MATH 533 ( Applied Managerial Statistics ) Final Exam Answers
MATH 533 Final Exam Set 1
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 |
0.5 | 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 |
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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
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 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.
(Points : 25)
Descriptive statistics 1st quartile: 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. Descriptive statistics 1st quartile: 98.900 the Chris Cross Manufacturing |
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) |
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: |
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
72 93 82 81 82 97 102 107 119
86 88 91 83 93 73 100 102
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
a. is considered “borderline.”
a. exactly 14 customers will pay for their purchases using credit cards.
Sample Size = 36
Sample Mean = 24.2 minutes
Sample Standard Deviation = 4.2 minutes
a. Compute the 90% confidence interval for the population mean refueling and baggage time.
a. Compute the 99% confidence interval for the population proportion of dentists who recommend the use of this toothpaste.
a. Formulate the null and alternative hypotheses.
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.
a. Formulate the null and alternative hypotheses.
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
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
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
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
a. Analyze the above output to determine the multiple regression equation.
MATH 533 Final Exam Set 3
MATH 533 Final Exam Set 4