Q 13 - The Practice of Statistics for AP

Question

Skittles Statistics teacher Jason Molesky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: “The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry, and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor.” 

(a) State appropriate hypotheses for a significance test of the company’s claim.

 (b) Find the expected counts for a bag of Skittles with 60 candies. 

(c) How large a C2 statistic would you need to get in order to have significant evidence against the company’s claim at the A 0.05 level? At the A 0.01 level? 

(d) Create a set of observed counts for a bag with 60 candies that gives a P-value between 0.01 and 0.05. Show the calculation of your chi-square statistic. 

Step-by-Step Solution

Answer

a)$H_0:p_{\text{Lemom }}=p_{\text{lime }}=p_{\text{Orange }}=p_{\text{Sinaterry }}=p_{\text{Ginape }}=0.20$

$H_a:\text{ At least one of the }{p}_i\text{ is incorrect }$

b)The expected values are:

$\begin{array}{|ll|}\hline \text{ Flavour }& \text{ Expected count }\\ \text{ Lemon }& 60(0.2)=12\\ \text{ Lime }& 60(0.2)=12\\ \text{ Orange }& 60(0.2)=12\\ \text{ Strawberry }& 60(0.2)=12\\ \text{ Grape }& 60(0.2)=12\\ \hline\end{array}$

c)$\alpha =0.05,{\chi }^2=9.49$

$\alpha =0.01,{\chi }^2=13.28$

d)The test statistic is:$10.1538$

Part (a) Step 1 Given Information

Given in the question that, The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry, and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor. 

For a significance test of the company's claim, we must formulate acceptable hypotheses.

Part (a) Step 2 Explanation

From the given information, the null and alternative hypotheses are:

$H_0:p_{\text{lemom }}=p_{\text{Lime }}=p_{\text{Orange }}=p_{\text{Siraberry }}=p_{\text{Girape }}=0.20$

$H_a\text{ : At least one of the }{p}_i\text{ is incorrect }$

Part (b) Step 1 Given Information

We need to calculate the expected counts for a bag of 60 Skittles.

Part (b) Step 2 Explanation

The expected values could be computed as:

$\begin{array}{|ll|}\hline \text{ Flavour }& \text{ Expected count }\\ \text{ Lemon }& 60(0.2)=12\\ \text{ Lime }& 60(0.2)=12\\ \text{ Orange }& 60(0.2)=12\\ \text{ Strawberry }& 60(0.2)=12\\ \text{ Grape }& 60(0.2)=12\\ \hline\end{array}$

Part (c) Step 1 Given Information

Given in the question that, the flavor blend is 20 percent of each flavor.

At the A 0.05 level, what size C2 statistic would you need to establish strong proof against the company's claim? At the level of A 0.01?

Part (c) Step 2 Explanation

We can calculate the degree of freedom as:

$\text{ Degree of freedom }=\text{ Number of categories }-1=5-1=4\text{. }$

Therefore, the critical values are:

$\alpha =0.05,{\chi }^2=9.49$ From table

$\alpha =0.01,{\chi }^2=13.28$ From table

Since the values are above the chi-square test statistic, we reject the null hypothesis and significant test.

Part (d) Step 1 Given Information

For a bag of 60 candies, we must produce a set of observed counts with a P-value between 0.01 and 0.05. Show your chi-square statistic computation.

Part (d) Step 2 Explanation

The test statistic can be calculated as:

$\begin{array}{|ccccc|}\hline \text{ Observed value }& \text{ Expected value }& (O-E)& (O-E{)}^2& (O-E{)}^2/E\\ 6& 10.4& -4.4& 19.36& 1.8615\\ 6& 10.4& -4.4& 19.36& 1.8615\\ 16& 10.4& -2& 4& 0.4\\ 16& 10.4& 5.6& 31.36& 3.0154\\ & 10.4& 5.6& 31.36& 3.0154\\ \hline\end{array}$

Therefore, the test statistic will be:

$$\begin{gathered} {\chi }^2=\sum \frac{(O-E{)}^2}{E} \\ =10.1538 \end{gathered}$$

This test statistic's p-value would be between $0.01$ and $0.05.$