Question solved using with R only e f g h and i please ACT GPA 21 3.897 14 3.885

Question

Question solved using with R

only e f g h and i please

ACT GPA

21 3.897

14 3.885

28 3.778

22 2.54

21 3.028

31 3.865

32 2.962

27 3.961

29 0.5

26 3.178

24 3.31

30 3.538

24 3.083

24 3.013

33 3.245

27 2.963

25 3.522

31 3.013

25 2.947

20 2.118

24 2.563

21 3.357

28 3.731

27 3.925

28 3.556

26 3.101

28 2.42

22 2.579

26 3.871

21 3.06

25 3.927

16 2.375

28 2.929

26 3.375

22 2.857

24 3.072

21 3.381

30 3.29

27 3.549

26 3.646

26 2.978

30 2.654

24 2.54

26 2.25

29 2.069

24 2.617

31 2.183

15 2

19 2.952

18 3.806

27 2.871

16 3.352

27 3.305

26 2.952

24 3.547

30 3.691

21 3.16

20 2.194

30 3.323

29 3.936

25 2.922

23 2.716

25 3.37

23 3.606

30 2.642

21 2.452

24 2.655

32 3.714

18 1.806

23 3.516

20 3.039

23 2.966

18 2.482

18 2.7

29 3.92

20 2.834

23 3.222

26 3.084

28 4

34 3.511

20 3.323

20 3.072

26 2.079

32 3.875

25 3.208

27 2.92

27 3.345

29 3.956

19 3.808

21 2.506

24 3.886

27 2.183

25 3.429

18 3.024

29 3.75

24 3.833

27 3.113

21 2.875

19 2.747

18 2.311

25 1.841

18 1.583

20 2.879

32 3.591

24 2.914

35 3.716

25 2.8

28 3.621

28 3.792

25 2.867

22 3.419

30 3.6

20 2.394

20 2.286

31 1.486

20 3.885

29 3.8

28 3.914

16 1.86

28 2.948

he director of admissions at a small college selected T 120 students at random from the new freshman class in a study to determine whether a student's grade point average (GPA) at the end of the freshman year can be predicted from their ACT test score. The data can be found at "GPA.txt" a. Fit a simple linear regression using ACT score as the explanatory variable, and GPA as the response variable. Verify all necessary model assum ptions and include all necessary p lots; b. E stimate the parameter ; c. Use a t-test to determine whether or not there a linear relationship between ACT score and GPA: d. Compute the ANOVA table corresponding to the model; e. From the table determine the mean square error f. Use the ANOVA F-test to determine whether or not there is a (MSE linear relationship between ACT score and GPA; g. How do the results in (e) compare to those in (b)? h. How do the results in (f) compare to those in (c)'? i. What proportion of the variation in GPA is explained by the re- gression model

