Question

In a small-scale experimental study of the relation between degree of brand liking (y) and moisture content (x1) and sweetness (x2) of the product, results were obtained from the experiment based on a completely randomized design. Fit a regression model of the form y = beta0 + beta1 x1 + beta2 x2 + e. Here, e is the random error term which is assumed to be normally distributed with mean 0 and constant variance.

y	x1	x2
64	4	2
73	4	4
61	4	2
76	4	4
72	6	2
80	6	4
71	6	2
83	6	4
83	8	2
89	8	4
86	8	2
93	8	4
88	10	2
95	10	4
94	10	2
100	10	4

Construct and interpret a 99% prediction interval for the brand liking score when the moisture content is 5 and the sweetness level is 4.

Which interval (the mean response or the prediction interval) is wider? Why?

What is the variance inflation factors associated with the two predictor variables? Is the multicollinearity assumption violated? Why or why not?

Answer 1

Construct and interpret a 99% prediction interval for the brand liking score when the moisture content is 5 and the sweetness level is 4.

The 99% prediction interval for the brand liking score when the moisture content is 5 and the sweetness level is 4 is between 68.481 and 86.069.

Which interval (the mean response or the prediction interval) is wider? Why?

The prediction interval is wider because prediction intervals must account for both the uncertainty in estimating the population mean, plus the random variation of the individual values.

What is the variance inflation factors associated with the two predictor variables? Is the multicollinearity assumption violated? Why or why not?

The variance inflation factor for the two predictor variables is 1.000. Therefore, the multicollinearity assumption is not violated because both variance inflation factors are less than 10.

The output is:

	R²	0.952
	Adjusted R²	0.945
	R	0.976
	Std. Error	2.693
	n	16
	k	2
	Dep. Var.	y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	1,872.7000	2	936.3500	129.08	2.66E-09
Residual	94.3000	13	7.2538
Total	1,967.0000	15
Regression output				confidence interval
variables	coefficients	std. error	t (df=13)	p-value	99% lower	99% upper	VIF
Intercept	37.6500
x1	4.4250	0.3011	14.695	1.78E-09	3.5179	5.3321	1.000
x2	4.3750	0.6733	6.498	2.01E-05	2.3468	6.4032	1.000
Predicted values for: y
			99% Confidence Interval	99% Prediction Interval
x1	x2	Predicted	lower	upper	lower	upper	Leverage
5	4	77.275	73.881	80.669	68.481	86.069	0.175