Concept explainers
(a)
The two-way table that explains the study results of stress and heart attacks.
(a)

Answer to Problem 25E
Solution: The two-way table that explains the study results of stress and heart attacks is obtained as:
Groups | Cardiac |
No cardiac events | Total |
Stress Management | 3 | 30 | 33 |
Exercise Program | 7 | 27 | 34 |
Usual Heart care | 12 | 28 | 40 |
Total | 22 | 85 | 107 |
Explanation of Solution
Calculation:
The newspaper article describes three groups and hence there are three rows for the two-way table. There are two types of events for the provided three groups. Hence, there are two columns for the two-way table. So, the two-way table that explains the study results of stress and heart attacks is obtained as:
Groups | Cardiac events | No cardiac events | Total |
Stress Management | 3 | 30 | 33 |
Exercise Program | 7 | 27 | 34 |
Usual Heart care | 12 | 28 | 40 |
Total | 22 | 85 | 107 |
(b)
The success rates for the three treatments to prevent cardiac events.
(b)

Answer to Problem 25E
Solution: The success rates for the three treatments to prevent cardiac events are obtained as:
Explanation of Solution
Calculation:
The success of the program is defined as “No cardiac events.” Hence, the success rate of three treatments is the percentage of no cardiac events in three groups. So, the success rate of stress management program is calculated as:
Now, calculate the success rate of Exercise program. So, the success rate of exercise program is calculated as:
And, the success rate of usual heart care is calculated as:
(c)
To find: The expected cells count for the three treatment groups and also verify that the obtained expected counts satisfy the guidelines to use chi-square test.
(c)

Answer to Problem 25E
Solution: The expected cell counts for the three treatment groups are obtained as:
Groups | Expected | |
Cardiac events | No cardiac events | |
Stress Management | 6.79 | 26.21 |
Exercise Management | 6.99 | 27.01 |
Usual heart care | 8.22 | 31.78 |
All the obtained expected counts are greater than 5, so it satisfies the guidelines to use chi-square test.
Explanation of Solution
Calculation:
The two-way table is obtained in part (a) which is shown below:
Groups | Cardiac events | No cardiac events | Total |
Stress Management | 3 | 30 | 33 |
Exercise Program | 7 | 27 | 34 |
Usual Heart care | 12 | 28 | 40 |
Total | 22 | 85 | 107 |
Now, calculate the expected cell counts of each cell. The expected cell count is defined by the formula:
The expected count for cardiac events for stress management group is calculated as:
The expected count for no cardiac events for stress management is calculated as:
Similarly calculate the expected counts for all cells and the table that shows the observed and expected counts is obtained as:
Groups | Observed | Expected | ||
Cardiac events | No cardiac events | Cardiac events | No cardiac events | |
Stress Management | 3 | 30 | 6.79 | 26.21 |
Exercise Management | 7 | 27 | 6.99 | 27.01 |
Usual heart care | 12 | 28 | 8.22 | 31.78 |
The requirement to satisfy to use the chi-square test is that all the expected counts should be greater than 5.
The obtained expected counts shows that all the counts are greater than 5. Hence, the guideline to use chi-square test is satisfied.
(d)
To Test: Whether there is a significant difference between the success rates for the three treatment groups.
(d)

