91Ó°ÊÓ

Given in the question to refer exercise 1 and 3.

The projected counts must be large enough for the chi-square distribution to be used. We have to calculate the degree of freedom as well.

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:

$$\begin{gathered} E\left(\text{Cashews}\right)=150×(0.52) \\ =78 \end{gathered}$$

$$\begin{gathered} E(Almonds)=150×(0.27) \\ =40.5 \end{gathered}$$

$$\begin{gathered} E\left(\text{Macadamia}\right)=150×(0.13) \\ =19.5 \end{gathered}$$

$$\begin{gathered} E\left(\text{Brazil}\right)=150×(0.08) \\ =12 \end{gathered}$$

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

The degree of freedom is as follows:

$$\begin{gathered} df=C-1 \\ =4-1 \\ =3 \end{gathered}$$

According to the information, we know that the test statistic is $6.5988$.

We must create a graph that displays the p value.

From Part (a), we observed that the degree of freedom is $3.$

As a result, the Chi-square distribution with three degrees of freedom must be used. The P-value is the possibility of winning the test statistic's value, or a number that is more extreme.

According to the information,${\mathrm{χ}}^2=6.5988$

Using a table and calculator, we must calculate the P-value.

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

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

$=P\left({\mathrm{χ}}^2>6.5988\right)$

$=0.086$

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

From the previous part, we know that the ${\mathrm{χ}}^2=6.5988$.

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 lack of evidence to dismiss the company's distribution claim for its deluxe mixed nuts.