in diameter (where this is to be assessed without the corruption of
measurement error). Give an ANOVA-based estimate of the long run standard deviation among
housing mean measurements.
2. Stat 231 concerns both PROBABILITY and STATISTICS. Define and contrast these two
subjects.
3. Pages 2 through 4 of the printout attached to this exam concern a study of the effectiveness of
armor plating. Armor-piercing bullets were fired at an angle of 40° against armor plate of
-2-
thickness B" (in Þ!!" inches) and Brinell hardness number B# , and the resulting "ballistic limit,"
C (in feet/second), was measured.
(a) There are two simple linear regressions included on the printout. Which of these seems to be
the most effective summarization of the data? Explain. Find the value of G: for this model.
G: œ _____________________
Best SLR Model: ______________
Explanation:
(b) Using the second of the two simple linear regressions, give a 90% two-sided confidence
interval for the increase in mean ballistic limit that accompanies a 1 unit increase in Brinell
hardness number. (No need to simplify.)
Henceforth use the MLR analysis beginning in the middle of page 3 of the printout.
(c) Give and interpret a :-value for testing whether B" and B# together provide some important
ability to explain or predict C.
:-value œ ___________
Interpretation:
(d) Give the value of and degrees of freedom for an F statistic used to test the hypothesis that
after accounting for B# , B" provides no important additional ability to predict C.
.0 œ ______ , ______
0 œ __________
(e) Tomorrow I plan to test a piece of armor plating Þ#&$ inch thick and with Brinell hardness
number %!(. Give me an interval that is in some sense "90% sure" to contain tomorrow's
ballistic limit.
4. In the manufacture of a certain glass product, defects of three different types (A, B and C) can
occur on the product. These differ in both importance and frequency of occurrence. Suppose
-3-
that on one item, \A , \B and \C are the numbers of occurrences of these defects and that these
can be modeled as independent Poisson random variables.
(a) Suppose that -A œ E\A œ Þ#Þ Find the probability that \A 2.
(b) I decide to assign "demerits" to an inspected item according to the expression
H œ \A +#\B  %\C
.
Suppose that -A œ E\A œ Þ#, -B œ E\B œ Þ1 and that -C œ E\C œ Þ05. Find the mean and
variance of H.
(c) For a different product (having a different product grading scheme) items have a mean
number of demerits of Þ& with a corresponding standard deviation of demerits of "Þ!.
Approximate the probability that a sample of "!! of these items has a total of at least 48
demerits. (Hint: >9>+6 ./7/<3>= %) is the same as B %Þ).)
5. End-of-the-line inspection is done on 100 items manufactured on a particular production line
with the results below (in terms of actual Good/Defective character and inspection result).
Inspection Result
Pass
Fail
Actual Part
Good
72
8
Condition
Defective
4
16
Consider first the random selection of a single item from the lot.
(a) What it the probability that the item is actually "Good" or has a "Pass" inspection result?
(b) What is the conditional probability that it is "Good" given that it "Passes" final inspection?
-4-
Consider now the random selection (without replacement) of two items from the lot.
Define random variables
\ œ the number of "Good&Pass" items in the sample,
and ] œ the number of "Defective&Fail" items in the sample.
(c) Below is a partially complete table giving the joint probability function for \ and ] . Finish
filling it in.
Bœ!
Bœ"
Bœ#
Þ"(%&
Þ&"'%
Cœ#
Cœ"
Þ!$))
Cœ!
Þ!"$$
(d) ^ œ \  ] is the number of items in the sample that are correctly classified by the
inspection process. Using your answer to (c) find T Ò^ œ #Ó.
6. A commonly used model in reliability work is that with
[ œ the life of a device of interest ,
] œ lnÐ[ Ñ has a normal distribution with some mean . and some standard deviation 5.
Suppose that the guarantee period for a device is .5 year and that its actual lifetime follows a
model like this where . œ ! and 5 œ Þ(. Find the probability that the device needs to be
replaced during its warranty period.
-5-
MTB > Oneway
'diameter'
'part'.
One-Way Analysis of Variance
Analysis
Source
part
Error
Total
of Variance on diameter
OF
SS
MS
8
211.33
26.42
18
70.67
3.93
26
282.00
F
6.73
p
0.000
Individual 95% CIs For Mean
Based on Pooled StDev
N
Level
1
2
3
4
5
6
7
8
9
Pooled
3
3
3
3
3
3
3
3
3
stOev
=
Mean
4.333
4.667
2.000
3.667
-0.000
2.667
2.667
2.333
10.667
stOev
4.509
1.155
2.000
2.082
1.000
1.155
1.155
1.155
0.577
-----+---------+---------+---------+(----*---)
(---*----)
(----*----)
(---*----)
(----*----)
(---*----)
(---*----)
(----*---)
(---*----)
-----+---------+---------+---------+1.981
-1-
MTB > print
c1-c3
Data Display
Row
y
xl
x2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
927
978
1028
906
1159
1055
1335
1392
1362
1374
1393
1401
1436
1327
950
998
1144
1080
1276
1062
253
258
259
247
256
246
257
262
255
258
253
252
246
250
242
243
239
242
244
234
317
321
341
350
352
363
365
375
373
391
407
426
432
469
257
302
331
355
385
426
MTB > Regress 'y' 1 'xl';
SUBC>
Constant.
Regression Analysis
The regression equation
y = - 886 + 8.27 xl
Predictor
Constant
xl
s
=
Coef
-886
8.267
180.2
Analysis
SOURCE
Regression
Error
Total
is
R~q
Stdev
1348
5.395
11.5%
t-ratio
-0.66
1.53
R-sq(adj)
p
0.519
0.143
=
6.6%
of Variance
OF
1
18
19
SS
76220
584368
660589
MS
76220
32465
-2-
F
2.35
P
0.143
MTB > Regress 'y' 1 'x2';
SUBC>
Constant.
Regression Analysis
The regression equation
y = 211 + 2.64 x2
Coef
211.0
2.6388
Predictor
Constant
x2
s
=
R-sq
135.0
Analysis
Stdev
228.5
0.6174
50.4%
t-ratio
0.92
4.27
R-sq(adj)
p
0.368
0.000
=
47.6%
of variance
SOURCE
Regression
Error
Total
Unusual
Obs.
15
20
is
OF
1
18
19
Observatir':1s
x2
Y
950.0
257
1062.0
426
R denotes
X denotes
MS
332705
18216
SS
332705
327883
660589
Fit
889.2
1335.1
p
0.000
F
18.26
Residual
60.8
-273.1
Stdev.Fit
74.3
47.4
an obs. with a large st. resid.
an obs. whose X value gives it large
influence.
MTB > Name c4 = 'SRES1' c5 = 'FITS1'
MTB > Regress 'y' 2 'xl' 'x2';
SUBC>
SResiduals
'SRES1';
Fits 'FITS1';
SUBC>
SUBC>
Constant;
SUBC>
Pure;
Predict 'xl' 'x2'.
SUBC>
Regression Analysis
The regression equation is
y = - 1674 + 7.61 xl + 2.59 x2
Predictor
Constant
xl
x2
s
=
Coef
-1674.1
7.612
2.5939
124.5
Analysis
SOURCE
Regression
Error
Total
R-sq
Stdev
947.4
3.730
0.5698
60.1%
t-ratio
-1. 77
2.04
4.55
R-sq(adj)
p
0.095
0.057
0.000
=
55.4%
of Variance
OF
2
17
19
SS
397230
263358
660589
MS
198615
15492
-3-
F
12.82
p
0.000
St.Resid
0.54 X
-2.16R
SOURCE
OF
xl
1
1
x2
Fit
1074.1
1122.5
1182.0
1114.0
1187.7
1140.1
1229.0
1293.0
1234.6
1304.1
1307.5
1349.2
1319.1
1445.5
834.7
959.0
1003.8
1088.9
1181.9
1212.2
Cannot
Stdev.Fit
41.9
49.6
47.0
31.1
37.4
31.3
38.7
53.4
34.0
43.2
37.7
44.2
48.9
64.5
73.5
52.1
52.5
40.6
37.0
74.4
do pure error
SEQ SS
76220
321010
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
95.0% C.!.
985.7,
1162.4)
1017.9,
1227.1)
1281. 2)
1082.8,
1048.3,
1179.7)
1108.8,
1266.6)
1074.1,
1206.1)
1147.3,
1310.7)
1180.4,
1405.7)
1162.8,
1306.4)
1212.9,
1395.2)
1228.1,
1387.0)
1255.9,
1442.5)
1215.9,
1422.3)
1581. 6)
1309.5,
679.6,
989.8)
849.1,
1069.0)
893.1,
1114.5)
1003.1,
1174.7)
1103.9,
1260.0)
1055.2,
1369.2)
test
-4-
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
(
95.0% P.!.
1351. 2)
797.0,
839.8,
1405.2)
901. 2, 1462.8)
843.2,
1384.7)
1461. 9)
913.4,
1410.9)
869.3,
953.9,
1504.1)
1578.8)
1007.2,
962.3,
1506.9)
1582.1)
1026.0,
1581. 9)
1033.1,
1070.5,
1627.9)
1601. 3)
1036.9,
1741.3)
1149.7,
529.7,
1139.7)
674.3,
1243.8)
1288.9)
718.8,
812.6,
1365.2)
1455.9)
907.9,
906.2,
1518.2)