91Ó°ÊÓ

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" $$\begin{gathered} E\left(\text{Red}\right)=200×\frac{18}{38} \\ =94.74 \end{gathered}$$

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

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

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 ${\mathrm{χ}}^2=4.0389$

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.

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

The graph is given below:

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

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

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.