Question

Use the distribution below to test that Plain M&Ms follow
the stated distribution. Discuss your choice of ?. Would a
different ? have changed your conclusion?

M&M states the following distribution for Plain M&Ms:

Red = 13%, Orange = 20%, Yellow = 13%, Green = 20%, Blue = 20%,
Brown = 14%

Total amount of Plain M&Ms: 678

Red=69 Orange=216 Yellow=75 Green=61 Blue=187 Brown=70

Answer #1

We would be using the chi square goodness of fit test here to test the claim whether the distribution follows the stated distribution.

The expected frequencies for each colour is computed here
as:

= Proportion of that colour expected * Total Sample size

E_{i} = p_{i}*678

The computations are thus made here as:

Colour | Observed Freq | Expected Freq | (O_i - E_i)^2/E_i |

Red | 69 | 88.14 | 4.1563 |

Orange | 216 | 135.6 | 47.6708 |

Yellow | 75 | 88.14 | 1.9589 |

Green | 61 | 135.6 | 41.0410 |

Blue | 187 | 135.6 | 19.4835 |

Brown | 70 | 94.92 | 6.5424 |

678 | 678 | 120.8530 |

The chi square test statistic now is computed as shown in the last column of the above table as:

For n - 1 = 5 degrees of freedom, the p-value here is obtained from the chi square distribution tables as:

As the p-value here is approx. negligible, therefore the test is
significant here and we can reject the null hypothesis here.
**Therefore we have sufficient evidence here that the given
observed frequencies are not in line with the stated
distribution.**

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17%
23%
7%
9%
8%
13%
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This question has several parts that must be completed
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