408 CHAPTER 6 Inferential Statistics
(The above calculation can be done on your TI-83, using 1−χ^2 cdf(0, 4 ,1).
The third argument, 1, is just the number of degrees of freedom.)
Since the P-value is .0455, one sees that there is a fair amount of
significance that can be attached to this outcome. We would—at the
5% level of significance—rejectH 0 and say that the coin is not fair.
Example 2. Let’s return to the case of the allegedly fair die and the
results of the 200 tosses. Theχ^2 test results in the value:
χ^2 =∑(ni−E(ni))^2
E(ni)=(33−^2006 )^2
200 / 6
+
(40−^2006 )^2
200 / 6
+
(39−^2006 )^2
200 / 6
+
(28−^2006 )^2
200 / 6
+
(36−^2006 )^2
200 / 6
+
(24−^2006 )^2
200 / 6
≈ 5. 98.
TheP-value corresponding to this measured value isP(χ^25 ≥ 5 .98)≈
0. 308 .(We sometimes writeχ^2 nfor the χ^2 random variable with nde-
grees of freedom.) This is really not small enough (not “significant
enough) for us to reject the null hypothesis of the fairness of the die.