Last Name _________________________ First Name _________________________ ID _________________________
Treatment of Experimental Data 85-222 Winter 2005
Faculty of Engineering
University of Windsor
Midterm Exam 2 Solution
Friday, April 1, 11:30 am – 1:30 pm, Odette OB 104
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and text (no notes).
Time available: 2 hour
Instructions:
 This solution has 9 pages including this cover page.
 Please be sure to put your name and student ID on each odd-numbered page.
 State answers up to four decimal places.
 Show your work.
Grading:
Question
Marks:
1
/10
2
/6
3
/6
4
/6
5
/6
6
/6
7
/6
8
/7
9
/12
Total:
/65
Name:_________________________________________________
ID:_________________________
Question 1: (10 points) Circle the most appropriate answer
1.1 Which distribution is often used to represent physical measurements?
a. Normal distribution
b. Poisson distribution
c. Exponential distribution
d. Chi-square distribution
1.2 Which distribution provides probabilities for the amount of time or space between successive
events occurring in a Poisson process?
a. Normal distribution
b. Exponential distribution
c. Chi-square distribution
d. F distribution
1.3 Which distribution represents a random variable that is the sum of the squares of several
independent normal random variables?
a. Student t distribution
b. Poisson distribution
c. Exponential distribution
d. Chi-square distribution
1.4 Identify situations for which it is possible to take a sample from a population in order to make an
estimate within the next few days.
a. The mean annual earnings are to be determined for electrical engineers graduating this
coming June
b. The preferred majors of the next year’s freshmen engineering students are to be found
c. The lifetimes of cathode ray tubes with an experimental phosphor coating are to be
determined
d. None of the above
1.5 Identify parameters for which an interval estimator is preferred over a point estimator
a. A ballistic coefficient is desired for a recoverable rocket engine to be reused in satellite
launches
b. A forecast value for the cost per unit of a particular ingredient gasoline must be established for
planning purposes
c. The mean strength of a new structural member is to be reported in an engineering
journal
d. None of the above
1.6 Consider a hypothesis test on whether a new design is safe. If the outcome of the test is to
approve an unsafe design, the following error is committed:
a. Type I error
b. Type II error
c. Both error
d. It’s not an error
2
Name:_________________________________________________
ID:_________________________
1.7 The following are consistent estimators
a. X and 
b. X and P
c. s 2
d. b and c
1.8 The sampling distribution of X
a. is approximately normal whether the population is normally distributed or not
b. is approximately uniform only if the population is uniformly distributed
c. is approximately normal only if the population is normally distributed
d. b and c
1.9 The limits of the interval estimate indicate
a. reliability
b. consistency
c. bias
d. precision
1.10
a.
b.
c.
d.
What is 0.50 ?
0.0
0.50
0.6915
1.00
Question 2: (6 points) 8-11
If a switching module of an electronic device malfunctions, the messages are misdirected. The
module must be replaced if the rate of such errors becomes too high. Let  represent the mean rate
of misdirected messages. The switch is operating according to specifications when   0.125 per
hour. Should  reach 0.50 error per hour, the module ought to be replaced. The exact value of  is
not known. A monitoring device may record the number of misdirections in an 8-hour test. Policy is to
replace any module that causes more than 1 error in the test, and otherwise to leave it in place.
a. (3 points) Compute the probability that a module needing replacement is retained.
P(Type II error) = P(Module retained | Module need replacement)
 P# error  1| rate of error,   0.50 per hour 
 P X  1 | t  0.508  4
 0.0916 (From Table C)
b. (3 points) Compute the probability that a module is replaced even though it is operating according
to specifications.
P(Type I error) = P(Module replaced | Module operates according to specification)
 P# error  2 | rate of error,   0.125 per hour 
 P X  2 | t  0.1258  1
 1  P X  1 | t  1  1  0.7358 (From Table C)  0.2642
3
Name:_________________________________________________
ID:_________________________
Question 3: (6 points) 8-20
The n power cells in a satellite will be arranged in parallel and will fail at a mean rate of 0.005 per
day. They have independent lifetimes.
a. (2 points) Find the probability that any specific cell will fail on or before 250 days.
  0.005 failures per day
t  250 days
PT  t   1  e t
So, PT  250  1  e 0.005250  1  e 1.25  0.7135
b. (2 points) Find the probability that at least one cell still working after 400 days, assuming n  5.
First, find the probability for the single cell case ( n  1). Then, consider the 5-cell case ( n  5 ).
PT  400  1  e t  1  e 0.005400  1  e 2  0.847
So, for the single cell case, P(the cell fails in 400 days) = 0.8647
P(at least one cell survives 400 days) = 1-P(all cells fail in 400 days)
 1  P X  5 | n  5, p  0.8647   1  b5;5,0.8647   1  0.8647   0.5167
5
c. (2 points) What should n be to provide a 95% chance that the satellite’s power source will survive
at least 400 days?
Required, P(at least one cell survives 400 days)  0.95
Or, 1-P(all cells fail in 400 days)  0.95
Or, 1  0.8647   0.95
n
Or, 0.8647   1  0.95  0.05
Or, n log 0.8647   log 0.05
Or, n 0.0631  1.3020
 1.3020
Or, n 
 20.60
 0.0631
Or, n  21.
n
Question 4: (6 points)
Many species of terrestrial frogs that hibernate at or near the ground surface can survive prolonged
exposure to low winter temperatures. In freezing conditions, the frog’s body temperature, called its
super-cooling temperature, remains relatively higher due to an accumulation of glycerol in its body
fluids. Studies have shown that super-cooling temperature of terrestrial frogs frozen at -60C has a
relative frequency distribution with a mean of -2.180C and a standard deviation of 0.32 0C (Science,
May 1983). Consider the mean super-cooling temperature, y , of a random sample of n  42
terrestrial frogs frozen at -60C.
a. (3 points) Find the probability that y exceeds -2.050C.
  2.18 0 C ,  y 

