{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Chapter 4 Answers - Chapter 4 Answers 1(a The predicted...

This preview shows pages 1–2. Sign up to view the full content.

Chapter 4 Answers 1. (a) The predicted average test score is 520 4 5 82 22 392 36 TestScore = . - . = . (b)The predicted change in the classroom average test score is ( 5 82 19) ( 5 82 23) 23 28 TestScore = - . - - . = . (c)Using the formula for 0 β in Equation (4.8), we know the sample average of the test scores across the 100 classrooms is 0 1 520 4 5 82 21 4 395 85 TestScore CS β β = + × = . - . × . = . . (d)Use the formula for the standard error of the regression (SER) in Equation (4.19) to get the sum of squared residuals: 2 2 ( 2) (100 2) 11 5 12961 SSR n SER = - = - × . = . Use the formula for 2 R in Equation (4.16) to get the total sum of squares: 2 2 12961 13044 1 1 0 08 SSR TSS R = = = . - - . The sample variance is 2 Y s = TSS 13044 1 99 131 8. n - = = . Thus, standard deviation is 2 11 5. Y Y s s = = . 2. The sample size 200. n = The estimated regression equation is 2 (2 15) 99 41 (0 31) 3 94 0 81 SER 10 2 Weight Height R = . - . + . . , = . , = . . (a)Substituting 70, 65, and 74 Height = inches into the equation, the predicted weights are 176.39, 156.69, and 192.15 pounds. (b) 3 94 3 94 1 5 5 91. Weight Height = . = . . = . (c)We have the following relations: 1 2 54 and1 0 4536 . in cm lb kg = . = . Suppose the regression equation in the centimeter-kilogram space is 0 1 ˆ ˆ Weight Height γ γ = + .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern