Chapter 12: Q. 12.4 (page 484)

The $t$ -table has entries for areas of $0.10, 0.05, 0.025, 0.01$ and $0.005$ . In contrast, the $χ^{2}$ -table has entries for those areas and for $0.995, 0.99, 0.975, 0.95$ and $0.90 .$ Explain why the $t$ -values corresponding to these additional areas can be obtained from the existing $t$ -table but must be provided explicitly in the $χ^{2}$ -table.

Short Answer

Expert verified

The $t$ -values corresponding to the additional areas can be obtained from the existing $t$ -table because $t$ -distributions are symmetric about $0$ . Therefore, $χ^{2}$ distribution of these values should be provided explicitly in the $χ^{2}$ -table.

Step by step solution

Step 1. Given information

The $t -$ table has entries for areas of $0.10, 0.05, 0.025, 0.01, 0.005$ . The $χ^{2}$ -table has entries for those areas of $0.995, 0.99, 0.975, 0.95, 0.90$ .

Step 2. Explanation for the additional areas which is obtained from the existing t-table:

If $t$ -distributions are symmetric about $0$ ,then the $t$ - table values corresponding to the additional areas can be obtained from the existing $t$ -table.

Similarly, if $0 < x < 1$ and $t_{x} = - t_{x}$ , then the values of $t_{0.995}, t_{0.99}, t_{0.975}, t_{0.95}, t_{0.90}$ can be obtained from $t_{0.005}, t_{0.01}, t_{0.025}, t_{0.05}, t_{0.10}$ . But $χ^{2}$ -curve is not symmetric. Hence, we cannot obtain the $χ^{2}$ -values of $χ_{0.995}^{2}$ from $χ_{0.005}^{2}$ etc.