Answer to Problem 25E
Solution: There is a significant difference between success rates for the three treatment groups at the significance level of 10%, but not significant at significance level of 5%.
Explanation of Solution
Calculation:
The steps followed for a significance test are as provided below:
Step 1: Formulate the hypotheses.
The null hypothesis
Step 2: Define the sampling distribution.
The chi-square test can be used if the expected cell counts are larger than 5. The obtained two-way table shows that there are three rows and two columns. Therefore, the degrees of freedom are obtained as:
Step 3: Find the expected cell counts.
The expected cell counts are obtained in part (c). So, the table that shows the observed counts and expected counts is obtained as:
Groups | Observed | Expected | ||
Cardiac events | No cardiac events | Cardiac events | No cardiac events | |
Stress Management | 3 | 30 | 6.79 | 26.21 |
Exercise Management | 7 | 27 | 6.99 | 27.01 |
Usual heart care | 12 | 28 | 8.22 | 31.78 |
Step 4: Determine the chi-square statistic.
Since all the expected counts are greater than 5. The chi-square test statistic can be used. The chi-square statistic is the measure of the distance of the observed counts from the expected counts in a two-way table. The formula for the χ2 statistic is defined as:
Substitute the obtained observed and expected counts for each cell to determine the chi-square statistic. So, the chi-square statistic is calculated as:
Thefore, the chi-square statistic is obtained as 4.851.
Step 5: Test the significance.
The degrees of freedom are obtained as two. Use Table 24.1 of critical values for chi-square test which is provided in the textbook. The obtained chi-square value is 4.851, which is smaller than the critical value of 5.99 at the significance level of 0.05 and greater than the critical value of 4.61 at the significance level of 0.10 for two degrees of freedom.
This shows that the result is significant at 10% level of significance, as the calculated 4.851 is less than the critical value 5.99, but not at 5% significance level as the calculated 4.851 is less than the critical value 4.61.
Conclusion:
From the report, it can be concluded that there is some difference between the success rates of the three experiment groups at 10% significance level and shows that there is no difference between the success rates at 5% significance level.
Want to see more full solutions like this?
Chapter 24 Solutions
Statistics: Concepts and Controversies - WebAssign and eBook Access
- The acidity or alkalinity of a solution is measured using pH. A pH less than 7 is acidic; a pH greater than 7 is alkaline. The accompanying data represent the pH in samples of bottled water and tap water. Complete parts (a) and (b). Click the icon to view the data table. (a) Determine the mean, median, and mode pH for each type of water. Comment on the differences between the two water types. Select the correct choice below and fill in any answer boxes in your choice. A. For tap water, the mean pH is (Round to three decimal places as needed.) B. The mean does not exist. Data table Тар 7.64 7.45 7.45 7.10 7.46 7.50 7.68 7.69 7.56 7.46 7.52 7.46 5.15 5.09 5.31 5.20 4.78 5.23 Bottled 5.52 5.31 5.13 5.31 5.21 5.24 - ☑arrow_forwardく Chapter 5-Section 1 Homework X MindTap - Cengage Learning x + C webassign.net/web/Student/Assignment-Responses/submit?pos=3&dep=36701632&tags=autosave #question3874894_3 M Gmail 品 YouTube Maps 5. [-/20 Points] DETAILS MY NOTES BBUNDERSTAT12 5.1.020. ☆ B Verify it's you Finish update: All Bookmarks PRACTICE ANOTHER A computer repair shop has two work centers. The first center examines the computer to see what is wrong, and the second center repairs the computer. Let x₁ and x2 be random variables representing the lengths of time in minutes to examine a computer (✗₁) and to repair a computer (x2). Assume x and x, are independent random variables. Long-term history has shown the following times. 01 Examine computer, x₁₁ = 29.6 minutes; σ₁ = 8.1 minutes Repair computer, X2: μ₂ = 92.5 minutes; σ2 = 14.5 minutes (a) Let W = x₁ + x2 be a random variable representing the total time to examine and repair the computer. Compute the mean, variance, and standard deviation of W. (Round your answers…arrow_forwardThe acidity or alkalinity of a solution is measured using pH. A pH less than 7 is acidic; a pH greater than 7 is alkaline. The accompanying data represent the pH in samples of bottled water and tap water. Complete parts (a) and (b). Click the icon to view the data table. (a) Determine the mean, median, and mode pH for each type of water. Comment on the differences between the two water types. Select the correct choice below and fill in any answer boxes in your choice. A. For tap water, the mean pH is (Round to three decimal places as needed.) B. The mean does not exist. Data table Тар Bottled 7.64 7.45 7.46 7.50 7.68 7.45 7.10 7.56 7.46 7.52 5.15 5.09 5.31 5.20 4.78 5.52 5.31 5.13 5.31 5.21 7.69 7.46 5.23 5.24 Print Done - ☑arrow_forward
- The median for the given set of six ordered data values is 29.5. 9 12 23 41 49 What is the missing value? The missing value is ☐.arrow_forwardFind the population mean or sample mean as indicated. Sample: 22, 18, 9, 6, 15 □ Select the correct choice below and fill in the answer box to complete your choice. O A. x= B. μεarrow_forwardWhy the correct answer is letter A? Students in an online course are each randomly assigned to receive either standard practice exercises or adaptivepractice exercises. For the adaptive practice exercises, the next question asked is determined by whether the studentgot the previous question correct. The teacher of the course wants to determine whether there is a differencebetween the two practice exercise types by comparing the proportion of students who pass the course from eachgroup. The teacher plans to test the null hypothesis that versus the alternative hypothesis , whererepresents the proportion of students who would pass the course using standard practice exercises andrepresents the proportion of students who would pass the course using adaptive practice exercises.The teacher knows that the percent confidence interval for the difference in proportion of students passing thecourse for the two practice exercise types (standard minus adaptive) is and the percent…arrow_forward
- Carpetland salespersons average $8,000 per week in sales. Steve Contois, the firm's vice president, proposes a compensation plan with new selling incentives. Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses.H 0: H a:arrow_forwardتوليد تمرين شامل حول الانحدار الخطي المتعدد بطريقة المربعات الصغرىarrow_forwardThe U.S. Postal Service will ship a Priority Mail® Large Flat Rate Box (12" 3 12" 3 5½") any where in the United States for a fixed price, regardless of weight. The weights (ounces) of 20 ran domly chosen boxes are shown below. (a) Make a stem-and-leaf diagram. (b) Make a histogram. (c) Describe the shape of the distribution. Weights 72 86 28 67 64 65 45 86 31 32 39 92 90 91 84 62 80 74 63 86arrow_forward
- (a) What is a bimodal histogram? (b) Explain the difference between left-skewed, symmetric, and right-skewed histograms. (c) What is an outlierarrow_forward(a) Test the hypothesis. Consider the hypothesis test Ho = : against H₁o < 02. Suppose that the sample sizes aren₁ = 7 and n₂ = 13 and that $² = 22.4 and $22 = 28.2. Use α = 0.05. Ho is not ✓ rejected. 9-9 IV (b) Find a 95% confidence interval on of 102. Round your answer to two decimal places (e.g. 98.76).arrow_forwardLet us suppose we have some article reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We will consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are 6, the sample mean force for lifting was 6.2 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.4 pounds with standard deviation 0.3 pounds. Assume that the standard deviations are known. Suppose that you wanted to detect a true difference in mean force of 0.25 pounds on the hands for these two activities. Under the null hypothesis, 40 = 0. What level of type II error would you recommend here? Round your answer to four decimal places (e.g. 98.7654). Use a = 0.05. β = i What sample size would be required? Assume the sample sizes are to be equal.…arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman





