Question

A study was performed on wear of a bearing y and its relationship to x₁ = oil viscosity and x₂ = load. The following data was obtained.

Y	X1	X2
293	1.6	851
230	15.5	816
172	22.0	1058
91	43.0	1201
113	33.0	1357
125	40.0	1115

USING MINITAB...

1. Fit a multiple linear regression model to these data.
2. Estimate σ².
3. Predict wear in which x₁ = 25 and x₂ = 1000.
4. Test for significance of regression using α = 0.05. What is the p-value for this test?
5. Find a 95% confidence interval and prediction interval when x₁ = 25 and x₂ = 1000.
6. Analyze R².
7. Analyze the residuals.

Answer 1

The multiple linear regression model is:

y = 383.8010 - 3.6381*X1 - 0.1117*X2

σ2 = 152.6191

The predicted value when x1 = 25 and x2 = 1000 is 181.167.

The hypothesis being tested is:

H0: β1 = β2 = 0

H1: At least one βi ≠ 0

The p-value is 0.0019.

Since the p-value (0.0019) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that the regression is significant.

The 95% confidence interval and prediction interval when x1 = 25 and x2 = 1000 are:

95% Confidence Interval	95% Prediction Interval
lower	upper	lower	upper
163.231	199.103	137.954	224.381

R² = 0.985

The residual plot is:

The output is:

	R²	0.985
	Adjusted R²	0.975
	R	0.992
	Std. Error	12.354
	n	6
	k	2
	Dep. Var.	Y
ANOVA table
Source	SS	df	MS	F	p-value
Regression	29,787.4761	2	14,893.7381	97.59	.0019
Residual	457.8572	3	152.6191
Total	30,245.3333	5
Regression output				confidence interval
variables	coefficients	std. error	t (df=3)	p-value	95% lower	95% upper
Intercept	383.8010
X1	-3.6381	0.5665	-6.423	.0077	-5.4408	-1.8354
X2	-0.1117	0.0434	-2.575	.0822	-0.2497	0.0264
Observation	Y	Predicted	Residual
1	293.0	282.9	10.1
2	230.0	236.3	-6.3
3	172.0	185.6	-13.6
4	91.0	93.2	-2.2
5	113.0	112.2	0.8
6	125.0	113.8	11.2
Predicted values for: Y
			95% Confidence Interval	95% Prediction Interval
X1	X2	Predicted	lower	upper	lower	upper	Leverage
25	1,000	181.167	163.231	199.103	137.954	224.381	0.208