Solved (Free): According to the Mars Company, & M’s Milk Chocolate candies are manufactured and distributed with the following color percentages

According to the Mars Company, & M's Milk Chocolate candies are manufactured and distributed with the following color percentages:

Blue - 24%; Orange - 20%; Green - 16%; Yellow - 14%; Red - 13%; Brown - 13%.

A large sample of this variety of M & M's was collected, and the frequencies of the colors were recorded (see the table below). Test the claim that the color distribution claimed by the company is not correct, using a 0.05 significance level.

M M's Milk Chocolate candies

Solution

The observed data is

Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

$H_0:p_{Blue}=0.24, p_{Orange} =0.20, p_{Green} =0.16, p_{Yellow}= 0.14, p_{Red} = 0.13, p_{Brown} = 0.13$

(i.e., The distribution of colors for plain M \& M's fits the distribution claimed by the Mars company.)

$H_1:$ The distribution of colors for plain M \& M' does not fits the distribution claimed by the Mars company.

Step 2 Test statistic

The test statistic for testing above hypothesis is

$$ \begin{equation*} \chi^2= \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ \end{equation*} $$

Step 3 Level of Significance

The level of significance is $\alpha =0.05$.

Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.05$. Degrees of freedom $df=k-1=6-1 =5$.

The critical value of $\chi^2$ for $df=5$ and $\alpha=0.05$ level of significance is $\chi^2 =11.0705$.

Step 5 Test Statistic

The expected frequencies can be calculated as

$$ \begin{equation*} E_{i} =N*p_i \end{equation*} $$

For example, $E_{1}$ is given by

$$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 2020*0.24\\ &=&484.8. \end{eqnarray*} $$

The test statistic is

$$ \begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(457-484.8)^2}{484.8}+\cdots + \frac{(258-262.6)^2}{262.6}\\ &=& 1.594 +\cdots + 0.081\\ &=& 10.866. \end{eqnarray*} $$

Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =10.866$ which falls $outside$ the critical region bounded by the critical value $11.0705$, we $\textit{fail to reject}$ the null hypothesis.

OR

Step 6 Decision ($p$-value approach)

The p-value is $P(\chi^2_{5}>10.866) =0.0541$.

As the p-value $0.0541$ is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

There is no sufficient evidence to support the claim that the color distribution does not fit the Mars Company's claimed distribution.

Further Reading

• Statistics
• Descriptive Statistics
• Probability Theory
• Probability Distribution
• Hypothesis Testing
• Confidence interval
• Sample size determination
• Non-parametric Tests
• Correlation Regression
• Statistics Calculators