Exam2

  • June 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Exam2 as PDF for free.

More details

  • Words: 353
  • Pages: 2
EXAM #2: Dummy variable – an indicator variable z1 = x1. (in this case, p=k=1) Simple first-order model with one predictor variable: Second-order model with one predictor variable: Second-order model with two predictor variables: • In this model, the variable z5 = x1x2 is added to account for the potential effects of the two variables acting together. This type of effect is called interaction Interaction Test Solution

Multiple Regression Approach to Experimental Design At a .05 level of significance, the critical value of F with k – 1 = 3 – 1 = 2 num. d.f. and nT – k = 15 – 3 = 12 denom. d.f. is 3.89 Because the observed value of F (9.55) is greater than the critical value of 3.89, we reject the Ho. Alternatively, we reject the Ho because the p-value of .003 < a = .05. Multiple regression with only one categorical variable In general, to test whether the addition of x2 to a model involving x1 (or the deletion of x2 from a model involving x1 and x2) is statistically significant we can perform an F Test. (SSE(reduced)-SSE(full))/ number of extra terms (SSE(x1 )-SSE(x1 ,x2 ))/ 1 F= F= MSE(full) (SSE(x1 , x2 ))/ (n − p − 1) This computed F value is then compared with critical value from F table with ‘number of extra terms’ numerator d.f. and (n-p-1) denominator d.f. and for a given level of significance. The p –value criterion can also be used to determine whether it is advantageous to add one or more dependent variables to a multiple regression model. The p –value associated with the computed F statistic can be compared to the level of significance a. It is difficult to determine the p –value directly from the tables of the F distribution, but computer software packages, such as Minitab provide the p-value.

H 0 : µ A = µ B = µC H 0 : µ A − µC = 0 H 0 : µ B − µC = 0

SSE( x1) − SSE( x1, x 2) 1 F= SSE( x1, x 2) n − p −1

Related Documents

Exam2
June 2020 10
Exam2
November 2019 26
Span1412- Exam2
June 2020 7
Prelim Exam2
June 2020 15
Solution Exam2
November 2019 19
Exam2 Solution
November 2019 21