n

0.32
42
 0.0494
4
Name:_________________________________________________
ID:_________________________

y   
 2.05   2.18 

P y  2.05  P z 
 P z 




0
.
0494


y 

 Pz  2.63  1  Pz  2.63  1  0.9957 (From Table D)  0.0043


b. (3 points) Find the probability that y falls between -2.200C and -2.100C.
  2.20  
 2.10   
 2.10   2.18 
  2.20   2.18
P  2.20  y  2.10  P
z
 P
z



y
y
0.0494
0.0494




 P 0.405  z  1.62  Pz  1.62  Pz  0.405  0.9474  0.34275 (From Table D)  0.5999


Question 5: (6 points)
A quality control supervisor in a cannery knows that the exact amount each can contains will vary,
since there are certain uncontrollable factors that affect the amount of fill. The mean fill per can is
important, but equally important is the variance  2 of the amount of fill. If  2 is large, some cans will
contain too little and others too much. In order to estimate the variation of fill at the cannery, the
supervisor randomly selects ten cans and weighs the contents of each. The following results are
obtained:
y  7.98 ounces and s  0.04 ounce
Construct a 90% confidence interval for the true variance in fill of cans at the cannery.
1    0.90,  / 2  1  0.90 / 2  0.05,1   / 2  1  0.05  0.95
n  10, d.f.  n  1  10  1  9
For d.f.  9,  2 / 2   02.05  16.9190,  12 / 2  3.325 (From Table H)
So, the 90% confidence interval,
n - 1s 2
 2 / 2
Or,

2

n  1s 2

 12 / 2
10 - 10.042
16.9190

2
2

10  10.04 

3.325
Or, 0.000851   2  0.004331
5
Name:_________________________________________________
ID:_________________________
Question 6: (6 points)
One desirable characteristics of water pipes is that the quality of water they deliver be equal to or
near the quality of water entering the system at the water treatment plant. A type of ductile iron pipe
has provided an excellent water delivery system for the St. Louis County Water Company. The
chlorine levels of water emerging from the South water treatment plant and at the Fire Station
(Fenton Zone 13) were measured at two-month intervals, with the results shown below:
Jan 21
Mar 21
May 21
July 21
Sep 21
Nov 21
South Plant
2.0
2.1
1.7
1.7
2.0
2.1
Fire Station
2.2
2.1
1.9
1.8
1.9
1.8
Difference
-0.2
0
-0.2
-0.1
0.1
0.3
Find a 95% confidence interval for the mean difference in monthly chlorine content between the two
locations.
1    0.95,  / 2  1  0.95 / 2  0.025, n  6, d.f.  n  1  6  1  5
For d.f.  5, t / 2  t 0.025  2.571 (From Table G)
d

i
Date
di
di  d
1
Jan 21
-0.2
-0.1833
0.0336
2
Mar 21
0
0.0167
0.0003
3
May 21
-0.2
-0.1833
0.0336
4
July 21
-0.1
-0.0833
0.0069
5
Sep 21
0.1
0.1167
0.0136
6
Nov 21
0.3
0.3167
0.1003
d
Sum
 d
i
d
n 1
 d
 0.01
i
d
d

