In: Statistics and Probability
A statistics professor would like to build a model relating student scores on the first test to the scores on the second test. The test scores from a random sample of 2121 students who have previously taken the course are given in the table.
| Student | First Test Grade | Second Test Grade |
|---|---|---|
| 1 | 6767 | 7070 |
| 2 | 9797 | 8181 |
| 3 | 4242 | 6161 |
| 4 | 6969 | 7272 |
| 5 | 5656 | 6363 |
| 6 | 6666 | 7272 |
| 7 | 9999 | 9090 |
| 8 | 7979 | 7373 |
| 9 | 8888 | 7878 |
| 10 | 6363 | 6767 |
| 11 | 6161 | 6565 |
| 12 | 8383 | 7777 |
| 13 | 8989 | 8585 |
| 14 | 4343 | 5757 |
| 15 | 8585 | 7676 |
| 16 | 4040 | 6464 |
| 17 | 4747 | 6666 |
| 18 | 9393 | 7979 |
| 19 | 9393 | 8080 |
| 20 | 5656 | 6767 |
| 21 | 5050 | 6767 |
Copy Data
Step 1 of 2 :
Using statistical software, estimate the parameters of the model
Second Test Grade=β0+β1(First Test Grade)+εiSecond Test
Grade=β0+β1(First Test Grade)+εi.
Enter a negative estimate as a negative number in the regression
model. Round your answers to 4 decimal places, if necessary.
i am using minitab to solve the problem.
steps for getting regression equation:-
stat
regression
regression
fit regression model
in responses select
final second Test Grade,in continuous predictors select First Test
Grade
ok
ok.
to round off any value to needed decimal places :-
right click on the value
decimal places
select the needed place(here, we
will select 4 )
the needed minitab output be:-
Coefficients
| Term | Coef | SE Coef | T-Value | P-Value | VIF |
| Constant | 43.8107 | 2.52 | 17.40 | 0.000 | |
| First test grade | 0.4024 | 0.0348 | 11.57 | 0.000 | 1.00 |
*** SOLUTION ***
the regression equation be:-

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