Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319055967
Author: Moore
Publisher: MAC HIGHER
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Chapter 5.1, Problem 6E
To determine
Whether the ideas of the provided poll can be applied to the populations and samples.
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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.
Solve please and thank you!
Chapter 5 Solutions
Introduction to the Practice of Statistics
Ch. 5.1 - Prob. 1UYKCh. 5.1 - Prob. 2UYKCh. 5.1 - Prob. 3UYKCh. 5.1 - Prob. 4UYKCh. 5.1 - Prob. 5UYKCh. 5.1 - Prob. 6ECh. 5.1 - Prob. 7ECh. 5.1 - Prob. 8ECh. 5.1 - Prob. 9ECh. 5.1 - Prob. 10E
Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.2 - Prob. 17UYKCh. 5.2 - Prob. 18UYKCh. 5.2 - Prob. 19UYKCh. 5.2 - Prob. 20UYKCh. 5.2 - Prob. 21UYKCh. 5.2 - Prob. 22UYKCh. 5.2 - Prob. 23UYKCh. 5.2 - Prob. 24UYKCh. 5.2 - Prob. 25ECh. 5.2 - Prob. 26ECh. 5.2 - Prob. 27ECh. 5.2 - Prob. 28ECh. 5.2 - Prob. 29ECh. 5.2 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.3 - Prob. 43UYKCh. 5.3 - Prob. 44UYKCh. 5.3 - Prob. 45UYKCh. 5.3 - Prob. 46UYKCh. 5.3 - Prob. 47UYKCh. 5.3 - Prob. 48UYKCh. 5.3 - Prob. 49UYKCh. 5.3 - Prob. 50UYKCh. 5.3 - Prob. 51UYKCh. 5.3 - Prob. 52UYKCh. 5.3 - Prob. 53UYKCh. 5.3 - Prob. 54UYKCh. 5.3 - Prob. 55UYKCh. 5.3 - Prob. 56UYKCh. 5.3 - Prob. 57ECh. 5.3 - Prob. 58ECh. 5.3 - Prob. 59ECh. 5.3 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90ECh. 5 - Prob. 91ECh. 5 - Prob. 92ECh. 5 - Prob. 93ECh. 5 - Prob. 94ECh. 5 - Prob. 95ECh. 5 - Prob. 96ECh. 5 - Prob. 97ECh. 5 - Prob. 98ECh. 5 - Prob. 99ECh. 5 - Prob. 100ECh. 5 - Prob. 101ECh. 5 - Prob. 102E
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- 7. 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_forwardSolve please and thank you!arrow_forward
- Question 1:We want to evaluate the impact on the monetary economy for a company of two types of strategy (competitive strategy, cooperative strategy) adopted by buyers.Competitive strategy: strategy characterized by firm behavior aimed at obtaining concessions from the buyer.Cooperative strategy: a strategy based on a problem-solving negotiating attitude, with a high level of trust and cooperation.A random sample of 17 buyers took part in a negotiation experiment in which 9 buyers adopted the competitive strategy, and the other 8 the cooperative strategy. The savings obtained for each group of buyers are presented in the pdf that i sent: For this problem, we assume that the samples are random and come from two normal populations of unknown but equal variances.According to the theory, the average saving of buyers adopting a competitive strategy will be lower than that of buyers adopting a cooperative strategy.a) Specify the population identifications and the hypotheses H0 and H1…arrow_forwardYou assume that the annual incomes for certain workers are normal with a mean of $28,500 and a standard deviation of $2,400. What’s the chance that a randomly selected employee makes more than $30,000?What’s the chance that 36 randomly selected employees make more than $30,000, on average?arrow_forwardWhat’s the chance that a fair coin comes up heads more than 60 times when you toss it 100 times?arrow_forward
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