A study concerning the bitterness of beer brewed with different types of hops in

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 when x1= 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 14

Explanation / Answer

a) Using a .05 significance level, test the utility of this multiple regression model.

from the given regression analysis

we can say for this regression F - test, value of F = 13.54 and respective p- value = 0.001, which is under significance level 0.05 so we can reject the null and can say that regression is statistically significant.

(b) Compute a 95% confidence interval for the value of 1 .tcritical = +- 2.201

1 (mean value ) = 0.166

Lower bound = 0.166 - 2.201 * 0.021 = 0.1198

upper bound = 0.166 + 2.201 * 0.021 =0.2122

so confidence interval = ( 0.1198 , 0.2122)

(c) Test the hypothesis H0 : 3 = 0 vs. 3 0 at the = .05 level.

Test - Statistic : t = 1.48 and P - value = 0.167 > 0.05

so we can not reject null and can say that 3 = 0 or can say the interaction effect of both explanatory variable is not statistically significant.

(D) Does the interaction term help predict the value of y ? Explain in a few sentences.

as the result got in problem c, that the interaction term is not statistically significant so it would not put a significant impact on y term. It may contain some residual values but nothing for using of practical purpose. We can also see that x2 have negative relationship with response variable y so interaction term may get used in manipulating its impact.

(E) If s = .27 when x1 = 10 and x2 = 4 , compute a 95% confidence interval for the mean value of y when x1= 10 and x2 = 4. Use y = + 1x1 + 2x2 + 3x3 + e for themodel.

y = + 1x1 + 2x2 + 3x3 + e

where x1= 10 and x2 = 4 and x3 = 40

so y = 0.09 + 0.166 x1 - 0.082 x2 + 0.037x1x2

y = 0.09 + 0.166 * 10 - 0.082 * 4 + 0.037 * 40 = 2.902

and sy = 0.27 as given here

so 95 % confidence interval

= 2.902 +- 2.201 * 0.27 = (2.3077, 3.4963)

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