Explanation / Answer

> ACT=c(21,
+ 14,
+ 28,
+ 22,
+ 21,
+ 31,
+ 32,
+ 27,
+ 29,
+ 26,
+ 24,
+ 30,
+ 24,
+ 24,
+ 33,
+ 27,
+ 25,
+ 31,
+ 25,
+ 20,
+ 24,
+ 21,
+ 28,
+ 27,
+ 28,
+ 26,
+ 28,
+ 22,
+ 26,
+ 21,
+ 25,
+ 16,
+ 28,
+ 26,
+ 22,
+ 24,
+ 21,
+ 30,
+ 27,
+ 26,
+ 26,
+ 30,
+ 24,
+ 26,
+ 29,
+ 24,
+ 31,
+ 15,
+ 19,
+ 18,
+ 27,
+ 16,
+ 27,
+ 26,
+ 24,
+ 30,
+ 21,
+ 20,
+ 30,
+ 29,
+ 25,
+ 23,
+ 25,
+ 23,
+ 30,
+ 21,
+ 24,
+ 32,
+ 18,
+ 23,
+ 20,
+ 23,
+ 18,
+ 18,
+ 29,
+ 20,
+ 23,
+ 26,
+ 28,
+ 34,
+ 20,
+ 20,
+ 26,
+ 32,
+ 25,
+ 27,
+ 27,
+ 29,
+ 19,
+ 21,
+ 24,
+ 27,
+ 25,
+ 18,
+ 29,
+ 24,
+ 27,
+ 21,
+ 19,
+ 18,
+ 25,
+ 18,
+ 20,
+ 32,
+ 24,
+ 35,
+ 25,
+ 28,
+ 28,
+ 25,
+ 22,
+ 30,
+ 20,
+ 20,
+ 31,
+ 20,
+ 29,
+ 28,
+ 16,
+ 28)
>
> GPA=c(3.897,
+ 3.885,
+ 3.778,
+ 2.54,
+ 3.028,
+ 3.865,
+ 2.962,
+ 3.961,
+ 0.5,
+ 3.178,
+ 3.31,
+ 3.538,
+ 3.083,
+ 3.013,
+ 3.245,
+ 2.963,
+ 3.522,
+ 3.013,
+ 2.947,
+ 2.118,
+ 2.563,
+ 3.357,
+ 3.731,
+ 3.925,
+ 3.556,
+ 3.101,
+ 2.42,
+ 2.579,
+ 3.871,
+ 3.06,
+ 3.927,
+ 2.375,
+ 2.929,
+ 3.375,
+ 2.857,
+ 3.072,
+ 3.381,
+ 3.29,
+ 3.549,
+ 3.646,
+ 2.978,
+ 2.654,
+ 2.54,
+ 2.25,
+ 2.069,
+ 2.617,
+ 2.183,
+ 2,
+ 2.952,
+ 3.806,
+ 2.871,
+ 3.352,
+ 3.305,
+ 2.952,
+ 3.547,
+ 3.691,
+ 3.16,
+ 2.194,
+ 3.323,
+ 3.936,
+ 2.922,
+ 2.716,
+ 3.37,
+ 3.606,
+ 2.642,
+ 2.452,
+ 2.655,
+ 3.714,
+ 1.806,
+ 3.516,
+ 3.039,
+ 2.966,
+ 2.482,
+ 2.7,
+ 3.92,
+ 2.834,
+ 3.222,
+ 3.084,
+ 4,
+ 3.511,
+ 3.323,
+ 3.072,
+ 2.079,
+ 3.875,
+ 3.208,
+ 2.92,
+ 3.345,
+ 3.956,
+ 3.808,
+ 2.506,
+ 3.886,
+ 2.183,
+ 3.429,
+ 3.024,
+ 3.75,
+ 3.833,
+ 3.113,
+ 2.875,
+ 2.747,
+ 2.311,
+ 1.841,
+ 1.583,
+ 2.879,
+ 3.591,
+ 2.914,
+ 3.716,
+ 2.8,
+ 3.621,
+ 3.792,
+ 2.867,
+ 3.419,
+ 3.6,
+ 2.394,
+ 2.286,
+ 1.486,
+ 3.885,
+ 3.8,
+ 3.914,
+ 1.86,
+ 2.948)

### a)

> LM=lm(GPA~ACT)
> summary(LM)

Call:
lm(formula = GPA ~ ACT)

Residuals:
Min 1Q Median 3Q Max
-2.74004 -0.33827 0.04062 0.44064 1.22737

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 2.11405 0.32089 6.588 1.3e-09 ***
ACT 0.03883 0.01277 3.040 0.00292 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.6231 on 118 degrees of freedom
Multiple R-squared: 0.07262, Adjusted R-squared: 0.06476
F-statistic: 9.24 on 1 and 118 DF, p-value: 0.002917

## b) estimate of paramater sigma is equal to Residual standard error: 0.6231.

### d) ANOVA table

> ANOVA_model=aov(GPA~ACT)
> summary(ANOVA_model)
Df Sum Sq Mean Sq F value Pr(>F)
ACT 1 3.59 3.588 9.24 0.00292 **
Residuals 118 45.82 0.388
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

## e) MSE is 0.388.

## f) The ANOVA table use only one regressor and the estimated p-value for F-test is 0.00292. Hence, ACT has the significant effect on GPA at 0.05 level of significance.

## g) The estimate of sigma from the ANOVA table is the square root of MSE=sqrt(0.388)=0.6229. It is same with the estimate of parameter sigma from linear model up to three after the decimal. In general, it is same.

## h) Result of f and c is compared on the significance of ACT on GPA.

## i) The proportion of the variation in GPA is explained by the regression model is 3.59/(3.59+45.82)=0.0727 that is 7.27%.

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