Probability and Statistics for Engineering and the Sciences
9th Edition
ISBN: 9781305251809
Author: Jay L. Devore
Publisher: Cengage Learning
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Textbook Question
Chapter 14, Problem 46SE
The NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16 teams in each region are then ranked (seeded) from 1 to 16. During the 12-year period from 1991 to 2002, the top-ranked team won its regional tournament 22 times, the second-ranked team won 10 times, the third-ranked team won 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3. Let Pij denote the probability that the team ranked i in its region is victorious in its game against the team ranked j. Once the Pij’s are available, it is possible to compute the probability that any particular seed wins its regional tournament (a complicated calculation because the number of outcomes in the
a. One model postulated Pij = .5 – (i−j)with
b. A more sophisticated model has game probabilities Pij = .5 + .2813625 (zi−zj), where the z’s are measures of relative strengths related to standard normal percentiles [percentiles for successive highly seeded teams are closer together than is the case for teams seeded lower, and .2813625 ensures that the
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4. (5 pts) Conduct a chi-square contingency test (test of independence) to assess whether
there is an association between the behavior of the elderly person (did not stop to talk,
did stop to talk) and their likelihood of falling. Below, please state your null and
alternative hypotheses, calculate your expected values and write them in the table,
compute the test statistic, test the null by comparing your test statistic to the critical
value in Table A (p. 713-714) of your textbook and/or estimating the P-value, and
provide your conclusions in written form. Make sure to show your work.
Did not stop walking to talk
Stopped walking to talk
Suffered a fall
12
11
Totals
23
Did not suffer a fall | 2
Totals
35
37
14
46
60
T
Question 2
Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples
of five parts each are collected, and the compressive strengths (in psi) are shown in Table 2.
Table 2: Strength Data for Question 2
Sample Number
x1
x2
23
x4
x5
R
1
83.0
2
88.6 78.3 78.8
3
85.7
75.8
84.3
81.2 78.7 75.7 77.0
71.0 84.2
81.0
79.1
7.3
80.2 17.6
75.2
80.4
10.4
4
80.8
74.4
82.5
74.1 75.7 77.5
8.4
5
83.4
78.4
82.6 78.2
78.9
80.3
5.2
File Preview
6
75.3
79.9
87.3 89.7
81.8
82.8
14.5
7
74.5
78.0 80.8
73.4
79.7
77.3
7.4
8
79.2
84.4 81.5 86.0
74.5
81.1
11.4
9
80.5
86.2
76.2 64.1
80.2
81.4
9.9
10
75.7
75.2
71.1 82.1
74.3
75.7
10.9
11
80.0 81.5
78.4 73.8
78.1
78.4
7.7
12
80.6
81.8
79.3
73.8
81.7 79.4
8.0
13
82.7
81.3
79.1
82.0 79.5 80.9
3.6
14
79.2
74.9
78.6 77.7
75.3
77.1
4.3
15
85.5 82.1
82.8 73.4
71.7
79.1
13.8
16
78.8 79.6
80.2 79.1
80.8 79.7
2.0
17
82.1
78.2
18
84.5
76.9
75.5
83.5 81.2
19
79.0 77.8
20
84.5
73.1
78.2 82.1
79.2 81.1 7.6
81.2 84.4 81.6 80.8…
Name:
Lab Time:
Quiz 7 & 8 (Take Home) - due Wednesday, Feb. 26
Contingency Analysis (Ch. 9)
In lab 5, part 3, you will create a mosaic plot and conducted a chi-square contingency test to
evaluate whether elderly patients who did not stop walking to talk (vs. those who did stop)
were more likely to suffer a fall in the next six months. I have tabulated the data below.
Answer the questions below. Please show your calculations on this or a separate sheet.
Did not stop walking to talk
Stopped walking to talk Totals
Suffered a fall
Did not suffer a fall
Totals
12
11
23
2
35
37
14
14
46
60
Quiz 7:
1. (2 pts) Compute the odds of falling for each group. Compute the odds ratio for those
who did not stop walking vs. those who did stop walking. Interpret your result verbally.
