Consider a simple linear regression model relating a high school student\'s ACT
ID: 3353114 • Letter: C
Question
Consider a simple linear regression model relating a high school student's ACT scores to his or her high school grade point average, GPA, y = BX, where,-ACT and X-GPA (Assignment using calculator) 1) Use the data on 8 students to obtain the OLS estimates for and A, and the standard errors for these estimates. 2) Comment on the direction of the relationship between GPA and ACT score. Perform a test of the null hypothesis that they are not related and draw your conclusion 3) What is the size of the R- for this regression equation? What does it mean?. (Compute using calculator 4) Please construct the 95% confidence interval for the "forecast of an individual's outcome" on the ACT using the above regression model for Xo2 or GPA-3.0 and at the sample mean of GPA The data for the problem: Student GPA 2.8 3.4 3.0 3.5 3.6 3.0 2.7 3.7 ACT 21 24 26 27 29 25 25 30 2 4 6 8Explanation / Answer
GPA = c(2.8,3.4,3,3.5,3.6,3,2.7,3.7)
> ACT = c(21,24,26,27,29,25,25,30)
> model = lm (ACT ~ GPA)
> summary(model)
Call:
lm(formula = ACT ~ GPA)
Residuals:
Min 1Q Median 3Q Max
-2.9344 -1.0106 0.6306 1.3369 2.0207
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 7.724 6.378 1.211 0.2714
GPA 5.650 1.973 2.863 0.0287 *
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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.001 on 6 degrees of freedom
Multiple R-squared: 0.5774, Adjusted R-squared: 0.507
F-statistic: 8.199 on 1 and 6 DF, p-value: 0.02868
1) b1^ = 7.724 , b2^ = 5.65
2) b2^ > 0 , there is positive relation between GPA and ACT.
since p-value of b2^ = 0.0287 < 0.05
we reject the null and conclude that the variable is significant
3)
R^2 = 0.5774
it means that 57.74 % of variability in ACT is explained using this model
4)
predict(model, data.frame(GPA=3), interval= "confidence")
fit lwr upr
1 24.67436 22.66173 26.68699
hence 95 % confidence level is (22.66173,26.68699)
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