Given:
The data table is given.
The equation of trend line is:
[tex]y=2x+15[/tex]
To find:
Values for residual table.
Solution:
Formula for residual:
[tex]Residual=\text{Observed value}-\text{Predicted value}[/tex]
The equation of trend line is:
[tex]y=2x+15[/tex] ...(i)
Substituting [tex]x=5[/tex] in (i), we get
[tex]y=2(5)+15[/tex]
[tex]y=10+15[/tex]
[tex]y=25[/tex]
The value of y from the table at [tex]x=5[/tex] is 25. So,
[tex]Residual=25-25[/tex]
[tex]Residual=0[/tex]
Similarly,
Substituting [tex]x=6[/tex] in (i), we get
[tex]y=2(6)+15[/tex]
[tex]y=12+15[/tex]
[tex]y=27[/tex]
The value of y from the table at [tex]x=6[/tex] is 28. So,
[tex]Residual=28-27[/tex]
[tex]Residual=1[/tex]
Substituting [tex]x=7[/tex] in (i), we get
[tex]y=2(7)+15[/tex]
[tex]y=14+15[/tex]
[tex]y=29[/tex]
The value of y from the table at [tex]x=7[/tex] is 29. So,
[tex]Residual=29-29[/tex]
[tex]Residual=0[/tex]
Substituting [tex]x=8[/tex] in (i), we get
[tex]y=2(8)+15[/tex]
[tex]y=16+15[/tex]
[tex]y=31[/tex]
The value of y from the table at [tex]x=8[/tex] is 30. So,
[tex]Residual=30-31[/tex]
[tex]Residual=-1[/tex]
Substituting [tex]x=9[/tex] in (i), we get
[tex]y=2(9)+15[/tex]
[tex]y=18+15[/tex]
[tex]y=33[/tex]
The value of y from the table at [tex]x=9[/tex] is 32. So,
[tex]Residual=32-33[/tex]
[tex]Residual=-1[/tex]
Therefore, the complete residual table is:
x : 5 6 7 8 9
Residuals : 0 1 0 -1 -1
