This document provides materials for the MATH 533 Applied Managerial Statistics course organized by week. It includes homework problems, discussion topics, quizzes, exams and a course project distributed across the 8 weeks of the course. The materials cover statistical concepts like descriptive statistics, probability, confidence intervals, hypothesis testing and their application to managerial decision making.

1. MATH 533 ( Applied Managerial Statistics ) Entire Course
https://homeworklance.com/downloads/math-533-applied-managerial-statistics-entire-course/
MATH 533 ( Applied Managerial Statistics ) Entire Course
(MATH 533 Applied Managerial Statistics – DeVry)
(MATH 533 Week 1)
MATH 533 Week 1 Homework Problems (MyStatLab)
MATH 533 Week 1 Graded Discussion Topics
MATH 533 Week 1 Quiz
(MATH 533 Week 2)
MATH 533 Week 2 Homework Problems (MyStatLab)
MATH 533 Week 2 Graded Discussion Topics
MATH 533 Week 2 Course Project – Part A (SALESCALL Inc.)
(MATH 533 Week 3)
MATH 533 Week 3 Homework Problems (MyStatLab)
MATH 533 Week 3 Graded Discussion Topics
(MATH 533 Week 4)
MATH 533 Week 4 Homework Problems (MyStatLab)
MATH 533 Week 4 Graded Discussion Topics
(MATH 533 Week 5)
MATH 533 Week 5 Homework Problems (MyStatLab)
MATH 533 Week 5 Quiz
MATH 533 Week 5 Graded Discussion Topics
(MATH 533 Week 6)
MATH 533 Week 6 Homework Problems (MyStatLab)
MATH 533 Week 6 Graded Discussion Topics
MATH 533 Week 6 Course Project – Part B (SALESCALL Inc.)
(MATH 533 Week 7)
MATH 533 Week 7 Course Project – Part C (SALESCALL Inc.)
MATH 533 Week 7 Graded Discussion Topics
(MATH 533 Week 8 Final Exam Answers)
MATH 533 ( Applied Managerial Statistics ) Final Exam Answers
MATH 533 Final Exam Set 1

2. 1. (TCO D) PuttingPeople2Work has a growing business placing out-of-work MBAs. They claim they can place
a client in a job in their field in less than 36 weeks. You are given the following data from a sample.
Sample size: 100
Population standard deviation: 5
Sample mean: 34.2
Formulate a hypothesis test to evaluate the claim. (Points : 10)
Ho: µ = 36; Ha: µ ≠ 36
Ho: µ ≥ 36; Ha: µ < 36
Ho: µ ≤ 34.2; Ha: µ > 34.2
Ho: µ > 36; Ha: µ ≤ 36
Ans. b.
H0 must always have equal sign, < 36 weeks
2. (TCO B) The Republican party is interested in studying the number of republicans that might vote in a particular
congressional district. Assume that the number of voters is binomially distributed by party affiliation (either republican
or not republican). If 10 people show up at the polls, determine the following:
Binomial distribution
10 n
0.5p
X P(X)
cumulative
probability
0 0.00098 0.00098
1 0.00977 0.01074
2 0.04395 0.05469
3 0.11719 0.17188
4 0.20508 0.37695
5 0.24609 0.62305
6 0.20508 0.82813
7 0.11719 0.94531
8 0.04395 0.98926
9 0.00977 0.99902
10 0.00098 1.00000
What is the probability that no more than four will be republicans? (Points : 10)
38%
12%
21%
62%
Ans. a
look at x=4, cumulative probability
3. (TCO A) Company ABC had sales per month as listed below. Using the Minitab output given,
determine:
(A) Range (5 points);
(B) Median (5 points); and
(C) The range of the data that would contain 68% of the results. (5 points).
Raw data: sales/month (Millions of $)

3. 23
45
34
34
56
67
54
34
45
56
23
19
Descriptive Statistics: Sales
Variable Total Count Mean StDev Variance Minimum Maximum Range
Sales 12 40.83 15.39 236.88 19.00 67.00 48.00
Stem-
and-
Leaf
Display:
Sales
Stem-
and-leaf
of Sales
N = 12
Leaf
Unit =
1.0
1 1 9
3 2 33
3 2
6 3 444
6 3
6 4
6 4 55
4 5 4
3 5 66
1 6
1 6 7
Reference:
(TCO A) Company ABC had sales per month as listed below. Using the MegaStat output given, determine:
(A) Range (5 points)
(B) Median (5 points)
(C) The range of the data that would contain 68% of the results. (5 points)
Raw data: sales/month (Millions of $)
19

