Question

Loss,x1,x2
372,45,162
206,55,233
175,61,232
154,66,231
136,71,231
112,71,237
55,81,224
45,86,219
221,53,203
166,60,189
164,64,210
113,68,210
82,79,196
32,81,180
228,56,200
196,68,173
128,75,188
97,83,161
64,88,119
249,59,161
219,71,151
186,80,165
155,82,151
114,89,128
341,51,161
340,59,146
283,65,148
267,74,144
215,81,134
148,86,127

I am asking the R studio Code, pleas leave your code and comment here, thanks a lot!

Q1.The data file abrasion contains the results from a small scale study (Davies, O.L. and Goldsmith, P.L. Statistical methods in Research and Production, 1972),  of the relation between rubber's resistance to abrasion (Y) and rubber hardness (X1) and rubber tensile strength (X2).

The data set abrasion is in Course Content -> Data Sets AL -> Ch04

• Import the data set into R.
• Obtain the scatter plot matrix and the correlation matrix.You can do this together using the commands in the file pairs.r

Upload the results here (one file in .png or .pdf formats) - Remember to include a title.

Q2.

Run the regression model. Obtain the estimates of the coefficients (round answer to 4 decimal places, it the answer is 7.5e-08 enter 0)

Coefficient	estimate	se	p-value
b0	__	__	__
b1	__	__	__
b2	__	__	__

Which variable is significant? __ (enter exactly on of the three options: x1, x2 or both)

Q3.

Enter here the coefficient of determination (adjusted R-squared). Round your answer to 4 decimal places.

Q4.

Enter here the estimate for σ, that is s or the residual standard error. Round your answer to 2 decimal places.

Q5.

Use your model to obtain the mean abrasion loss for rubber with hardness 71 an tensile strength 201. Round your answer to 2 decimal places.

Q6.

Use your model to obtain a 98% confidence interval for the mean abrasion loss for rubber with hardness 71 an tensile strength 201.

Enter here the Lower Bound for the confidence interval. Round your answer to 2 decimal places.

Q7.

After the scatter plots, the correlation between the variables, the summary of the model, R-squared and s, and the F-test, briefly comment on the adequacy of the model fit.

Answer 1

code

datar<-"Loss,x1,x2
372,45,162
206,55,233
175,61,232
154,66,231
136,71,231
112,71,237
55,81,224
45,86,219
221,53,203
166,60,189
164,64,210
113,68,210
82,79,196
32,81,180
228,56,200
196,68,173
128,75,188
97,83,161
64,88,119
249,59,161
219,71,151
186,80,165
155,82,151
114,89,128
341,51,161
340,59,146
283,65,148
267,74,144
215,81,134
148,86,127"

data <-read.table(textConnection(object=datar),
                  header=TRUE,
                  sep=",",
                  stringsAsFactors = FALSE)

                  stringsAsFactors = FALSE)
pairs(data)
cor(data)

> cor(data)
           Loss         x1         x2
Loss  1.0000000 -0.7377107 -0.2983939
x1   -0.7377107  1.0000000 -0.2992345
x2   -0.2983939 -0.2992345  1.0000000

2)

model = lm (Loss ~ . , data = data )
summary(model)

Call:
lm(formula = Loss ~ ., data = data)

Residuals:
    Min      1Q  Median      3Q     Max 
-79.385 -14.608   3.816  19.755  65.981 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept) 885.1611    61.7516  14.334 3.84e-14 ***
x1           -6.5708     0.5832 -11.267 1.03e-11 ***
x2           -1.3743     0.1943  -7.073 1.32e-07 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 36.49 on 27 degrees of freedom
Multiple R-squared:  0.8402,    Adjusted R-squared:  0.8284 
F-statistic:    71 on 2 and 27 DF,  p-value: 1.767e-11

both are significant as their p-value < alpha

Q3

adjusted R^2 = 0.8284

Q4

Residual standard error: 36.49

Please rate

Please post rest parts again