Concept explainers
(a)
To find: The
To find: The least-squares regression line for all four data sets.
To find: The predicted value for
(a)

Answer to Problem 5.42E
The correlation for the data set A is 0.816.
The correlation for the data set B is 0.816.
The correlation for the data set C is 0.816.
The correlation for the data set D is 0.8176.
The least-squares regression line for the data set A is
The least-squares regression line for the data set B is
The least-squares regression line for the data set C is
The least-squares regression line for the data set D is
The predicted value for
The predicted value for
The predicted value for
The predicted value for
Explanation of Solution
Given info:
The four data sets are used to exploring the
Calculation:
Correlation for Data set A:
Software procedure:
Step-by-step procedure to find the correlation between the x and y for data set A by using the MINITAB software:
- Select Stat >Basic Statistics > Correlation.
- In Variables, select x and y.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the correlation between the x and y for data set A is 0.816.
Correlation for Data set B:
Software procedure:
Step-by-step procedure to find the correlation between the x and y for data set B by using the MINITAB software:
- Select Stat >Basic Statistics > Correlation.
- In Variables, select x and y.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the correlation between the x and y for data set B is 0.816.
Correlation for Data set C:
Software procedure:
Step-by-step procedure to find the correlation between the x and y for data set C by using the MINITAB software:
- Select Stat >Basic Statistics > Correlation.
- In Variables, select x and y.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the correlation between the x and y for data set C is 0.816.
Correlation for Data set D:
Software procedure:
Step-by-step procedure to find the correlation between the x and y for data set D by using the MINITAB software:
- Select Stat >Basic Statistics > Correlation.
- In Variables, select x and y.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the correlation between the x and y for data set D is 0.817.
Equation of the least-squares line for Data set A:
Software procedure:
Step-by-step procedure to find the equation of the least-squares line by using the MINITAB software:
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the least-squares line for predicting y from x for data set A is
Equation of the least-squares line for Data set B:
Software procedure:
Step-by-step procedure to find the equation of the least-squares line by using the MINITAB software:
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the least-squares line for predicting y from x for data set B is
Equation of the least-squares line for Data set C:
Software procedure:
Step-by-step procedure to find the equation of the least-squares line by using the MINITAB software:
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the least-squares line for predicting y from x for data set C is
Equation of the least-squares line for Data set D:
Software procedure:
Step-by-step procedure to find the equation of the least-squares line by using the MINITAB software:
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the least-squares line for predicting y from x for data set D is
Predicted value for
Software procedure:
Step-by-step procedure to find the predicted value for
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- In option, enter 10 under prediction.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the predicted value for
Predicted value for
Software procedure:
Step-by-step procedure to find the predicted value for
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- In option, enter 10 under prediction.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the predicted value for
Predicted value for
Software procedure:
Step-by-step procedure to find the predicted value for
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- In option, enter 10 under prediction.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the predicted value for
Predicted value for
Software procedure:
Step-by-step procedure to find the predicted value for
- Choose Stat > Regression > Regression.
- In Responses, enter the column of y.
- In Predictors, enter the column of x.
- In option, enter 10 under prediction.
- Click OK.
Output using the MINITAB software is given below:
From the MINITAB output, the predicted value for
From the results, it can be observed that the correlation for all four data sets, the least-squares regression line and the predicted value for
(b)
To construct: The
(b)

Answer to Problem 5.42E
Scatterplot for Data set A:
Output using the MINITAB software is given below:
Scatterplot for Data set B:
Output using the MINITAB software is given below:
Scatterplot for Data set C:
Output using the MINITAB software is given below:
Scatterplot for Data set D:
Output using the MINITAB software is given below:
Explanation of Solution
Calculation:
Scatterplot:
Software procedure:
Step-by-step procedure to construct scatterplot for x and y for all four data sets by using the MINITAB software:
- Choose Graph > Scatter plot.
- Choose With Regression, and then click OK.
- Under Y variables, enter a column of y.
- Under X variables, enter a column of x.
- Click OK.
Observation:
The scatterplot shows that the predicted values are passed through the regression line of the model. Moreover, there is outlier that appears in the x and y directions for the data set A, B, and C. Also, the scatterplot for the data set D shows that the most of the points are plotted around 8.
(c)
To identify: Which of the four cases would you be willing to use the regression line to describe the dependence of y on x.
(c)

Answer to Problem 5.42E
The data set A would use the regression line to describe the dependence of y on x.
Explanation of Solution
From the scatterplots for all data sets, it can be observed that the points for data set A are scattered around the straight line when compared to the other data sets. Hence, the data set A would use the regression line to describe the dependence of y on x.
Want to see more full solutions like this?
Chapter 5 Solutions
Basic Practice of Statistics (Instructor's)
- 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_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_forwardHow 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_forward
- Consider 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_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 α = 0.05. β = 0.0594 What sample size would be required? Assume the sample sizes are to be…arrow_forward
- Consider 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_forwardPlease solve 6.31 by using the method of sections im stuck and need explanationarrow_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





