
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
Launchpad For Moore's Statistics: Concepts And Controversies (twelve Month Access)
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forward
- I need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Statistics: Engineering Probabilities)arrow_forward3. Consider the following regression model: Yi Bo+B1x1 + = ···· + ßpxip + Єi, i = 1, . . ., n, where are i.i.d. ~ N (0,0²). (i) Give the MLE of ẞ and σ², where ẞ = (Bo, B₁,..., Bp)T. (ii) Derive explicitly the expressions of AIC and BIC for the above linear regression model, based on their general formulae.arrow_forward
- How does the width of prediction intervals for ARMA(p,q) models change as the forecast horizon increases? Grows to infinity at a square root rate Depends on the model parameters Converges to a fixed value Grows to infinity at a linear ratearrow_forwardConsider the AR(3) model X₁ = 0.6Xt-1 − 0.4Xt-2 +0.1Xt-3. What is the value of the PACF at lag 2? 0.6 Not enough information None of these values 0.1 -0.4 이arrow_forwardSuppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a. Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.) I worked out the Upper Limit, but I can't seem to arrive at the correct answer for the Lower Limit. What is the Lower Limit?…arrow_forward
- Let 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 α = 0.05. β = 0.0594 What sample size would be required? Assume the sample sizes are to be…arrow_forwardConsider the hypothesis test Ho: 0 s² = = 4.5; s² = 2.3. Use a = 0.01. = σ against H₁: 6 > σ2. Suppose that the sample sizes are n₁ = 20 and 2 = 8, and that (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = 1.96 The critical value is f = 6.18 Conclusion: fail to reject the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/2/622 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) 035arrow_forwardUsing the method of sections need help solving this please explain im stuckarrow_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





