Question

Degrees of Freedom

For Example 1 on page 431, we used df\( = \)smaller of\({n_1} - 1\)and\({n_2} - 1\), we got\(df = 11\), 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\( \pm 2.093\). How is using the critical values of\(t = \pm 2.201\)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:

\({\rm{df}} = 11\), 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\({\rm{df}} = 19.63\), 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.