2
 0.1883
d   d i / n  0.01 / 6  0.0167
Average
sd 
i
i

0.1883
 0.1941
6 1
The 95% confidence interval
 A   B  d  t / 2
sd
 0.0167  t 0.025
n
Or,  A   B   0.2204,0.1870
0.1941
6
 0.0167  2.571
6
0.1941
6
 0.0167  0.2037
Name:_________________________________________________
ID:_________________________
Question 7: (6 points) 10-30
An industrial engineer’s assistant made 60 random observations of the upholstery installation team in
an automobile assembly plant. During 15 of the observations the workers were arranging materials
beside their workstation.
a. (4 points) Construct a 99% confidence interval estimate of the proportion of time installers spend
arranging materials.
p
15
 0.25,1    0.99,  / 2  1  0.99 / 2  0.005, n  60, z / 2  z 0.005  2.575
60
The 99% confidence interval of proportion
p1  p 
0.251  0.25
 0.25  2.575
 0.25  2.5750.0559   0.25  0.1439
n
60
 0.1060,0.3939 
 p  z / 2
b. (2 points) A total of 500 cars passed the upholstery installers during the 8-hour shift, during which
time the line was in operation 450 minutes. Using your answers to (a), determine an interval
estimate for the mean time per car spent by the installation team just arranging materials.
The 99% confidence interval of mean time per car
450
450 

  0.1060
,0.3939
  0.0954,0.3546 
500
500 

So, 0.0954    0.3546
Question 8: (7 points) 10-42, 10-38
Dr. Shafiq Chowdhury, an electrical engineer, wishes to find the difference between the mean time
between failures (MTBF) for transformers obtained from two different vendors. The following data
were obtained from independent high-temperature testing:
Vendor A
Vendor B
n A  20
nB  25
X A  1,225 hr
X B  1,320 hr
s A  150
s B  200
Construct a 90% confidence interval for the difference between MTBF for the two vendors. Assume
unequal population variance and small sample size.
7
Name:_________________________________________________
ID:_________________________
1    0.90,  / 2  1  0.90 / 2  0.05
sD 
s A2 s B2
150 2 200 2



 1125  1600  52.20
n A nB
20
25
2
 s A2 s B2 



n A n B 
1125  16002  42.85  43

d.f. 

2
2
11252  16002
s A2 / n A
s B2 / n B

20  1
25  1
nA 1
nB  1

 

Table G does not give t / 2  t 0.05 for d.f. = 43. The nearest is d.f.=40, for which t / 2  t 0.05  1.684 .
An alternative is to interpolate between d.f.=40 and d.f.=60.
The 90% confidence interval
 A   B  X A  X B  t / 2 s D  1225  1320  t 0.05 52.20   95  1.68452.20   95  87.9048
  182.9048,7.0952
Question 9: (12 points) 11-23, known variance
Mr. Ramadan Barakat, a quality control inspector for a microwave transmitter manufacturer assesses
shipments of 500 crystal controls each. The actual broadcast frequency depends on the resonant
frequency, which will vary slightly from crystal to crystal, but the mean level should achieve the rated
target of 0.56 mHz. Suppose that the population standard deviation for the individual crystal
frequencies is known to be   502 Hz. A random sample of 45 crystals from the shipment is tested
and the resonant frequency determined for each. The entire shipment is rejected if the observed
mean is slightly above or below the rated level and accepted otherwise. The inspector wants a 2%
chance of rejecting a shipment in which the mean frequency exactly matches the rated level.
This problem has small/finite population N  500, large sample n  45,
standard deviation,   502 Hz = 0.000502 mHz.
and known population
a. (2 points) Formulate the inspector’s hypotheses.
H O :   0.56 mHz
H A :   0.56 mHz
b. (4 points) Assuming that X will serve as the test statistic and using a significance level of
  0.02 , find the critical values for the sample mean and determine the decision rule (assume
recurring decision problem).
8
Name:_________________________________________________
ID:_________________________
1    0.98,  / 2  1  0.98 / 2  0.01, z / 2  z 0.01  2.33 (From Table D)

n

X 
0.000502
45

n
 0.00007483
N n
500  45
 0.00007483
 0.000074830.9549  0.00007146
N 1
500  1
The 98% confidence interval
   H  z / 2 X  0.56  2.330.00007146  0.56  0.00017  0.559833,0.560166
)
c. (2 points) Should a shipment be accepted or rejected if X  0.5602 mHz?
If X  0.5602  0.560166 mHz, reject H 0 :   0.56 mHz. So, reject shipment.
d. (4 points) Determine the Type II error probability when   0.5601 mHz.
P(Type II error) = P(accept H O | H O false) = P(accept H O |   0.5601 mHz)

 P 0.559833  X  0.560167 |   0.5601 mHz

 0.559833  
0.560167   
 P
z


X
X


0.560167  0.5601 
 0.559833  0.5601
 P
z

0.00007146 
 0.00007146
 P 3.7294  z  0.9306 
 Pz  0.9306   Pz  3.7294 
 0.8238  0 (From table D)
 0.8238
9