4. 34
23
34
56
45
35
36
46
47
19
23
count 12
mean 34.75
sample variance 146.20
sample standard deviation 12.09
minimum 19
maximum 56
range 37
Stem and Leaf plot for # 1
stem unit = 10
leaf unit = 1
count 12.00000
mean 34.75000
sample variance 146.20455
sample standard
deviation
12.09151
minimum 19.00000
maximum 56.00000
range 37.00000
1st quartile 23.00000
median 34.50000
3rd quartile 45.25000
interquartile range 22.25000
mode 19.00000
4. (TCO C,
D) Tesla Motors
needs to buy axles
for their new car.
They are
considering using
Chris Cross
Manufacturing as a
vendor. Tesla’s
requirement is that
95% of the axles are

5. 100 cm ± 2 cm. The
following data is
from a test run from
Chris Cross
Manufacturing.
Should Tesla select
them as a vendor?
Explain your
answer.
Descriptive
statistics
count 16
mean 99.850
sample
variance
4.627
sample
standard
deviation
2.151
minimum 96.9
maximum 104
range 7.1
population
variance
4.338
population
standard
deviation
2.083
standard
error of the
mean
0.538
tolerance
interval
95.45%
lower
95.548
tolerance
interval
95.45%
upper
104.152
margin of
error
4.302
1st quartile 98.850
median 99.200
3rd quartile 100.550
interquartile
range
1.700
mode 103.000
(Points : 25)

6. Reference: Chegg
Tesla Motors needs
to buy axles for
their new car. They
are considering
using Chris Cross
Manufacturing as a
vendor. Tesla’s
requirement is that
95% of the axles are
100 cm ± 5 cm. The
following data is
MegaStat output
from a test run from
Chris Cross
Manufacturing.
Descriptive
statistics
count: 16
mean: 99.938
sample
variance: 2.313
sample standard
deviation: 1.521
minimum: 97
maximum: 102.9
range: 5.9
population variance:
2.169
population standard
deviation: 1.473
standard error of the
mean: 0.380
tollerance interval
95.45% lower:
96.896
tolerance interval
95.45%
upper: 102.979
half-width: 3.042
1st quartile: 98.900
median: 99.850
3rd quartile:
100.475
interquartile range:
1.575

7. mode: 98.900
Question: Should
Tesla select them as
a vendor? Explain
your answer.
Answers (1)
· Given that,
Tesla Motors needs
to buy axles for
their new car.
They are
considering using
Chris Cross
Manufacturing as a
vendor.
Tesla’s requirement
is that 95% of the
axles are 100 cm ±
5 cm.
The following data
is MegaStat output
from a test run from
Chris
Cross
Manufacturing:
Descriptive
statistics
count: 16
mean: 99.938
sample variance:
2.313
sample standard
deviation: 1.521
minimum: 97
maximum: 102.9
range: 5.9
population variance:
2.169
population standard
deviation: 1.473
standard error of the
mean: 0.380
tollerance interval
95.45% lower:
96.896
tolerance interval

8. 95.45% upper:
102.979
half-width: 3.042
1st quartile: 98.900
median: 99.850
3rd quartile:
100.475
interquartile range:
1.575
mode: 98.900
Now, we have to
construct 95%
confidence interval
for the data from
the Chris Cross
Manufacturing
1. (TCO D) A PC manufacturer claims that no more than 2% of their machines are defective. In a random
sample of 100 machines, it is found that 4.5% are defective. The manufacturer claims this is a fluke of the sample. At
a .02 level of significance, test the manufacturer’s claim, and explain your answer.
Test and CI for One Proportion
Test of p = 0.02 vs p > 0.02
SampleXN Sample p
98%
Lower
Bound
Z-
Value
P-
Value
1 4 1000.0400000.0000001.43 0.077
Reference:
Set up the hypotheses:
H0: p <= 0.02
Ha: p > 0.02
This is a one tailed test, since we will only reject for high proportions.
Since we are using a 0.02 level of significance (it’s just chance that the hypotheses happen to have the same
value as this), we’ll reject the null hypothesis if our P Value is less than 0.02.
The computed P value from Megastat was 0.0371.
This is higher than the significance level.
Therefore, we do not reject H0:.
We can say that the proportion is still less than or equal to 2%, and this was a fluke.
Final Page 2
1. (TCO B) The following table gives
the number of visits to recreational
facilities by kind and geographical
region.
(Points : 30)
a.
Total
people
=
2459
South
+
West
=
1368
probability
— divide
these:
1773/2459
= approx
0.721
b.
Total
Midwest
= 298
Midwest
local
park =
29
Divide:

