Exhibit 134 Below you are given a partial Excel output based
Exhibit 13-4
 Below you are given a partial Excel output based on a sample of 16 observations.
ANOVA
df
SS
MS
F
Regression
4,853
2,426.5
Residual
485.3
Coefficients
Standard Error
Intercept
12.924
4.425
x1
-3.682
2.630
x2
45.216
12.560
 
 Refer to Exhibit 13-4. The t value obtained from the table which is used to test an individual parameter at the 1% level is
2.65
2.921
2.977
3.012
| ANOVA | ||||
| df | SS | MS | F | |
| Regression | 4,853 | 2,426.5 | ||
| Residual | 485.3 | |||
| Coefficients | Standard Error | |||
| Intercept | 12.924 | 4.425 | ||
| x1 | -3.682 | 2.630 | ||
| x2 | 45.216 | 12.560 | 
Solution
here we find that value of t distribution under which 99% of the all values lie such that t ditribution has (n-1) df.
So here t distrubution with 15 df as n=16
Now we get using minitab or r or any othe statistical tool
t value= a)2.65


