The following data are for intelligence-test (IT) scores, reading rates (RR), an

Question

The following data are for intelligence-test (IT) scores, reading rates (RR), and grade-point averages (GPA) of 8 at-risk students. IT 184 202 202 167 202 210 199 181 RR 34 30 42 34 22 45 22 25 GPA||2.4 | 1.812.012.3] 1.813.1 || 1.7[20 Part a: Calculate the line of best fit that predicts the GPA on the basis of RR scores. Part b: Calculate the line of best fit that predicts the GPA on the basis of IT scores. Part c: Which of the two lines calculated in parts a and b best fits the data? Justify your answer.

Explanation / Answer

the complete R snippet is as follows

IT <- c(184,202,202,167,202,210,199,181)
RR <- c(34,30,42,34,22,45,22,25)
GPA <- c(2.4,1.8,2.9,2.3,1.8,3.1,1.7,2)

#3 part a

fit <- lm(GPA~IT)
summary(fit)

## part B

fit1 <- lm(GPA~RR)
summary(fit1)

The results are

summary(fit)

Call:
lm(formula = GPA ~ IT)

Residuals:
Min 1Q Median 3Q Max
-0.58545 -0.50435 0.01853 0.31107 0.74524

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.031423 2.833475 0.364 0.728
IT 0.006302 0.014617 0.431 0.681

Residual standard error: 0.56 on 6 degrees of freedom
Multiple R-squared: 0.03005, Adjusted R-squared: -0.1316
F-statistic: 0.1859 on 1 and 6 DF, p-value: 0.6814

> summary(fit1)

Call:
lm(formula = GPA ~ RR)

Residuals:
Min 1Q Median 3Q Max
-0.34887 -0.00992 0.03881 0.09157 0.14008

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.41516 0.24089 1.723 0.135583
RR 0.05779 0.00735 7.863 0.000224 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.1691 on 6 degrees of freedom
Multiple R-squared: 0.9115, Adjusted R-squared: 0.8968
F-statistic: 61.83 on 1 and 6 DF, p-value: 0.0002239

we see that the r square value is highest in case of second model , which is with RR

the r2 value is high for model 2

model 2 = 0.9115

model 1 = 0.03

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