Question

M&Ms/Mars Company, the maker of Skittles, state that each bag of Skittles has the same number of each flavor in it. Use 0.05 level of significance to test the company's claim. Below is the observed number in a bag of Skittles: Observed Flavors: Lime 8, Lemon 14, Orange 20, Strawberry 10, Grape 8 Expected Flavors: All the categories of flavors are the same.

Answer 1

HYPOTHESIS TEST-

We have to perform Chi-square test for goodness of fit.

We have to test for null hypothesis

HO : All the categories of flavors are the same

against the alternative hypothesis $\tiny H_1:\text{All the categories of flavors are not the same.}$

Our Chi-square test statistic is given by

$\tiny \chi^2=\sum_{i=1}^{n}\frac{\left(f_{i}-e_{i}\right)^2}{e_{i}}$

Here,

Number of categories

n = 5

Corresponding calculations are as follows.

Grade (f-e)^2/e Lime Lemon Orange Strawberry Grape Total Observed Expected frequency (f) frequency (e) 8 12 12 20 12 10 12 8 12 60 60 1.3333 0.3333 5.3333 0.3333 1.3333 8.6667

$\tiny \therefore \chi^2_{calculated}=\sum_{i=1}^{5}\frac{\left(f_{i}-e_{i}\right)^2}{e_{i}}=8.6667$

Degrees of freedom $\tiny v=n-1=4$

$\tiny \therefore \text{p-value}=P\left(\chi^2_{4}>8.6667\right)=0.06999227$ [Using R-code '1-pchisq(8.6667,4)']

Level of significance $\tiny \alpha=0.05$

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$

Here, we observe that $\tiny \text{p-value}=0.06999227\nless0.05=\alpha$

So, we cannot reject our null hypothesis.

Hence, based on the given data we can conclude that there is no significant evidence that all the categories of flavors are not the same.

ANSWERS-

49.

Test statistic is $\tiny \chi^2=8.67$

50.

$\tiny \text{p-value}=0.07$

51.

We fail to reject the null hypothesis.