Chapter 15, Problem 86CE

To determine

Test hypothesis.

Explanation of Solution

The given hypothesis and alternative hypothesis are as follows:

$\begin{array}{c}{\text{H}}_{\text{0}}:{\text{P}}_{\text{1}}=\frac{1}{3},{\text{P}}_{\text{2}}=\frac{1}{3},{\text{P}}_{\text{3}}=\frac{1}{3}\\ {\text{H}}_{\text{1}}:{\text{ At least one P}}_{\text{i}}\ne \text{ its specified value}\text{.}\end{array}$

To test the hypothesis, construct a table as follows:

Cell	Frequency (F)	Hypothesized value(Pi)	Expected value (E)	F-E	$\frac{{\left(\text{F}-\text{E}\right)}^2}{\text{E}}$
1	14	0.333	10	4	1.60
2	10	0.333	10	0	0.00
3	6	0.333	10	-4	1.60

$\begin{array}{c}{\text{χ}}^{\text{2}}={\displaystyle \sum \frac{{\left(\text{F}-\text{E}\right)}^2}{\text{E}}}\\ =3.20\end{array}$

Thus, ${\text{χ}}^{\text{2}}=3.20$ and the p-value will be 0.2019. There is no enough evidence to infer that the game is unfair.