
Concept explainers
(a)
To find: The row and column totals.
(a)

Answer to Problem 29RE
Solution: The result obtained is as follows:
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 721 | 173 | 412 | 1306 |
Left | 22 | 26 | 28 | 76 |
Total | 743 | 199 | 440 | Grand Total = 1382 |
Explanation of Solution
Calculation:
Compute the row and column total as follows:
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 721 | 173 | 412 | 721+173+412 = 1306 |
Left | 22 | 26 | 28 | 22+26+28 = 76 |
Total | 721+22 = 743 | 173+26 = 199 | 412+28 = 440 | Grand Total = 1382 |
Interpretation: The row total of people who stayed HMO is 1306 and who left HMO is 76. The column total of people with no complaint is 743, people with medical complaint is 199 and people with nonmedical complaint is 440. The grand total is 1382.
(b)
To find: The percentage of each group who left.
(b)

Answer to Problem 29RE
Solution: The percentage of people who left HMO with no complaint, with medical complaint and with nonmedical complaint is 2.96%, 13.07%, and 6.36%, respectively.
Explanation of Solution
Calculation:
Compute the percentage of each group who left as follows:
Similarly,
(c)
To find: The expected counts.
(c)

Answer to Problem 29RE
Solution: Yes, chi-square test can be used safely and the table of expected counts is as follows:
Excepted Counts Table | ||||
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 702.14 | 188.06 | 415.8 | 1306 |
Left | 40.86 | 10.94 | 24.20 | 76 |
Total | 743 | 199 | 440 | 1382 |
Explanation of Solution
Calculation:
Calculate the expected count using the formula as follows:
Now, compute the expected count for all the rows and column total as follows:
Excepted Counts Table | ||||
No complaint | Medical complaint | Non-medical complaint | Total | |
Stayed | 1306 | |||
Left | 76 | |||
Total | 743 | 199 | 440 | 1382 |
Interpretation: From the above table, it can be seen that all the expected counts are greater than 5. Therefore, it can be concluded that chi-square test can be used safely.
(d)
Section 1:
The null and alternate hypothesis.
(d)
Section 1:

Answer to Problem 29RE
Solution: The null and alternate hypothesis is as follows:
Explanation of Solution
Section 2:
To find: The degrees of freedom of the test.
Section 2:

Answer to Problem 29RE
Solution: The degrees of freedom is 2.
Explanation of Solution
Calculation:
Compute the degree of freedom as follows:
Therefore, it can be concluded that the degrees of freedom of the test are 2.
Section 3:
To Test: The significance of the chi-square test statistic.
Solution: Yes, the result is significant.
Explanation:
Calculation:
The
Conclusion:
Since, the null hypothesis is rejected, it can be concluded that the test is significant.
Section 4:
To explain: The relationship between complaints and people leaving HMO.
Solution: There is a significant relationship between complaints and people leaving HMO.
Explanation: The null and alternate hypothesis stated in section 1 of part (d) is as follows:
In addition, according to the results in section 3 of part (d), the null hypothesis was rejected.
Since the null hypothesis is rejected, it can be stated that it is wrong to say that there is no relationship between members complaining and leaving HMO.
Want to see more full solutions like this?
Chapter 24 Solutions
Statistics: Concepts and Controversies
- ons 12. A sociologist hypothesizes that the crime rate is higher in areas with higher poverty rate and lower median income. She col- lects data on the crime rate (crimes per 100,000 residents), the poverty rate (in %), and the median income (in $1,000s) from 41 New England cities. A portion of the regression results is shown in the following table. Standard Coefficients error t stat p-value Intercept -301.62 549.71 -0.55 0.5864 Poverty 53.16 14.22 3.74 0.0006 Income 4.95 8.26 0.60 0.5526 a. b. Are the signs as expected on the slope coefficients? Predict the crime rate in an area with a poverty rate of 20% and a median income of $50,000. 3. Using data from 50 workarrow_forward2. The owner of several used-car dealerships believes that the selling price of a used car can best be predicted using the car's age. He uses data on the recent selling price (in $) and age of 20 used sedans to estimate Price = Po + B₁Age + ε. A portion of the regression results is shown in the accompanying table. Standard Coefficients Intercept 21187.94 Error 733.42 t Stat p-value 28.89 1.56E-16 Age -1208.25 128.95 -9.37 2.41E-08 a. What is the estimate for B₁? Interpret this value. b. What is the sample regression equation? C. Predict the selling price of a 5-year-old sedan.arrow_forwardian income of $50,000. erty rate of 13. Using data from 50 workers, a researcher estimates Wage = Bo+B,Education + B₂Experience + B3Age+e, where Wage is the hourly wage rate and Education, Experience, and Age are the years of higher education, the years of experience, and the age of the worker, respectively. A portion of the regression results is shown in the following table. ni ogolloo bash 1 Standard Coefficients error t stat p-value Intercept 7.87 4.09 1.93 0.0603 Education 1.44 0.34 4.24 0.0001 Experience 0.45 0.14 3.16 0.0028 Age -0.01 0.08 -0.14 0.8920 a. Interpret the estimated coefficients for Education and Experience. b. Predict the hourly wage rate for a 30-year-old worker with four years of higher education and three years of experience.arrow_forward
- 1. If a firm spends more on advertising, is it likely to increase sales? Data on annual sales (in $100,000s) and advertising expenditures (in $10,000s) were collected for 20 firms in order to estimate the model Sales = Po + B₁Advertising + ε. A portion of the regression results is shown in the accompanying table. Intercept Advertising Standard Coefficients Error t Stat p-value -7.42 1.46 -5.09 7.66E-05 0.42 0.05 8.70 7.26E-08 a. Interpret the estimated slope coefficient. b. What is the sample regression equation? C. Predict the sales for a firm that spends $500,000 annually on advertising.arrow_forwardCan you help me solve problem 38 with steps im stuck.arrow_forwardHow do the samples hold up to the efficiency test? What percentages of the samples pass or fail the test? What would be the likelihood of having the following specific number of efficiency test failures in the next 300 processors tested? 1 failures, 5 failures, 10 failures and 20 failures.arrow_forward
- The battery temperatures are a major concern for us. Can you analyze and describe the sample data? What are the average and median temperatures? How much variability is there in the temperatures? Is there anything that stands out? Our engineers’ assumption is that the temperature data is normally distributed. If that is the case, what would be the likelihood that the Safety Zone temperature will exceed 5.15 degrees? What is the probability that the Safety Zone temperature will be less than 4.65 degrees? What is the actual percentage of samples that exceed 5.25 degrees or are less than 4.75 degrees? Is the manufacturing process producing units with stable Safety Zone temperatures? Can you check if there are any apparent changes in the temperature pattern? Are there any outliers? A closer look at the Z-scores should help you in this regard.arrow_forwardNeed help pleasearrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 4. One-Way ANOVA: Analyze the customer satisfaction scores across four different product categories to determine if there is a significant difference in means. (Hints: The null can be about maintaining status-quo or no difference among groups) H0 = H1=arrow_forward
- Please conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points 2. Two-Sample T-Test: Compare the average sales revenue of two different regions to determine if there is a significant difference. (Hints: The null can be about maintaining status-quo or no difference among groups; if alternative hypothesis is non-directional use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null) H0 = H1=arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points 3. Paired T-Test: A company implemented a training program to improve employee performance. To evaluate the effectiveness of the program, the company recorded the test scores of 25 employees before and after the training. Determine if the training program is effective in terms of scores of participants before and after the training. (Hints: The null can be about maintaining status-quo or no difference among groups; if alternative hypothesis is non-directional, use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting the null) H0 = H1= Conclusion:arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. The data for the following questions is provided in Microsoft Excel file on 4 separate sheets. Please conduct these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to…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





