Question

Skittles® Statistics teacher Jason Mole sky 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, green apple, 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 random sample of $60$ candies.

c. How large a $\chi 2$test statistic would you need to have significant evidence against the company’s claim at the α=0.05 level? At the $\alpha =0.01$ level?

d. Create a set of observed counts for a random sample of $60$ candies that gives a $P-$value between $0.01$ and $0.05$ Show the calculation of your chi-square test statistic.

Short answer

Part (a)$H_0:p_{lemon}=0.20,p_{lime}=0.20,p_{orange}=0.20,p_{strawberry}=0.20,p_{graph}=0.20$

$H_a:Atleastoneofthe{p}_i’sisincorrect.$

Part (b) Expected count for each flavor is $12$

Part (c) If chi square statistic is greater than above critical values at specified level of significance then reject $H_0$

Part (d)

Step-by-step solution

Part (a) Step 1: Given information

The flavor blend is $20\%$ of each flavor.

Sample size is $60$ candies.

Part (a) Step 2: Calculation

The null and alternative hypotheses:

$H_0:p_{lemon}=0.20,p_{lime}=0.20,p_{orange}=0.20,p_{strawberry}=0.20,p_{grape}=0.20$

$H_0:Atleastoneofthe{p}_i’sisincorrect.$

Part (b) Step 1: Calculation

To get the predicted count, multiply each frequency of flavor by $60$As a result, the projected count is,

Part (c) Step 1: Calculation

The degrees of freedom $=df=c-1$

The $df=5-1=4$

For each degree of significance, various critical values will be used.

When significance level$=a=0.05$

$$\begin{gathered} {\chi }^2=9.49\dots U\sin gexcelformula,=CHIINV(0.05,4) \\ {\chi }^2=13.28\dots U\sin gexcelformula,=CHIINV(0.01,4) \end{gathered}$$

Reject $H_0$ if the chi square statistic is greater than the crucial values at the selected level of significance.

Part (d) Step 1: Calculation

The frequency of the random sample is shown in the table below:

Using excel,

Therefore, the$p-$value is between $0.01$ and $0.05$