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.

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.

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

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

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

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

The expected values could be computed as:

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

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?

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:

localid="1653415588029" $\alpha =0.05,{\chi }^2=9.49$From table

localid="1653415592837" $\alpha =0.01,{\chi }^2=13.28$From table

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

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.

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 $0.05.$