Finding Lower Critical F Values For hypothesis tests that are two-tailed, the methods of Part 1 require that we need to find only the upper critical value. Let’s denote the upper critical value by \({F_R}\), where the subscript indicates the critical value for the right tail. The lower critical value \({F_L}\)(for the left tail) can be found as follows: (1) Interchange the degrees of freedom used for finding \({F_R}\), then (2) using the degrees of freedom found in Step 1, find the F value from Table A-5; (3) take the reciprocal of the F value found in Step 2, and the result is \({F_L}\). Find the critical values \({F_L}\)and \({F_R}\) for Exercise 16 “Blanking Out on Tests.”

Two samples are considered showing the anxiety scores dueto the arrangement of questions on a test paper. One sample represents anxiety scores due to the arrangement of questions from easy to difficult. Another sample representsanxiety scores due to the arrangement of questions from difficult to easy.

The lower and the upper critical values need to be determined.

The numerator degrees of freedom is computed below:

\(\begin{array}{c}{n_1} - 1 = 25 - 1\\ = 24\end{array}\)

The denominator degrees of freedom is computed below:

\(\begin{array}{c}{n_2} - 1 = 16 - 1\\ = 15\end{array}\)

Thus, the degrees of freedom is equal to (24,15)

The upper critical value can be obtained using the F distribution table with numerator degrees of freedom equal to 24 and denominator degrees of freedom equal to 15, and the significance level\(\alpha = 0.05\)for a right-tailed test.

Thus, the value of \({F_R}\) is equal to 2.2878.

The degrees of freedom is obtained by interchanging the numerator degrees of freedom and the denominator degrees of freedom used above.

Referring to the F distribution table, the lower critical value\(\left( {{F_L}} \right)\)can be obtained by taking the reciprocal of the upper critical value with the numerator degrees of freedom equal to 15, the denominator degrees of freedom equal to 24and the significance level\(\alpha = 0.05\)for a right-tailed test.

Thus,\({F_L}\)is equal to:

\(\begin{array}{c}{F_L} = \frac{1}{{{F_{R\left( {15,24;0.05} \right)}}}}\\ = \frac{1}{{2.1077}}\\ = 0.4745\end{array}\)

Therefore, \({F_L}\) is equal to 0.4745.