Mind on Statistics
5th Edition
ISBN: 9781285463186
Author: Jessica M. Utts, Robert F. Heckard
Publisher: Brooks Cole
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 4, Problem 4.68E
To determine
To Find:
The Chi-Square statistics and find the statistical relationship between bird ownership and lung cancer
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
A major company in the Montreal area, offering a range of engineering services from project preparation to construction execution, and industrial project management, wants to ensure that the individuals who are responsible for project cost estimation and bid preparation demonstrate a certain uniformity in their estimates. The head of civil engineering and municipal services decided to structure an experimental plan to detect if there could be significant differences in project evaluation.
Seven projects were selected, each of which had to be evaluated by each of the two estimators, with the order of the projects submitted being random. The obtained estimates are presented in the table below.
a) Complete the table above by calculating: i. The differences (A-B) ii. The sum of the differences iii. The mean of the differences iv. The standard deviation of the differences
b) What is the value of the t-statistic?
c) What is the critical t-value for this test at a significance level of 1%?…
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
Chapter 4 Solutions
Mind on Statistics
Ch. 4 - Prob. 4.1ECh. 4 - Prob. 4.2ECh. 4 - Each fall, auditions for the band and orchestra...Ch. 4 - Prob. 4.4ECh. 4 - For each pair of variables, indicate whether or...Ch. 4 - For each pair of variables, indicate whether or...Ch. 4 - Suppose a study on the relationship between gender...Ch. 4 - Prob. 4.8ECh. 4 - Prob. 4.9ECh. 4 - Do grumpy old men have a greater risk of having...
Ch. 4 - Prob. 4.11ECh. 4 - Prob. 4.12ECh. 4 - Prob. 4.13ECh. 4 - Prob. 4.14ECh. 4 - Prob. 4.15ECh. 4 - Prob. 4.16ECh. 4 - For Exercise 4.17 and 4.18: A study is done to...Ch. 4 - For Exercise 4.17 and 4.18: A study is done to...Ch. 4 - Prob. 4.19ECh. 4 - a. For a relative risk of 2.1, what is the percent...Ch. 4 - a. For a relative risk of 1.53, what is the...Ch. 4 - Science News (February 25, 1995, p. 124) reported...Ch. 4 - Prob. 4.23ECh. 4 - Prob. 4.24ECh. 4 - Exercise 4.10 described a study in which men were...Ch. 4 - Prob. 4.26ECh. 4 - Prob. 4.27ECh. 4 - Suppose a newspaper article states that drinking...Ch. 4 - Prob. 4.29ECh. 4 - Prob. 4.30ECh. 4 - Prob. 4.31ECh. 4 - Case Study 1.5 (p. 4) was called Does Prayer Lower...Ch. 4 - Prob. 4.33ECh. 4 - Prob. 4.34ECh. 4 - Prob. 4.35ECh. 4 - Prob. 4.36ECh. 4 - Prob. 4.37ECh. 4 - Prob. 4.38ECh. 4 - Prob. 4.39ECh. 4 - Researchers studied a random sample of North...Ch. 4 - Refer to Exercise 4.40. a. Write a two-way table...Ch. 4 - In a national survey, n=1000 randomly selected...Ch. 4 - Prob. 4.43ECh. 4 - Prob. 4.44ECh. 4 - Prob. 4.45ECh. 4 - Prob. 4.46ECh. 4 - Prob. 4.47ECh. 4 - Prob. 4.48ECh. 4 - Prob. 4.49ECh. 4 - Prob. 4.50ECh. 4 - Prob. 4.51ECh. 4 - Prob. 4.52ECh. 4 - Refer to the Minitab output for Exercise 4.52. The...Ch. 4 - Prob. 4.54ECh. 4 - Prob. 4.55ECh. 4 - Prob. 4.56ECh. 4 - Prob. 4.57ECh. 4 - Prob. 4.58ECh. 4 - Exercise 4.2 (p.136) gave the following data from...Ch. 4 - In Example 4.15 (p. 130), a statistically...Ch. 4 - Prob. 4.61ECh. 4 - Prob. 4.62ECh. 4 - Prob. 4.63ECh. 4 - Prob. 4.64ECh. 4 - Prob. 4.65ECh. 4 - Prob. 4.66ECh. 4 - Prob. 4.67ECh. 4 - Prob. 4.68ECh. 4 - Prob. 4.69ECh. 4 - Prob. 4.70ECh. 4 - Prob. 4.71ECh. 4 - Prob. 4.72ECh. 4 - Prob. 4.73ECh. 4 - Prob. 4.74ECh. 4 - Prob. 4.75ECh. 4 - Prob. 4.76ECh. 4 - Prob. 4.77ECh. 4 - Prob. 4.78ECh. 4 - Prob. 4.79ECh. 4 - Prob. 4.80ECh. 4 - Prob. 4.81ECh. 4 - Prob. 4.82ECh. 4 - Prob. 4.83ECh. 4 - Prob. 4.84E
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- 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] 1arrow_forward4. (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_forward
- 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.arrow_forwardSolve 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_forward
- Solve please and thank you!arrow_forwardSolve please and thank you!arrow_forwardQuestion 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_forward
- You 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_forwardSuppose that you have a normal population of quiz scores with mean 40 and standard deviation 10. Select a random sample of 40. What’s the chance that the mean of the quiz scores won’t exceed 45?Select one individual from the population. What’s the chance that his/her quiz score won’t exceed 45?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin Harcourt
Holt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
How to make Frequency Distribution Table / Tally Marks and Frequency Distribution Table; Author: Reenu Math;https://www.youtube.com/watch?v=i_A6RiE8tLE;License: Standard YouTube License, CC-BY
Frequency distribution table in statistics; Author: Math and Science;https://www.youtube.com/watch?v=T7KYO76DoOE;License: Standard YouTube License, CC-BY
Frequency Distribution Table for Grouped/Continuous data | Math Dot Com; Author: Maths dotcom;https://www.youtube.com/watch?v=ErnccbXQOPY;License: Standard Youtube License