9. Ans.
EastSouthMidwestWestTotals
Local
Park
55 328 29 52 464
National
Park
233 514 204 251 1202
State
Park
100 526 65 102 793
Totals 388 1368 298 405 2459
(A) Referring to the above table, if a
visitor is chosen at random, what is the
probability that he or she is either from
the South or from the West? (15
points)
(B) Referring to the above table, given
that the visitor is from the Midwest,
what is the probability that he or she
visited a local park? (15 points)
+ 405
=
1773
1. (TCO B, F) The length of time Americans exercise each week is normally distributed with a mean of 15.8
minutes and a standard deviation of 2.2 minutes
X P(X≤x) P(X≥x) Mean
Std
dev
11.0146 .9854 15.8 2.2
15.3581 .6419 15.8 2.2
21.9910 .0090 15.8 2.2
24.9999 .0001 15.8 2.2
p(lower)p(upper)
(A) Analyze the output above to determine what percentage of Americans will exercise between 11 and 21 minutes
per week. (15 points)
(B) What percentage of Americans will exercise less than 15 minutes? If 1000 Americans were evaluated, how many
would you expect to have exercised less than 15 minutes? (15 points) (Points : 30)
MATH 533 Final Exam Set 2
1. (TCO A) Seventeen salespeople reported the following number of sales calls completed last month.
72 93 82 81 82 97 102 107 119
86 88 91 83 93 73 100 102
1. Compute the mean, median, mode, and standard deviation, Q1, Q3, Min, and Max for the above sample
data on number of sales calls per month.
b. In the context of this situation, interpret the Median, Q1, and Q3. (Points : 33)
1. (TCO B) Cedar Home Furnishings has collected data on their customers in terms of whether they reside in
an urban location or a suburban location, as well as rating the customers as either “good,” “borderline,” or “poor.” The
data is below.
2. Urban Suburban Total
Good 60 168 228

10. Borderline 36 72 108
Poor 24 40 64
Total 120 280 400
If you choose a customer at random, then find the probability that the customer
1. is considered “borderline.”
1. (TCO B) Historically, 70% of your customers at Rodale Emporium pay for their purchases using credit cards.
In a sample of 20 customers, find the probability that
1. exactly 14 customers will pay for their purchases using credit cards.
1. (TCO C) An operations analyst from an airline company has been asked to develop a fairly accurate
estimate of the mean refueling and baggage handling time at a foreign airport. A random sample of 36 refueling and
baggage handling times yields the following results.
Sample Size = 36
Sample Mean = 24.2 minutes
Sample Standard Deviation = 4.2 minutes
1. Compute the 90% confidence interval for the population mean refueling and baggage time.
1. (TCO C) The manufacturer of a certain brand of toothpaste claims that a high percentage of dentists
recommend the use of their toothpaste. A random sample of 400 dentists results in 310 recommending their
toothpaste.
1. Compute the 99% confidence interval for the population proportion of dentists who recommend the use of
this toothpaste.
1. (TCO D) A Ford Motor Company quality improvement team believes that its recently implemented defect
reduction program has reduced the proportion of paint defects. Prior to the implementation of the program, the
proportion of paint defects was .03 and had been stationary for the past 6 months. Ford selects a random sample of
2,000 cars built after the implementation of the defect reduction program. There were 45 cars with paint defects in
that sample. Does the sample data provide evidence to conclude that the proportion of paint defects is now less than
.03 (with a = .01)? Use the hypothesis testing procedure outlined below.
1. Formulate the null and alternative hypotheses.
1. (TCO D) A new car dealer calculates that the dealership must average more than 4.5% profit on sales of
new cars. A random sample of 81 cars gives the following result.
Sample Size = 81
Sample Mean = 4.97%
Sample Standard Deviation = 1.8%
Does the sample data provide evidence to conclude that the dealership averages more than 4.5% profit on sales of
new cars (using a = .10)? Use the hypothesis testing procedure outlined below.
1. Formulate the null and alternative hypotheses.
1. (TCO E) Bill McFarland is a real estate broker who specializes in selling farmland in a large western state.
Because Bill advises many of his clients about pricing their land, he is interested in developing a pricing formula of
some type. He feels he could increase his business significantly if he could accurately determine the value of a
farmer’s land. A geologist tells Bill that the soil and rock characteristics in most of the area that Bill sells do not vary

