For an $F$-curve with $\mathrm{df}=(6,10)$, find

a. $F_{0.05}$-

b. $F_{0.01}$.

c. $F_{002s}$-

Given in the question that, An $F-$curve has $df=(12,5)$. We need to flnd$F_{0.05}$.

Given: An F-curve has$df=(6,10)$

The area to the right of $0.05$F-value

Calculation:

The numerator's degrees of freedom are $=6$.

The denominator's degrees of freedom are $=10$.

$\alpha =0.05$significance level

The F-value is calculated using Table VIII. Scroll down to'6'in the dfd column. After that, concentrate on the $\alpha$0.05 row. Cross that row to the '10' column, and the point earned will be $3.22$.

Therefore,$F_{0.05}=3.22$

Given in the question that, An $F$ - curve has $df=(12,5)$. We need to flnd $F_{0.01}$.

Given: $df=(6,10)$on an F-curve

The area immediately to the right of the $F$-value $=0.01$.

The numerator's degrees of freedom are $=6$.

The denominator's degrees of freedom are $=10$.

$\alpha =0.01$significance level

The F-value is calculated using Table VIII. Scroll down to'6'in the dfd column. After that, concentrate on the $\alpha$row labelled 0.01. Cross that row to the '10' column, and the point earned will be $5.39$

Therefore,

$F_{0.01}=5.39$

Given in the question that, An $F$ - curve has $df=(12,5)$. We need to find $F_{0.025}$.

Given: $df=(6,10)$on an F-curve

The area to the right of $=0.025$F-value

The numerator's degrees of freedom are $=6$.

The denominator's degrees of freedom are $=10$.

$\alpha =0.025$significance level

The F-value is calculated using Table VIII. Scroll down to'6'in the dfd column. Next, concentrate on the $0.025$column for $\alpha$. The point acquired will be $4.07$if you go across that row to the column labeled'10 '.

Therefore,

$F_{0.025}=4.07$