A researcher developed a regression model to predict the tear rating of a bag of coffee...
A researcher developed a regression model to predict the tear rating of a bag of coffee based on the plate gap in bag-sealing equipment. Data were collected on
2828
bags in which the plate gap was varied. An analysis of variance from the regression showed that
b1equals=0.79240.7924
and
Upper S Subscript b 1Sb1equals=0.20940.2094.
a. At the 0.05 level of significance, is there evidence of a linear relationship between the plate gap of the bag-sealing machine and the tear rating of a bag of coffee?
b. Construct a 95% confidence interval estimate of the population slope,
betaβ1.
Homework Answers
Answer:
a)
Given,
b1 = 0.7924
Sb1 = 0.2094
Now,
t = b1/Sb1
substitute the known values
t = 0.7924/.2094
test statistic = 3.784
Degree of freedom = n - 2
= 28 - 2
= 26
So the critical value is tcrit = 2.056
Here t > tcrit , so there is a linear relationship b/w plate of the bag-sealing machine and the tear rating of a bag of coffee.
b)
To determine the 95% confidence interval
Now CI = b1 +/- t0.05,26 Sb1
substitute the known values
= 0.7924 +/- 2.056*0.2094
= 0.7924 +/- 0.4305264
CI = (0.3619 , 1.2229)
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