Q.6 - The Practice of Statistics for AP

Question

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 

Step-by-Step Solution

Answer

(a) The Chi-square distribution with $2$ degrees of freedom must be used.

(b)

(c) The value of $p$ is: $0.133$.

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

Part (a) Step 1: Given Information

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.

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: 

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

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

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

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: 

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

Part (b) Step 1: Given Information

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

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

Part (b) Step 2: Explanation

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:

Part (c) Step 1: Given Information

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:

Part (d) Step 1: Given Information

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

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 insufficient evidence to dismiss the company's claim.