Q. 5 - The Practice of Statistics for AP

Question

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. 

Step-by-Step Solution

Answer

(a) The degree of freedom is 3 and we will use chi- square distribution.

(b)

(c) The value of $p$ is $0.086$

(d) There is insufficient evidence to dismiss the company's claim.

Part (a) Step 1: Given Information

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.

Part (a) Step 2: Explanation

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: 

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

Part (b) Step 1: Given Information

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

We must create a graph that displays the p value.

Part (b) Step 2: Explanation

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. 

Part (c) Step 1: Given Information

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

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

Part (c) Step 2: Explanation

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:

Part (d) Step 1: Given Information

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

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

Part (d) Step 2: Explanation

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.