Chapter 14 Solutions
Probability and Statistics for Engineering and the Sciences
Ch. 14.1 - What conclusion would be appropriate for an...Ch. 14.1 - The article Racial Stereotypes in Childrens...Ch. 14.1 - It is hypothesized that when homing pigeons are...Ch. 14.1 - The article Application of Methods for Central...Ch. 14.1 - An information-retrieval system has ten storage...Ch. 14.1 - The article The Gap Between Wine Expert Ratings...Ch. 14.1 - Criminologists have long debated whether there is...Ch. 14.1 - The article Psychiatric and Alcoholic Admissions...Ch. 14.1 - Prob. 9ECh. 14.1 - Prob. 10E
Ch. 14.1 - Prob. 11ECh. 14.2 - Consider a large population of families in which...Ch. 14.2 - A study of sterility in the fruit fly (Hybrid...Ch. 14.2 - Prob. 14ECh. 14.2 - A certain type of flashlight is sold with the four...Ch. 14.2 - Let X = the number of adult police contacts for a...Ch. 14.2 - In a genetics experiment, investigators looked at...Ch. 14.2 - The article A Probabilistic Analysis of Dissolved...Ch. 14.2 - Each headlight on an automobile undergoing an...Ch. 14.2 - Prob. 20ECh. 14.2 - Prob. 21ECh. 14.2 - The article A Method for the Estimation of Alcohol...Ch. 14.2 - The article Nonbloated Burned Clay Aggregate...Ch. 14.3 - The accompanying two-way table was constructed...Ch. 14.3 - In an investigation of alcohol use among college...Ch. 14.3 - Contamination of various food products is an...Ch. 14.3 - The article Human Lateralization from Head to...Ch. 14.3 - Prob. 28ECh. 14.3 - The accompanying data on degree of spirituality...Ch. 14.3 - Three different design configurations are being...Ch. 14.3 - A random sample of smokers was obtained, and each...Ch. 14.3 - Eclosion refers to the emergence of an adult...Ch. 14.3 - Prob. 33ECh. 14.3 - Prob. 34ECh. 14.3 - Prob. 35ECh. 14.3 - Consider the accompanying 2 3 table displaying...Ch. 14 - The article Birth Order and Political Success...Ch. 14 - Does the phase of the moon have any bearing on...Ch. 14 - Prob. 39SECh. 14 - The authors of the article Predicting Professional...Ch. 14 - The accompanying two-way frequency table appears...Ch. 14 - Prob. 42SECh. 14 - Prob. 43SECh. 14 - Prob. 44SECh. 14 - Prob. 45SECh. 14 - The NCAA basketball tournament begins with 64...Ch. 14 - Have you ever wondered whether soccer players...Ch. 14 - Do the successive digits in the decimal expansion...Ch. 14 - Prob. 49SE
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- Solve please and thank you!arrow_forward7. In a 2011 article, M. Radelet and G. Pierce reported a logistic prediction equation for the death penalty verdicts in North Carolina. Let Y denote whether a subject convicted of murder received the death penalty (1=yes), for the defendant's race h (h1, black; h = 2, white), victim's race i (i = 1, black; i = 2, white), and number of additional factors j (j = 0, 1, 2). For the model logit[P(Y = 1)] = a + ß₁₂ + By + B²², they reported = -5.26, D â BD = 0, BD = 0.17, BY = 0, BY = 0.91, B = 0, B = 2.02, B = 3.98. (a) Estimate the probability of receiving the death penalty for the group most likely to receive it. [4 pts] (b) If, instead, parameters used constraints 3D = BY = 35 = 0, report the esti- mates. [3 pts] h (c) If, instead, parameters used constraints Σ₁ = Σ₁ BY = Σ; B = 0, report the estimates. [3 pts] Hint the probabilities, odds and odds ratios do not change with constraints.arrow_forwardSolve please and thank you!arrow_forward
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