A study concerning the bitterness of beer brewed with different types of hops in
ID: 3219616 • Letter: A
Question
A study concerning the bitterness of beer brewed with different types of hops investigated the relationship between y = Bavarian Bitterness score of the beer and x1 = alpha level and x2 = beta level of the type of hops used. The following computer output was used in the analysis of the observed data for the interaction model,
y = + 1x1 + 2x2 + 3x3 + e , where x3 = x1x2 .
a) Using a .05 significance level, test the utility of this multiple regression model.
b) Compute a 95% confidence interval for the value of 1 .
c) Test the hypothesis H0 : 3 = 0 vs. 3 0 at the = .05 level.
d) Does the interaction term help predict the value of y ? Explain in a few sentences.
e) If s = .27 when x1 = 10 and x2 = 4 , compute a 95% confidence interval for the mean value of y whenx1= 10 and x2 = 4. Use y = + 1x1 + 2x2 + 3x3 + e for the model.
f) If s = .27 when x1 = 10 and x2 = 4 , compute a 95% prediction interval for the value of y when x1 =10 and x2 = 4. Use y = + 1x1 + 2x2 + 3x3 + e for the model.
The regression equation is Y 09 037 x3 Predictor Stdev t-ratio Coef 092 Constant 054 1.70 118 166 021 000 7.90 X1 037 048 082 2.23 X2 037 167 025 1.48 X3 R-sq 78.7 R-sg (adj) 72.9% S 1.221 Analysis of variance DF SS MS SOURCE Regression 60.5 20.17 13.54 0.001 16.4 1.49 11 Error Total 76.9 14Explanation / Answer
a)
Ho: model is not significant
H1: model is significant
With F = 13.54 and p-value < 0.05, I reject ho and conclude that model is significant
b)
95% CI = beta +- t(a/2,n-k-1)*SE_beta
lower
=0.166-ABS(T.INV.2T(0.05,15-3-1))*(0.166/7.9)
0.119751
upper
=0.166+ABS(T.INV.2T(0.05,15-3-1))*(0.166/7.9)
0.212249
c)
H0 : 3 = 0 vs. 3 0
With t = 1.48 and p-value > 0.05, I fail to reject ho and conclude that 3 = 0
d)
Since the interaction effect is not significant, I can say that interaction term doesn’t help to predict the value of y
lower
=0.166-ABS(T.INV.2T(0.05,15-3-1))*(0.166/7.9)
0.119751
upper
=0.166+ABS(T.INV.2T(0.05,15-3-1))*(0.166/7.9)
0.212249
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