Student Solutions Manual for Basic Business Statistics
13th Edition
ISBN: 9780321926708
Author: David M. Levine; Mark L. Berenson; Timothy C. Krehbiel; Kathryn A. Szabat; David F. Stephan
Publisher: Pearson Education
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Chapter 15, Problem 17PS
Refer to Problem 14.5 on page 542. Perform a multiple
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Compute the relative risk of falling for the two groups (did not stop walking vs. did stop). State/interpret your result verbally.
Microsoft Excel include formulas
Question 1
The data shown in Table 1 are and R values for 24 samples of size n = 5 taken from a process producing
bearings. The measurements are made on the inside diameter of the bearing, with only the last three decimals
recorded (i.e., 34.5 should be 0.50345).
Table 1: Bearing Diameter Data
Sample Number
I
R
Sample Number
I R
1
34.5
3
13
35.4
8
2
34.2
4
14
34.0
6
3
31.6
4
15
37.1
5
4
31.5
4
16
34.9
7
5
35.0
5
17
33.5
4
6
34.1
6
18
31.7
3
7
32.6
4
19
34.0
8
8
33.8
3
20
35.1
9
34.8
7
21
33.7
2
10
33.6
8
22
32.8
1
11
31.9
3
23
33.5
3
12
38.6 9
24
34.2
2
(a) Set up and R charts on this process. Does the process seem to be in statistical control? If necessary,
revise the trial control limits. [15 pts]
(b) If specifications on this diameter are 0.5030±0.0010, find the percentage of nonconforming bearings pro-
duced by this process. Assume that diameter is normally distributed. [10 pts]
1
Chapter 15 Solutions
Student Solutions Manual for Basic Business Statistics
Ch. 15 - The following is the quadratic regression equation...Ch. 15 - Business actively recruit business student with...Ch. 15 - A national chain of consumer electronics stores...Ch. 15 - Is the number of calories in a beer related to the...Ch. 15 - In the production of printed circuit boards,...Ch. 15 - A production manager wishes to examine the...Ch. 15 - Researchers wanted to investigate the relationship...Ch. 15 - Prob. 8PSCh. 15 - Prob. 9PSCh. 15 - Prob. 10PS
Ch. 15 - Using the data of Problem 15.4 on page 600, stored...Ch. 15 - Prob. 12PSCh. 15 - Prob. 13PSCh. 15 - If the coefficient of determination between two...Ch. 15 - If the coefficient of determination between two...Ch. 15 - Prob. 16PSCh. 15 - Refer to Problem 14.5 on page 542. Perform a...Ch. 15 - Refer to Problem 14.6 on page 542. Perform a...Ch. 15 - Refer to Problem 14.7 on page 542. Perform a...Ch. 15 - Refer to Problem 14.8 on page 542. Perform a...Ch. 15 - Prob. 21PSCh. 15 - Prob. 22PSCh. 15 - The file FTMBA contains a sample of 2012...Ch. 15 - You need to develop a model to predict the...Ch. 15 - Accounting Today identified top public accounting...Ch. 15 - How can you evaluate whether collinearity exists...Ch. 15 - Prob. 27PSCh. 15 - Prob. 28PSCh. 15 - A specialist in baseball analytics has expanded...Ch. 15 - In the production of printed circuit boards,...Ch. 15 - Hemlock Farms is a community located in the Pocono...Ch. 15 - Prob. 32PSCh. 15 - Prob. 33PSCh. 15 - Prob. 34PSCh. 15 - You are a real estate broker who wants to compare...Ch. 15 - You are a real estate broker who wants to compare...Ch. 15 - Financial analysts engage in business valuation to...Ch. 15 - Prob. 38PSCh. 15 - A molding machine that contains different cavities...Ch. 15 - In problem 15.32-15.36 you developed multiple...
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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 Tarrow_forwardQuestion 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…arrow_forwardName: 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.arrow_forward
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