Chapter 12, Problem 1RE

Roulette Wheel A pit boss suspects that a roulette wheel is out of balance. A roulette wheel has 18 black slots, 18 red slots, and 2 green slots. The pit boss spins the wheel 500 times and records the following frequencies:

Outcome	Frequency
Black	233
Red	237
Green	30

Is the wheel out of balance? Use the α = 0.05 level of significance.

To determine

To test: Whether the wheel is balanced at $\alpha =0.05$ level of significance.

Answer to Problem 1RE

There is not enough evidence to conclude that the wheel is balanced at $\alpha =0.05$ level of significance.

Explanation of Solution

Given info:

The data represent the frequencies recorded when a pit boss spins the wheel 500 times.

Calculation:

The test hypotheses are given below:

Null hypothesis:

$H_0:$ The wheel is balanced.

Alternative hypothesis:

$H_1:$ The wheel is not balanced.

Decision rule:

If ${\chi }_0^2>{\chi }_{\alpha }^2$, then reject the null hypothesis.

The requirements that must be satisfied to perform a goodness-of-fit test are as follows:

• All the expected frequencies are greater than or equal to 1. That is, all $E_i\ge 1$ where $E_i$ represents the expected counts of the category i.
• Not more than 20% of the expected frequencies are less than 5.

The roulette wheel has 18 black slots, 18 red slots, and 2 green slots. The proportion of each slot is obtained by dividing each of the number of slots with the total number of slots.

The total number of slots is 38. The expected proportion of black is $\frac{18}{38}=\frac{9}{19}$. The expected proportion of red is $\frac{18}{38}=\frac{9}{19}$. The expected proportion of green is $\frac{2}{38}=\frac{1}{19}$.

The expected count for each category is obtained by multiplying the obtained proportion for each color with the total number of spins.

The expected count of black is as follows:

$\begin{array}{c}{E}_i=500\left(\frac{9}{19}\right)\\ =236.842\end{array}$

The expected count of red is as follows:

$\begin{array}{c}{E}_i=500\left(\frac{9}{19}\right)\\ =236.842\end{array}$

The expected count of green is as follows:

$\begin{array}{c}{E}_i=500\left(\frac{1}{19}\right)\\ =26.316\end{array}$

Since all $E_i\ge 1$, the requirements for the goodness-of-fit test are satisfied.

The test statistic for the goodness-of-fit test is given as follows:

${\chi }_0^2={\displaystyle \sum \frac{{\left(O_i-E_i\right)}^2}{E_i}}$

The test statistic can be obtained as follows:

Outcome	$O_i$	$E_i$	${\left(O_i-E_i\right)}^2$	$\frac{{\left(O_i-E_i\right)}^2}{E_i}$
Red	233	236.842	14.76096	0.062324
Black	237	236.842	0.024964	0.000105
Green	30	26.316	13.57186	0.515726

Therefore, the test statistic is as follows:

$\begin{array}{c}{\chi }_0{}^2=0.062324+0.000105+0.515726\\ =0.578156\\ =0.578\end{array}$

Thus, the value of the test statistic is 0.578.

Degrees of freedom:

Here, there are three categories.

The degree of freedom is as follows:

$\begin{array}{c}k-1=3-1\\ =2\end{array}$

Thus, the degree of freedom for the test statistic is 2.

Critical value:

The level of significance is $\alpha =0.05$.

Therefore, the critical value is obtained from Table VIII of ‘Chi-square Distribution Area to the Right of critical value’.

Procedure:

• Locate 2 in the column of degrees of freedom in Table VIII.
• Take the value corresponding to $\alpha =0.05$.

From the table, the critical value is 5.991.

Thus, the critical value using $\alpha =0.05$ is 5.991.

Conclusion:

Here, the chi-square test statistic is 0.578, and the critical value is 5.991.

The chi-square test statistic is less than the critical value.

That is ${\chi }_0^2\left(=0.578\right)<{\chi }_{\alpha }^2\left(=5.991\right)$.

Therefore, the null hypothesis is not rejected.

Thus, there is not enough evidence to conclude that the wheel is balanced at $\alpha =0.05$ level of significance.