Finding Critical t Values When finding critical values, we often need significance levels other than those available in Table A-3. Some computer programs approximate critical t values by calculating\(t = \sqrt {df \times \left( {{e^{{A^2}/df}} - 1} \right)} \)where df = n-1, e = 2.718,\(A = z\left( {8 \times df + 3} \right)/\left( {8 \times df + 1} \right)\), and z is the critical z score. Use this approximation to find the critical t score for Exercise 12 “Tornadoes,” using a significance level of 0.05. Compare the results to the critical t score of 1.648 found from technology. Does this approximation appear to work reasonably well?

A sample of 500 tornadoes is considered. It is claimed that the mean tornado length is greater than 2.5 miles.

Since\(\sigma \)is unknown, the test statistic used to test the given claim is the t-score which follows Student’s t distribution with\(\left( {n - 1} \right)\)degrees of freedom.

Let n be the sample size.

The formula to compute the value of the critical score is given below:

\(t = \sqrt {df \times \left( {{e^{{A^2}/df}} - 1} \right)} \)

Where,

dfis the degree of freedom which is equal to \(n - 1\)

e is equal to 2.718

The formula to compute A is given below:

\(A = \frac{{z\left( {8\left( {df} \right) + 3} \right)}}{{\left( {8\left( {df} \right) + 1} \right)}}\)

Here, n=500.

Thus, the value of the degrees of freedom is computed below:

\(\begin{array}{c}df = n - 1\\ = 500 - 1\\ = 499\end{array}\)

Let the level of significance be equal to\(\alpha = 0.05\)

Since the claim involves a greater than sign, the test would be right-tailed.

The value of the z-score for;\(\alpha = 0.05\)a right-tailed test is equal to 1.645.

Thus, the value of A is computed below:

\(\begin{array}{c}A = \frac{{z\left( {8\left( {df} \right) + 3} \right)}}{{\left( {8\left( {df} \right) + 1} \right)}}\\ = \frac{{1.645\left( {8\left( {499} \right) + 3} \right)}}{{\left( {8\left( {499} \right) + 1} \right)}}\\ = 1.646\end{array}\)

The critical score using the formula stated above is computed as follows:

\(\begin{array}{c}t = \sqrt {df \times \left( {{e^{{A^2}/df}} - 1} \right)} \\ = \sqrt {df \times \left( {{e^{\frac{{{A^2}}}{{df}}}} - 1} \right)} \\ = \sqrt {499 \times \left( {{e^{\frac{{{{\left( {1.646} \right)}^2}}}{{499}}}} - 1} \right)} \\ = 1.648\end{array}\)

Thus, the value of the critical score is equal to 1.648.

The value of the critical score at\(\alpha = 0.05\)using the formula stated in the question is equal to 1.648.

The value of the critical score at\(\alpha = 0.05\)using the t-distribution table/technology is equal to 1.648.

Since the two critical values using the two methods are approximately equal, the given formula to approximate the critical score works considerably well.