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

Color | Frequency |
---|---|

Blue | 457 |

Orange | 389 |

Green | 371 |

Yellow | 268 |

Red | 277 |

Brown | 258 |

#### Solution

The observed data is

Color | Obs. Freq.$(O)$ | Prop. |
---|---|---|

Blue | 457 | 0.24 |

Orange | 389 | 0.2 |

Green | 371 | 0.16 |

Yellow | 268 | 0.14 |

Red | 277 | 0.13 |

Brown | 258 | 0.13 |

##### 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

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

Color | Obs. Freq.$(O)$ | Prop. $p_i$ | Expe.Freq.$(E)$ | $(O-E)^2/E$ |
---|---|---|---|---|

Blue | 457 | 0.24 | 484.8 | 1.594 |

Orange | 389 | 0.2 | 404 | 0.557 |

Green | 371 | 0.16 | 323.2 | 7.069 |

Yellow | 268 | 0.14 | 282.8 | 0.775 |

Red | 277 | 0.13 | 262.6 | 0.79 |

Brown | 258 | 0.13 | 262.6 | 0.081 |

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