
Find the 95% confidence interval for the group means or group proportions for the health characteristics.

Answer to Problem 3CS
The 95% confidence intervals for the each of the characteristics are as follows:
Characteristic | 95% confidence interval |
Age | |
Systolic blood pressure | |
Diastolic blood pressure | |
Treatment for hypertension | |
Atrial fibrillation | |
Diabetes | |
Cigarette smoking | |
Coronary bypass surgery |
Explanation of Solution
Calculation:
The data represents the means and standard deviations corresponding to three characteristics of standard treatment and new treatment.
Furthermore, the data corresponding to the percentage with the characteristics for the 5 characteristics for standard treatment and new treatment is given.
Here, it is given that the
95% confidence interval for the variable “Age”:
Software Procedure:
Step by step procedure to find 95% confidence interval forthe difference between the two population means using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Sample t.
- Choose Summarized data.
- In first sample, enter the sample size as 731, sample mean as 64 and sample standard deviation as 11.
- In second sample, enter the sample size as 1,089, sample mean as 65 and sample standard deviation as 11.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the mean age of new treatment and standard treatment is
95% confidence interval for the variable “Systolic blood pressure”:
Software Procedure:
Step by step procedure to find 95% confidence interval forthe difference between the two population means using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Sample t.
- Choose Summarized data.
- In first sample, enter the sample size as 731, sample mean as 121 and sample standard deviation as 18.
- In second sample, enter the sample size as 1,089, sample mean as 124 and sample standard deviation as 17.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the mean systolic blood pressure of new treatment and standard treatmentis
95% confidence interval for the variable “Diastolic blood pressure”:
Software Procedure:
Step by step procedure to find 95% confidence interval for the difference between the two population means using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Sample t.
- Choose Summarized data.
- In first sample, enter the sample size as 731, sample mean as 71 and sample standard deviation as 10.
- In second sample, enter the sample size as 1,089, sample mean as 72 and sample standard deviation as 10.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the mean diastolic blood pressure of new treatment and standard treatmentis
95% confidence interval for the variable “Treatment for hypertension”:
The sample size of standard treatment is
Hence, the specified characteristics of treatment for hypertension under new treatment is
The sample size of new treatment is
Hence, the specified characteristics of treatment for hypertension under new treatment is
Software Procedure:
Step by step procedure to find 95% confidence interval for the difference between the two population proportions using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Proportions.
- Choose Summarized data.
- In first sample, enter the number of trials as 731 and number of events as 462.
- In second sample, enter the number of trials as 1,089 and number of events as 694.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the proportion of treatment for hypertension under new treatment and standard treatmentis
95% confidence interval for the variable “Atrial fibrillation”:
The sample size of standard treatment is
Hence, the specified characteristics of treatment for Atrial fibrillation under standard treatment is
The sample size of new treatment is
Hence, the specified characteristics of Atrial fibrillation under new treatment is
Software Procedure:
Step by step procedure to find 95% confidence interval for the difference between the two populationproportions using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Proportions.
- Choose Summarized data.
- In first sample, enter the number of trials as 731 and number of events as 92.
- In second sample, enter the number of trials as 1,089 and number of events as 121.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the proportion of Atrial fibrillation under new treatment and standard treatment is
95% confidence interval for the variable “Diabetes”:
The sample size of standard treatment is
Hence, the specified characteristics of diabetes under standard treatment is
The sample size of new treatment is
Hence, the specified characteristics of diabetes under new treatment is
Software Procedure:
Step by step procedure to find 95% confidence interval for the difference between the two population proportions using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Proportions.
- Choose Summarized data.
- In first sample, enter the number of trials as 731 and number of events as 221.
- In second sample, enter the number of trials as 1,089 and number of events as 333.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the95% confidence interval for the difference between the proportion of Diabetes under new treatment and standard treatment is
95% confidence interval for the variable “Cigarette smoking”:
The sample size of standard treatment is
Hence, the specified characteristics of cigarette smoking under standard treatment is
The sample size of new treatment is
Hence, the specified characteristics of cigarette smoking under new treatment is
Software Procedure:
Step by step procedure to find 95% confidence interval for the difference between the two population proportions using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Proportions.
- Choose Summarized data.
- In first sample, enter the number of trials as 731 and number of events as 94.
- In second sample, enter the number of trials as 1,089 and number of events as 124.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the proportion of Cigarette smoking under new treatment and standard treatment is
95% confidence interval for the variable “Coronary bypass surgery”:
The sample size of standard treatment is
Hence, the specified characteristics of Coronary bypass surgery under standard treatment is
The sample size of new treatment is
Hence, the specified characteristics of coronary bypass surgery under new treatment is
Software Procedure:
Step by step procedure to find 95% confidence interval for the difference between the two population proportions using the MINITAB software:
- Choose Stat > Basic Statistics > 2-Proportions.
- Choose Summarized data.
- In first sample, enter the number of trials as 1,089 and number of events as 317.
- In second sample, enter the number of trials as 731 and number of events as 208.
- Choose Options.
- In Confidence level, enter 95.
- Enter 0 in Hypothesized value.
- In Alternative, select not equal to
- Click OK in all the dialogue boxes.
Output using the MINITAB software is given below:
From, the MINITAB output the 95% confidence interval for the difference between the proportion of Coronary bypass surgery under new treatment and standard treatment is
Want to see more full solutions like this?
Chapter 9 Solutions
Essential Statistics
- 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





