Mock exam 3 solutions
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University of Illinois, Urbana Champaign *
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Course
300
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
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8
Uploaded by DeanLightning6647
Mock Exam 3
IE 300: Analysis of Data
Spring 2023
04/25/2023
Name:
Solution Key
Please do not turn the page until instructed to do so.
This exam has 5 questions for a total of 100 points.
The points and expected time of each question are shown next to it.
Some questions have an asterisk. This implies that they may require some extra thinking.
There are 8 pages in the exam.
Extra empty paper to write your answers is provided in page 4.
You have 50 minutes to complete all questions.
The exam is
open books, open notes
.
You
are allowed
to use a scientific calculator.
All tables/values you will need for the exam are in the last two pages.
Should you end up needing a different value, make an assumption and use the most suitable value
from the ones available.
You are
not
allowed to use a cell phone instead of a calculator.
You are
not
allowed to discuss the questions with anyone other than me.
Answer every question
to the best of your knowledge.
Make sure to show your work for
partial credit
.
1
Mock Exam 3
Question 1 (10 minutes, 20 points)
Answer the following questions.
(a) Consider two confidence intervals with confidence 1
-
α
1
and 1
-
α
2
.
If the first confidence
interval contains the second (that is, the values in the second interval are also in the first interval),
then this means that:
α
1
≤
α
2
a)
α
1
≥
α
2
b)
We cannot tell.
c)
(b) In a hypothesis test, rejecting the null hypothesis using
α
implies that with probability 1
-
α
the null hypothesis is not true.
True
a)
False
b)
(c) In a hypothesis test, failing to reject the null hypothesis using
α
implies that with probability
1
-
α
the null hypothesis is true.
True
a)
False
b)
(d) For any degrees of freedom
n <
∞
, we have
t
α,n
≤
z
α
.
True
a)
False
b)
(e) Assume
z
α
to be the critical value of the standard normal distribution. What is Φ(
z
α
)?
α/
2
a)
α
b)
1
-
α
c)
1
-
α/
2
d)
(f) Assume that
α
= 0
.
5 and
n
1
=
n
2
. What is
f
α,n
1
-
1
,n
2
-
1
?
1
a)
0
.
5
b)
2
c)
We cannot tell.
d)
Answer: a
,
a
,
b
,
b
,
c
,
a
.
2
Mock Exam 3
Question 2 (10 minutes, 20 points)
A social scientist wants to investigate the number of friends that a person has on Facebook. They
have collected a sample of
n
= 12 Facebook users and have found that the average number of
friends is
X
= 200 with sample standard deviation
s
= 10. Assume that the number of friends that
a user has on Facebook is normally distributed. Answer the following questions.
(a) Build a 95% confidence interval around the unknown mean.
(b) Build a 95% confidence interval around the unknown variance.
(c) How big should
n
be in order for the confidence interval to have a precision error of at most 5
friends? You may assume that the true standard deviation is indeed
σ
= 10 for this part.
(d) The scientist is now interested in the proportion of users that have more friends than “average”.
Based on their results, they define “average” as having 200 friends: hence, they are looking for users
with more than 200 friends. Assume that there are 26 users (out of the total of 64) that have more
than 200 friends. Build a 95% confidence interval around the proportion of users with more than
200 friends.
Answer:
(a) We then have:
X
-
t
α/
2
,n
-
1
s
√
n
,
X
+
t
α/
2
,n
-
1
s
√
n
=
200
-
2
.
201
·
10
√
12
,
200 + 2
.
201
·
10
√
12
= [193
.
6
,
206
.
4]
.
(b) For the variance, we have:
"
(
n
-
1)
s
2
χ
2
α/
2
,n
-
1
,
(
n
-
1)
s
2
χ
2
1
-
α/
2
,n
-
1
#
=
1100
21
.
92
,
1100
3
.
816
= [50
.
18
,
288
.
3]
.
Note that we can also bound the standard deviation by taking square roots.
(c)
n
=
z
α/
2
σ
E
2
= 15
.
37
→
16
.
(d) We have ˆ
p
=
26
64
= 0
.
