Refer to Exercises 1 and 3.

(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 the company’s claimed distribution for its deluxe mixed nuts? Justify your answer.

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:

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

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

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

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

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,${\chi }^2=6.5988$

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

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

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

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

$=0.086$

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

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

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.