Question

Observation	x1	y1		x2	y2		x3	y3		x4	y4
1	10	8.04		10	9.14		10	7.46		8	6.58
2	8	6.95		8	8.14		8	6.77		8	5.76
3	13	7.58		13	8.74		13	12.74		8	7.71
4	9	8.81		9	8.77		9	7.11		8	8.84
5	11	8.33		11	9.26		11	7.81		8	8.47
6	14	9.96		14	8.1		14	8.84		8	7.04
7	6	7.24		6	6.13		6	6.08		8	5.25
8	4	4.26		4	3.1		4	5.39		19	12.5
9	12	10.84		12	9.13		12	8.15		8	5.56
10	7	4.82		7	7.26		7	6.42		8	7.91
11	5	5.68		5	4.74		5	5.73		8	6.89

• Fit a simple linear regression model to each set of (x, y) data, i.e., one model fit to (x₁, y₁), one model fit to (x₂, y₂), one model fit to (x₃, y₃), and one model fit to (x₄, y₄).

• Write down the estimated regression equation for each fitted model, together with the values of the coefficient of determination, r2, and the standard error of the estimate, s=MSE‾‾‾‾‾√.

• For each set of (x, y) data, create a scatterplot of y (vertical) versus x (horizontal) with the estimated regression line added to the plot.

• For each set of (x, y) data, create a scatterplot of the residuals (vertical) versus (horizontal). Based on each plot, do the zero mean and constant variance assumptions about the simple linear regression model error seem reasonable?

• For each set of (x, y) data, create a normal probability plot of the standardized residuals. Based on each plot, does the normality assumption about the simple linear regression model error seem reasonable?

• For each set of (x, y) data, are there any outliers?

• For each set of (x, y) data, are there any high leverage points?

• For each set of (x, y) data, are there any influential points?

  Post a summary of your group’s analysis. What important “big picture” conclusions can you draw from your analysis?

Answer 1

I used R software to solve this question.

For data (x1,y1)

R codes and output:

> x1=scan('clipboard');x1
Read 11 items
[1] 10 8 13 9 11 14 6 4 12 7 5
> y1=scan('clipboard');y1
Read 11 items
[1] 8.04 6.95 7.58 8.81 8.33 9.96 7.24 4.26 10.84 4.82 5.68
> plot(x1,y1)
> fit=lm(y1~x1)
> summary(fit)

Call:
lm(formula = y1 ~ x1)

Residuals:
Min 1Q Median 3Q Max
-1.92127 -0.45577 -0.04136 0.70941 1.83882

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 3.0001 1.1247 2.667 0.02573 *
x1 0.5001 0.1179 4.241 0.00217 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.237 on 9 degrees of freedom
Multiple R-squared: 0.6665, Adjusted R-squared: 0.6295
F-statistic: 17.99 on 1 and 9 DF, p-value: 0.00217

> res=fit$residuals
> plot(x1,res)
> std_res=rstandard(fit) # standardized residuals
> qqnorm(std_res)
> qqline(std_res)

Estimated regression equation:

Y1 = 3.0001 + 0.5001 X1

Coefficient of determination: R2 = 0.6665

Standard error of the estimate = 1.237

Scatter plot:

Residual plot:

Residual plot shows random pattern hence residuals are independent.

Normal probability plot for standardized residuals:

Points des not lie on straight line, hence normality assumption is not satisfied.