
Concept explainers
a.
Find the equation of the least-squares line, if y is expressed in kilograms instead of pounds.
a.

Answer to Problem 28E
The equation of the least-squares line, if y is expressed in kilograms instead of pounds is ˆy=−427+3.90x_.
Explanation of Solution
Calculation:
The least-squares regression equation relating the shear strength (pounds or lb), y of steel to weld diameter, x is given as ˆy=−936.22+8.577x. It is also mentioned that 1 lb=0.4536 kg, so that, when y is expressed in kilograms instead of pounds, new y=0.4536(old y).
Data transformation:
Software procedure:
Step-by-step procedure to transform the data using the MINITAB software:
- Choose Calc > Calculator.
- Enter the column of y1 under Store result in variable.
- Enter the 0.4536*‘y’ in Expression.
- Click OK in all dialogue boxes.
The transformed data, where each new y=0.4536(old y), is stored in the column y1.
The least-squares equation can be obtained using software.
Least-squares equation:
Software procedure:
Step-by-step procedure to obtain the least-squares equation using the MINITAB software:
- Choose Stat > Regression > Regression > Fit Regression Model.
- Enter the column of y1 under Responses.
- Enter the column of x under Continuous predictors.
- Choose Results and select Coefficients, Regression Equation.
- Click OK in all dialogue boxes.
Output obtained using MINITAB is given below:
Thus, the equation of the least-squares line, if y is expressed in kilograms instead of pounds is ˆy=−427+3.90x_.
b.
Find the new slope, intercept, and equation of the least-squares line, if a constant, c is multiplied with every observation on y.
b.

Answer to Problem 28E
The new slope is –936.22 c.
The new intercept is 8.577 c.
The new equation of the least-squares line is ˆy=−936.22c+8.577cx_.
Explanation of Solution
Calculation:
Changing the unit of y from pounds to kilograms causes a change in the scale of y. Now, if the scale of the response variable in a regression, y, is changed by multiplying with a constant c, each least-squares coefficient is also multiplied by c.
Thus, each coefficient in the regression equation, that is, the intercept –936.22 and the slope corresponding to x, 8.577, will be multiplied by c, in order to get the regression equation in this situation. The regression equation is new(ˆy)=−936.22c+8.577cx.
Thus, the new slope is –936.22 c; the new intercept is 8.577 c; the new equation of the least-squares line is ˆy=−936.22c+8.577cx_.
These can be verified by using the formulae for a and b. The general formulae for a and b, for a simple least-squares regression equation, ˆy=a+bx, are as follows:
a=ˉy−bˉx;b=∑xy−(∑x)(∑y)n∑x2−(∑x)2n.
Now, new y=c(old y), replace each value of y by cy in the above formulae. Note that, if each observation on y is multiplied by c, then the mean of y is also multiplied by c, that is, new ˉy=cˉy.
Therefore,
newb=∑x(cy)−(∑x)(∑cy)n∑x2−(∑x)2n=c∑xy−c(∑x)(∑y)n∑x2−(∑x)2n=c[∑xy−(∑x)(∑y)n∑x2−(∑x)2n]=cb.
newa=cˉy−(newb)ˉx=cˉy−cbˉx=c(ˉy−bˉx)=ca.
Substitute the values of new a and newb in the general form of the regression equation.
newˆy=newa+newbx=ca+cbx.
In this problem, a = –936.22 and b = 8.577. As a result, the new slope is –936.22 c; the new intercept is 8.577 c.
Substituting these values in the equation for newˆy, the new equation of the least-squares line is ˆy=−936.22c+8.577cx_.
Want to see more full solutions like this?
Chapter 5 Solutions
Introduction to Statistics and Data Analysis
- 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
- Big Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageLinear Algebra: A Modern IntroductionAlgebraISBN:9781285463247Author:David PoolePublisher:Cengage Learning

