Question

Degrees of Freedom

For Example 1 on page 431, we used df smaller of $n_1-1$and $n_2-1$, we got , and the corresponding critical values are$t=\pm 2.201.$ If we calculate df using Formula 9-1, we get$df=19.063$, and the corresponding critical values are $t=\pm 2.201$. How is using the critical values of more “conservative” than using the critical values of $\pm 2.093$.

Short answer

The formula used is simpler and less accurate as compared to the formula 9-1.

Step-by-step solution

Given information

The formula used for degree of freedom of mean is:

$df=\min \left(n_1-1,n_2-1\right)$ , the results are:

$\text{df}=11$, critical values$t=\pm 2.201$

The formula used for degree of freedom for comparison of mean is:

$df=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_1-1}}$, the results are:

When $\text{df}=19.63$, critical values$t=\pm 2.093$

Explanation of the statement

The formula used to obtain the critical value 2.093 is obtained accurately:

The simpler formula $df=\min \left(n_1-1,n_2-1\right)$can be used more easily and flexibly but may not be that accurate to give results.

Hence, the critical value obtained using $df=\min \left(n_1-1,n_2-1\right)$is more conservative than the ones computed with another formula.