Refer to Exercises 2 and 4.

(a) Confirm that the expected counts are large enough to use a chi-square distribution. Which distribution (specify the degrees of freedom) should you use?

(b) Sketch a graph like Figure 11.4 (page 683) that shows the P-value.

(c) Use Table C to find the P-value. Then use your calculator’s C2cdf command.

(d) What conclusion would you draw about whether the roulette wheel is operating correctly? Justify your answer

Given in the question that,

The total number of balls chosen is $18,18,$and $2.$

The whole cost is $200.$

$4.0389$is the test statistic.

To calculate the degree of freedom, use the following formula:

Freedom of degree = number of categories $-1$

The predicted counts can be calculated as follows:

role="math" localid="1652867464684" $$\begin{gathered} E\left(\text{Red}\right)=200\times \frac{18}{38} \\ =94.74 \end{gathered}$$

$$\begin{gathered} E\left(\text{Black}\right)=200\times \frac{18}{38} \\ =94.74 \end{gathered}$$

$$\begin{gathered} E\left(\text{Green}\right)=200\times \frac{2}{38} \\ =10.53 \end{gathered}$$

If ALL predicted counts are at least $5,$, the expected counts are large enough to utilize a chi-square distribution.

The number of categories has been reduced by one degree of freedom:

$$\begin{gathered} df=c-1 \\ =3-1 \\ =2 \end{gathered}$$

Given in the question that, the degree of freedom is $2.$

We must create a graph that displays the $p$value.

From the previous exercise, we observed that the ${\chi }^2=4.0389$

The $df$is: 2

As a result, the Chi-square distributions with 2 $df$must be used.

The $P-$value is the possibility of winning the test statistic's value, or a number that is more extreme.

$P=P\left({\chi }^2>4.0389\right)=0.13273$

The graph is given below:

Using the table, the $p$value at $2$degrees of freedom is:

$$\begin{gathered} P-\text{value}=P\left({\chi }^2>\left|{\chi }_{\text{Statistic}}^2\right|\right) \\ =P\left({\chi }^2>4.039\right) \\ =0.133 \end{gathered}$$

Let's use the Ti-83 calculator to find the $p$value:

Given in the question that, the value of $p=0.133$.

We must reach a conclusion regarding the company's claim.

The significance level is exceeded by the $P-$-value. The null hypothesis is un rejectable. As a result, there is insufficient evidence to dismiss the company's claim.