Degrees of FreedomIn Exercise 20 “Blanking Out on Tests,” using the “smaller of\({n_1} - 1\) and \({n_2} - 1\)” for the number of degrees of freedom results in df = 15. Find the number of degrees of freedom using Formula 9-1. In general, how are hypothesis tests and confidence intervals affected by using Formula 9-1 instead of the “smaller of \({n_1} - 1\)and \({n_2} - 1\)”?

Data are given on the anxiety scores for the two different arrangements of questions on a test.

The following formula is used to compute the value of the degrees of freedom:

\(df = \frac{{{{\left( {A + B} \right)}^2}}}{{\frac{{{A^2}}}{{{n_1} - 1}} + \frac{{{B^2}}}{{{n_2} - 1}}}}\)

Here,

\(A = \frac{{s_1^2}}{{{n_1}}}\)

\(B = \frac{{s_2^2}}{{{n_2}}}\)

The sample size\(\left( {{n_1}} \right)\)is equal to 25.

The sample size\(\left( {{n_2}} \right)\)is equal to 16.

The sample mean of the first sample\(\left( {{{\bar x}_1}} \right)\)is equal to:

\(\begin{array}{c}{{\bar x}_1} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_i}} }}{{{n_1}}}\\ = \frac{{26.64 + 39.29 + ..... + 30.72}}{{25}}\\ = 27.12\end{array}\)

The sample mean of the second sample\(\left( {{{\bar x}_2}} \right)\)is equal to:

\(\begin{array}{c}{{\bar x}_2} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_i}} }}{{{n_2}}}\\ = \frac{{33.62 + 34.02 + ..... + 32.54}}{{16}}\\ = 31.73\end{array}\)

The sample variance for the first sample \(\left( {s_1^2} \right)\) is equal to:

\(\begin{array}{c}{s_1}^2 = \frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}\\ = \frac{{{{\left( {24.64 - 27.12} \right)}^2} + {{\left( {39.29 - 27.12} \right)}^2} + ....... + {{\left( {30.72 - 27.11} \right)}^2}}}{{25 - 1}}\\ = 47.06\end{array}\)

The sample variance for the second sample \(\left( {s_2^2} \right)\)

\(\begin{array}{c}s_2^2 = \frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ = \frac{{{{\left( {33.62 - 31.73} \right)}^2} + {{\left( {34.02 - 31.73} \right)}^2} + ........ + {{\left( {32.54 - 31.73} \right)}^2}}}{{16 - 1}}\\ = 18.15\end{array}\)

Thevalue of A is equal to:

\(\begin{array}{c}A = \frac{{{s_1}^2}}{{{n_1}}}\\ = \frac{{47.06}}{{25}}\\ = 1.88\end{array}\)

Thevalue of B is equal to:

\(\begin{array}{c}B = \frac{{s_2^2}}{{{n_2}}}\\ = \frac{{18.15}}{{16}}\\ = 1.13\end{array}\)

The value of the degrees of freedom is equal to:

\(\begin{array}{c}df = \frac{{{{\left( {A + B} \right)}^2}}}{{\frac{{{A^2}}}{{{n_1} - 1}} + \frac{{{B^2}}}{{{n_2} - 1}}}}\\ = \frac{{{{\left( {1.88 + 1.13} \right)}^2}}}{{\frac{{{{\left( {1.88} \right)}^2}}}{{25 - 1}} + \frac{{{{\left( {1.13} \right)}^2}}}{{16 - 1}}}}\\ = 38.99\\ \approx 39\end{array}\)

Thus, the value of the degrees of freedom is equal to 39.

The result of the test using the method of “smaller of\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\)” to compute the degrees of freedom is approximately the same as the result obtained when the formula is used to compute the degrees of freedom.

In general, the conclusion of the hypothesis test is rarely affected by the choice of the method of computing the degrees of freedom.