Question: 4 . Confidence intervals for predictions - an application Suppose a data set of 4 , 0 0 0 observations ( nn = 4 ,

4. Confidence intervals for predictions - an application
Suppose a data set of 4,000 observations (nn=4,000) was analyzed using OLS to examine the factors influencing students' college grade point averages. The regression results are as follows, with standard errors in parentheses:
colgpa=1.550(0.09)+0.0010(0.00006)sat0.03(0.0007)hsperc0.05(0.01)hsize+0.007(0.003)hsize2colgpa^=1.550(0.09)+0.0010(0.00006)sat0.03(0.0007)hsperc0.05(0.01)hsize+0.007(0.003)hsize2
where
colgpacolgpa= GPA in collegesatsat= score on the SAT examhsperchsperc= percentile in high school graduating classhsizehsize= size of graduating class, in hundreds
nn=4,000R2R2=0.254R2R2=0.253^=0.45
The predicted college GPA for someone who scored 1,120 on the SAT and graduated in the 27th percentile of their 700-person high school graduating class, is.
Suppose you would like to construct a 95% confidence interval around the expectedcollege GPA for someone with sat=1,120sat=1,120, hsperc=27hsperc=27, and hsize=7hsize=7.
colgpa=1.853(0.01)+0.0010(0.00006)sat00.03(0.0007)hsperc00.05(0.01)hsize0+0.007(0.003)hsize02colgpa^=1.853(0.01)+0.0010(0.00006)sat00.03(0.0007)hsperc00.05(0.01)hsize0+0.007(0.003)hsize02
where
colgpacolgpa= GPA in collegesat0sat0= sat1,120sat1,120hsperc0hsperc0= hsperc27hsperc27hsize0hsize0= hsize7hsize7
nn=4,000R2R2=0.254R2R2=0.253^=0.45
Hint: The 95% critical value of the tdistribution is 1.96.
The confidence interval around this expected value of college GPA would be1.961.96.
Now suppose you would like to construct a 95% confidence interval around the college GPA for someone with sat=1,120sat=1,120, hsperc=27hsperc=27, and hsize=7hsize=7, and not just the expected value of college GPA for these values of the explanatory variables.
This confidence interval around this value of college GPA would be1.961.96.

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