MATH 533 ( Applied Managerial
Statistics ) Entire Course
Follow Below Link to Download Tutorial
Email us At: Support@homeworklance.com or lancehomework@gmail.com
(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 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
- (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
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
|
|
0.5
|
p
|
|
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 $) 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)
(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
34
23
34
56
45
35
36
46
47
19
23
19
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
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
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 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
(Points : 25)
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 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 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
|
- (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
|
|
||||||
|
|
|
||||||
|
Sample
|
X
|
N
|
Sample p
|
98% Lower Bound
|
Z-Value
|
P-Value
|
|
|
1
|
4
|
100
|
0.040000
|
0.000000
|
1.43
|
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)
Ans.
(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) |
a. Total people = 2459
|
South + West = 1368 + 405 = 1773
|
probability — divide these:
|
1773/2459 = approx 0.721
|
|
b.
|
Total Midwest = 298
|
Midwest local park = 29
|
Divide:
|
- (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)
(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
- (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
86 88 91 83 93 73 100 102
- 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)
- (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.
|
Urban
|
Suburban
|
Total
|
|
|
Good
|
60
|
168
|
228
|
|
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
- is considered “borderline.”
- (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
- exactly 14 customers will pay for their purchases using
credit cards.
- (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
Sample Mean = 24.2 minutes
Sample Standard Deviation = 4.2 minutes
- Compute the 90% confidence interval for the population
mean refueling and baggage time.
- (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.
- Compute the 99% confidence interval for the population
proportion of dentists who recommend the use of this toothpaste.
- (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.
- Formulate the null and alternative hypotheses.
- (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%
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.
- Formulate the null and alternative hypotheses.
- (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 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
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
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
Source
DF SS
MS F P
Regression 1 229757 229757 4414.11 0.000
Residual Error 28 1457 52
Total 29 231215
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
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 50
2 250
- Analyze the above output to determine the regression
equation.
- (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
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
Residual Error 9 331.9 36.9
Total 11 4822.3
Regression 2 4490.3 2245.2 60.88 0.000
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)
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
1 8.00 1.00
Correlations: Y, X1, X2
Y X1
X1 -0.800
0.002
Y X1
X1 -0.800
0.002
X2 0.933 -0.660
0.000 0.020
0.000 0.020
Cell Contents: Pearson correlation
P-Value
P-Value
- Analyze the above output to determine the multiple
regression equation.
MATH 533 Final Exam Set 3
MATH 533 Final Exam Set 4
No comments:
Post a Comment