Question: From a random sample of 6 college students, a school counselor obtained the students' high school GPAs(x) and their college GPAs(y) (show in the table
From a random sample of 6 college students, a school counselor obtained the students' high school GPAs(x) and their college GPAs(y) (show in the table below). She proposed to use the simple linear regression to develop a model that will allow her to predict a student's college GPA from his or her own high school GPA.
Obsx y
1 2.78 2.95
2 3.82 3.64
3 3.16 3.55
4 3.53 3.89
5 2.96 3.06
6 3.37 3.31
You can copy the following R code to fit the simple linear regression model to the data in R.
y=c(2.95,3.64,3.55,3.89,3.06,3.31) x=c(2.78,3.82,3.16,3.53,2.96,3.37) result=lm(y~x)
summary(result)
anova(result)
Based on the output in R, answer the following questions. Please provide R output in your solutions.
(a) Write out the estimated regression line1.
(b) Find a 95% confidence interval for .(Note: t critical value can be found by R function
qt(.975,dfe) )
(c) Find SSR, SSE, and TSS.
(d) Based on the ANOVA F test, is the simple linear regression model useful to fit the data at the 0.05 significance level? Justify your answer clearly.
(e) Find a 95% confidence interval for the mean college GPA for all students who have 3.5 high school GPA. (Note: x and Sxx can be calculated by R functions mean(x)=3.27; var(x)*(6- 1)=0.7284, respectively.
(e) Find a 95% prediction interval for a student's college GPA who has 3.5 high school GPA. (You will notice the prediction interval is wider than the confidence interval)
The R-output
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