11. much. Thus the price of land should depend greatly on acreage. Bill selects a sample of 30 plots recently sold. The
data is found below (in Minitab), where X=Acreage and Y=Price ($1,000s).
PRICE ACREAGE PREDICT
60 20.0 50
130 40.5 250
25 10.2
300 100.0
85 30.0
182 56.5
115 41.0
24 10.0
60 18.5
92 30.0
77 25.6
122 42.0
41 14.0
200 70.0
42 13.0
60 21.6
20 6.5
145 45.0
61 19.2
235 80.0
250 90.0
278 95.0
118 41.0
46 14.0
69 22.0
220 81.5
235 78.0
50 16.0
25 10.0
290 100.0
Correlations: PRICE, ACREAGE
Pearson correlation of PRICE and ACREAGE = 0.997
P-Value = 0.000
Regression Analysis: PRICE versus ACREAGE
The regression equation is
PRICE = 2.26 + 2.89 ACREAGE
Predictor Coef SE Coef T P
Constant 2.257 2.231 1.01 0.320
ACREAGE 2.89202 0.04353 66.44 0.000
S = 7.21461 R-Sq = 99.4% R-Sq(adj) = 99.3%
Analysis of Variance

12. Source DF SS MS F P
Regression 1 229757 229757 4414.11 0.000
Residual Error 28 1457 52
Total 29 231215
Predicted Values for New Observations
New Obs Fit SE Fit 95% CI 95% PI
1 146.86 1.37 (144.05, 149.66) (131.82, 161.90)
2 725.26 9.18 (706.46, 744.06) (701.35, 749.17)XX
XX denotes a point that is an extreme outlier in the predictors.
Values of Predictors for New Observations
New Obs ACREAGE
1 50
2 250
1. Analyze the above output to determine the regression equation.
1. (TCO E) An insurance firm wishes to study the relationship between driving experience (X1, in years),
number of driving violations in the past three years (X2), and current monthly auto insurance premium (Y). A sample
of 12 insured drivers is selected at random. The data is given below (in MINITAB):
Y X1 X2 Predict X1 Predict X2
74 5 2 8 1
38 14 0
50 6 1
63 10 3
97 4 6
55 8 2
57 11 3
43 16 1
99 3 5
46 9 1
35 19 0
60 13 3
Regression Analysis: Y versus X1, X2
The regression equation is
Y = 55.1 – 1.37 X1 + 8.05 X2
Predictor Coef SE Coef T P
Constant 55.138 7.309 7.54 0.000
X1 -1.3736 0.4885 -2.81 0.020
X2 8.053 1.307 6.16 0.000
S = 6.07296 R-Sq = 93.1% R-Sq(adj) = 91.6%
Analysis of Variance
Source DF SS MS F P
Regression 2 4490.3 2245.2 60.88 0.000

13. Residual Error 9 331.9 36.9
Total 11 4822.3
Predicted Values for New Observations
New Obs Fit SE Fit 95% CI 95% PI
1 52.20 2.91 (45.62, 58.79) (36.97, 67.44)
Values of Predictors for New Observations
New Obs X1 X2
1 8.00 1.00
Correlations: Y, X1, X2
Y X1
X1 -0.800
0.002
X2 0.933 -0.660
0.000 0.020
Cell Contents: Pearson correlation
P-Value
1. Analyze the above output to determine the multiple regression equation.
MATH 533 Final Exam Set 3
MATH 533 Final Exam Set 4