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
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62f6355a-05e6-4234-9d59-a73a1309b418%2Fb0a7a054-ab18-49b5-999f-6a041d349973%2Fr46hybk_processed.jpeg&w=3840&q=75)
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.
Expert Solution
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?
Step by step
Solved in 2 steps
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