40625 out of
n
= 64. We can build the two-sided confidence interval as:
"
ˆ
p
-
z
α/
2
r
ˆ
p
(1
-
ˆ
p
)
n
,
ˆ
p
+
z
α/
2
r
ˆ
p
(1
-
ˆ
p
)
n
#
=
"
0
.
40625
-
1
.
96
r
0
.
40625 (1
-
0
.
40625)
64
,
0
.
40625 + 1
.
96
r
0
.
40625 (1
-
0
.
40625)
64
#
=
[0
.
2859
,
0
.
5266]
.
3
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Mock Exam 3
Question 3 (10 minutes, 20 points)
A power supply manufacturer is interested in the output voltage of their newest product.
The
output voltage is assumed to be normally distributed with standard deviation
σ
= 0
.
25
V
. Ideally,
the output should be equal to
μ
= 5
V
.
The manufacturer is performing quality control using a
sample of
n
= 8 products. Answer the following questions.
(a) The “fail to reject” region for the quality control using
n
= 8 products is whether the output
voltage is between 4
.
75
V
and 5
.
25
V
. What is
α
?
(b) Assume that 5
.
4
V
is enough to be cause for a fire hazard concern. What is
β
for the previous
test using a sample size of
n
= 8 products, if indeed the true voltage is equal to 5
.
4
V
?
(c) A sample of
n
= 8 has given us an average of 5
.
1
V
.
Should we accept the hypothesis that
μ
= 5
V
or reject it in favor
μ
6
= 5
V
? What is the
P
-value? Use
α
= 0
.
1, and not the limits from
part (a).
(d) The same sample from before (with an average of 5
.
1
V
) is tested for the null hypothesis of
μ
= 5
.
25
V
. Should we accept the hypothesis that
μ
= 5
.
25
V
or reject it in favor
μ <
5
.
25
V
? What
is the
P
-value? Use
α
= 0
.
1 again.
Answer:
(a) 1
-
α
=
P
(4
.
75
≤
X
≤
5
.
25) assuming that
X
∼ N
5
,
0
.
25
2
8
=
⇒
X
∼
N
(5
,
0
.
0078125). Hence:
1
-
α
=
P
(4
.
75
≤
X
≤
5
.
25) =
P
(
X
≤
5
.
25)
-
P
(
X
≤
4
.
75) =
= Φ(
5
.
25
-
5
√
0
.
0078125
)
-
Φ(
4
.
75
-
5
√
0
.
0078125
) = Φ(2
.
83)
-
Φ(
-
2
.
83) =
= 2Φ(2
.
83)
-
1 = 2
·
0
.
9977
-
1 = 0
.
9954 =
⇒
=
⇒
α
= 1
-
0
.
9954 = 0
.
0046
.
(b)
β
=
P
(4
.
75
≤
X
≤
5
.
25) assuming that
X
∼ N
5
.
4
,
0
.
25
2
8
=
⇒
X
∼ N
(5
.
4
,
0
.
0078125).
Hence:
β
=
P
(4
.
75
≤
X
≤
5
.
25) =
P
(
X
≤
5
.
25)
-
P
(
X
≤
4
.
75) =
= Φ(
5
.
25
-
5
.
4
√
0
.
0078125
)
-
Φ(
4
.
75
-
5
.
4
√
0
.
0078125
) = Φ(
-
1
.
7)
-
Φ(
-
7
.
35) =
= 0
.
04457
-
0 = 0
.
04457 = 4
.
46%
.
(c) We are doing a hypothesis test on the mean with known standard deviation. Hence:
Z
0
=
X
-
μ
0
σ/
√
n
=
0
.
1
0
.
088388348
= 1
.
13
.
As
|
Z
0
|
is less than or equal to
z
α/
2
= 1
.
645, we should fail to reject the null hypothesis.
The
P
-value is equal to 2(1
-
Φ(
Z
0
)) = 2
·
(1
-
Φ(1
.
13)) = 2
·
(1
-
0
.
8708) = 0
.
2584
.
(d) Again, this is a hypothesis test on the mean with known standard deviation. Hence:
Z
0
=
X
-
μ
0
σ/
√
n
=
-
0
.
