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 GPA| 2.4 | 1.8[2.912.3 1.813.1 | 1.7120 42 34 22 45 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

sOLUTIONa:

Here y-GPA

x-RR

Code in R:
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)
GPA_RR <- lm(GPA~RR)
summary(GPA_RR)

output:

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

line of best fit is

GPA=0.41516+0.05779(RR)

Solutionb:

here yis GPA

x is IT

Rcode:

GPA <- c(2.4,1.8,2.9,2.3,1.8,3.1,1.7,2)
IT <- c(184,202,202,167,202,210,199,181)
GPA_IT <- lm(GPA~IT)
summary(GPA_IT)

output:

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

the line of best fit is

GPA=1.031423+0.006302(IT)

Solutionc:

For 1 st model

R sq=0.9115

91.15% variation in GPA is explained by RR

Good model.

and also

F-statistic: 61.83 on 1 and 6 DF, p-value: 0.0002239

p<0.05

Model is significant.

For 2nd model

Multiple R-squared: 0.03005

0.03005*100=3.005% variation in GPA is explained by IT.

not a good model.

F-statistic: 0.1859 on 1 and 6 DF, p-value: 0.6814

p>0.05

Model is not significant.

GPA on RR best fits the data.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.