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using the attached data table a. should we reject the null or fail to reject and justify using 95% confidece intervals of beta 1. b.  interpret the decision in the contect of can we predict oxidation based on tempertature?

using the attached data table a. should we reject the null or fail to reject and justify using 95% confidece intervals of beta 1. b.  interpret the decision in the contect of can we predict oxidation based on tempertature?

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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using the attached data table

a. should we reject the null or fail to reject and justify using 95% confidece intervals of beta 1.

b.  interpret the decision in the contect of can we predict oxidation based on tempertature?

Bivariate Fit of oxidation By temperature oxidation ♡ 2. N 1 -2 -1 Linear Fit Linear Fit oxidation=2-0.7*temperature Summary of Fit RSquare RSquare Adj Source Model Error C. Total 0 temperature Root Mean Square Error Mean of Response Observations (or Sum Wgts) Analysis of Variance Term Intercept temperature 0.816667 0.755556 0.60553 DF 1 3 4 6.0000000 Parameter Estimates 2 5 4.9000000 1.1000000 • 1 Sum of Squares Mean Square F Ratio 4.90000 13.3636 0.36667 Prob > F 0.0354 2 G 10 < Estimate Std Error t Ratio Prob>lt] Lower 95% Upper 95% 7.39 0.0051" 1.1381895 2 0.270801 2.8618105 -0.7 0.191485 -3.66 0.0354" -1.309392 -0.090608 0.
Transcribed Image Text:Bivariate Fit of oxidation By temperature oxidation ♡ 2. N 1 -2 -1 Linear Fit Linear Fit oxidation=2-0.7*temperature Summary of Fit RSquare RSquare Adj Source Model Error C. Total 0 temperature Root Mean Square Error Mean of Response Observations (or Sum Wgts) Analysis of Variance Term Intercept temperature 0.816667 0.755556 0.60553 DF 1 3 4 6.0000000 Parameter Estimates 2 5 4.9000000 1.1000000 • 1 Sum of Squares Mean Square F Ratio 4.90000 13.3636 0.36667 Prob > F 0.0354 2 G 10 < Estimate Std Error t Ratio Prob>lt] Lower 95% Upper 95% 7.39 0.0051" 1.1381895 2 0.270801 2.8618105 -0.7 0.191485 -3.66 0.0354" -1.309392 -0.090608 0.
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Step 1

The question is about testing of slope

Given :

From given output :

95 % CI of " temperature " : ( -1.309392, -0.090608 )

 

To find :

a ) reject the null or fail to reject and justify CI of beta 1.

b )  interpret the decision in the context that predict oxidation based on temperature?

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