15
0
.
088388348
=
-
1
.
7
.
As
|
Z
0
|
is greater than
z
α
= 1
.
282, we reject the null hypothesis. The
P
-value is equal to Φ(
Z
0
) =
Φ(
-
1
.
7) = 1
-
Φ(1
.
7) = 1
-
0
.
9554 = 0
.
0446
.
4
Mock Exam 3
Question 4 (10 minutes, 20 points)
A soft drink company is having trouble with their bottling mechanism. The mechanism is assumed
to fill contents into bottles following a normal distribution with a mean of 16 fl oz and a standard
deviation of 0.5 fl oz. To verify that the mechanism is functioning properly, an inspector observed
a sample of
n
= 100 bottles that revealed a sample average fill of 16.1 fl oz and a sample standard
deviation of 0.45 fl oz. Answer the following questions.
(a) Using
α
= 0
.
05, do you have enough evidence to claim that the mechanism is broken and fills
more than 6 fl oz of contents into each bottle?
(b) What is the corresponding
P
-value for part (a)?
(c) Assume that bottles break when the contents are equal to 16.15 fl oz or more. What is the
β
error for your hypothesis, assuming that the true fill mean of the mechanism is 16.15 fl oz?
(d) The company wants to also test the hypothesis of
H
0
:
σ
2
= 0
.
25 against
H
1
:
σ
2
6
= 0
.
25. Based
on the sample observed, would you accept or reject the null hypothesis, under
α
= 0
.
05?
Answer:
(a)
H
0
:
μ
= 16
, H
1
:
μ >
16.
(b) The test statistic is:
Z
0
=
16
.
1
-
16
0
.
5
√
100
=
1
0
.
5
= 2
.
Comparing that with
z
α
=
z
0
.
05
= 1
.
645, we should reject the null hypothesis, and deduce that the
mechanism is not working properly. The
P
-value is:
P
= 1
-
Φ(2) = 0
.
0228
.
(c) Assuming the true mean is 16.15 fl oz, then the fill contents of a sample of size
n
= 100 are
normally distributed with mean 16.15 fl oz and standard deviation 0
.
5
/
√
100 = 0
.
05. The upper
bound of acceptance in our original hypothesis test is
U
= 16 + 1
.
645
·
0
.
05 = 16
.
08225
.
We then
have:
β
=
P
(
X
≤
16
.
08225) = Φ
16
.
08225
-
16
.
15
0
.
05
= Φ (
-
1
.
355) = 0
.
0877 = 8
.
77%
.
(d) We have:
H
0
:
σ
2
= 0
.
25
, H
1
:
σ
2
6
= 0
.
25. The test statistic is
χ
2
0
=
99
·
0
.
2025
0
.
25
= 80
.
19
.
Comparing that with the upper and lower bounds of
χ
2
α/
2
,
99
= 128
.
42 and
χ
2
1
-
α/
2
,
99
= 73
.
36
.
Because 73
.
36
≤
80
.
19
≤
128
.
42, we fail to reject the hypothesis.
5
Mock Exam 3
Question 5 (10 minutes, 20 points)
The concentration of an active ingredient in laundry detergent is dependent on the selection of
catalyst used in the process. The concentration is assumed to be normally distributed. We have
collected ten observations per catalyst for two types of catalysts. The data is as follows:
Catalyst 1
57.9
66.2
65.4
65.4
65.2
62.6
67.6
63.7
68.2
71.0
Catalyst 2
66.4
71.7
70.3
69.3
64.8
69.6
68.6
69.4
66.3
68.8
Answer the following questions.
(a) Under the assumption that the standard deviations of the two catalysts are equal, should
you accept or reject the hypothesis that the two catalysts produce the same active ingredient
concentration? Use
α
= 0
.
05.
(b) Should you accept or reject the hypothesis that the two catalysts have equal standard devia-
tions? Use
α
= 0
.
05.
(c) We are interested in the proportion of observations that are above 65.
Using
α
= 0
.
05, is
there enough evidence to deduce that the second catalyst leads to a higher proportion of active
ingredients above 65?
Answer:
(a) Our hypothesis is formulated as:
H
0
:
μ
1
-
μ
2
= 0 vs.
H
1
:
μ
1
-
μ
2
6
= 0
.
From the two samples, we get:
•
X
1
= 65
.
32
, s
1
= 3
.
52
, n
1
= 10.
•
X
2
= 68
.
52
, s
2
= 2
.
09
, n
2
= 10.
Now, first we estimate the pooled estimator for the standard deviation:
s
P
=
s
(
n
1
-
1)
s
2
1
+ (
n
2
-
1)
s
2
2
n
1
+
n
2
-
2
= 2
.
89
.
Now, we are ready to calculate our
T
statistic:
T
0
=
(
X
1
-
X
2
)
-
0
s
P
q
1
n
1
+
1
n
2
=
-
2
.
47
.
Compare to
t
α/
2
,n
1
+
n
2
-
2
=
t
0
.
025
,
18
= 2
.
101. As
|
T
0
|
> t
α/
2
,n
1
+
n
2
-
2
, we reject the null hypothesis
and deduce that the two catalysts produce different active ingredient concentrations.
(b) Our hypothesis now becomes:
H
0
:
σ
2
1
=
σ
2
2
vs.
H
1
:
σ
2
1
6
=
σ
2
2
.
We calculate our
F
statistic as:
F
0
=
s
2
1
s
2
2
=
3
.
52
2
2
.
09
2
= 2
.
85
.
We now need two critical values:
1.
f
α/
2
,n
1
-
1
,n
2
-
1
=
f
0
.
025
,
9
,
9
= 4
.
10 (
this value is not in the table, so I used
n
1
-
1 = 10
instead.
)
6
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Mock Exam 3
2.
f
1
-
α/
2
,n
1
-
1
,n
2
-
1
=
f
0
.
975
,
9
,
9
=
1
f
0
.
025
,
9
,
9
=
1
4
.
10
= 0
.
244
.
Because
f
1
-
α/
2
,n
1
-
1
,n
2
-
1
≤
F
0
≤
f
α/
2
,n
1
-
1
,n
2
-
1
, we fail to reject the null hypothesis.
(c) We now formulate the hypothesis as:
H
0
:
p
1
-
p
2
= 0 vs.
H
1
:
p
1
-
p
2
<
0
.
First, count the number of observations in each sample that are above 65. We check that there are
7 such observations in the first sample, and 9 such observations in the second sample for observed
proportions of ˆ
p
1
= 0
.
7 and ˆ
p
2
= 0
.
9. We go ahead and use this to estimate the pooled proportion:
ˆ
p
=
n
1
ˆ
p
1
+
n
2
ˆ
p
2
n
1
+
n
2
= 0
.
8
.
Finally, we are ready to calculate the
Z
statistic:
Z
0
=
(ˆ
p
1
-
ˆ
p
2
)
-
Δ
0
r
ˆ
p
(1
-
ˆ
p
)
1
n
1
+
1
n
2
=
-
0
.
2
0
.
179
=
-
1
.
12
.
Comparing to
z
α
=
z
0
.
05
= 1
.
645, we would fail to reject the null hypothesis.
Good luck!
7
Mock Exam 3
NORMAL CUMULATIVE DISTRIBUTION FUNCTION (Φ (
z
))
z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.0
0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1
0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2
0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3
0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4
0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5
0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6
0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7
0.7580 0.7611 0.7642 0.7673 0.7703 0.7734 0.7764 0.7794 0.7823 0.7852
0.8
0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9
0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1.0
0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1
0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830
1.2
0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
1.3
0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177
1.4
0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319
1.5
0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441
1.6
0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545
1.7
0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633
1.8
0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9
0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2.0
0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817
2.1
0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857
2.2
0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890
2.3
0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916
2.4
0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936
2.5
0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952
2.6
0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964
2.7
0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974
2.8
0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981
2.9
0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
3.0
0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990
3.1
0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993
3.2
0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.9995
3.3
0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.9997
3.4
0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998
3.5
0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998 0.9998
3.6
0.9998 0.9998 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999
3.7
0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999
3.8
0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999 0.9999
3.9
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
8