Question

Given the following Hypotheses:

Null: An equal number of fruits is distributed per consumer box

Alt: An unequal number of fruits is distributed in each box

Coconut | 21 |

Raspberry | 34 |

Banana | 33 |

Avocado | 0 |

Orange | 30 |

Starfruit | 14 |

Total: 132

Mean: 22

SD: 2.6

Find the degrees of freedom and chi Square analysis. What is the value of chi-square needed for statistical significance at a 0.05 level for the data?

I am struggling to find the correct chi square for this, I am unsure if I am using the right expected value in the chi-square test.

Answer #1

Category | Observed Frequency (O) | Proportion, p | Expected Frequency (E) | (O-E)²/E |

Coconut | 21 | 1/6 | 132 * 1/6 = 22 | (21 - 22)²/22 = 0.0455 |

Raspberry | 34 | 1/6 | 132 * 1/6 = 22 | (34 - 22)²/22 = 6.5455 |

Banana | 33 | 1/6 | 132 * 1/6 = 22 | (33 - 22)²/22 = 5.5 |

Avocado | 0 | 1/6 | 132 * 1/6 = 22 | (0 - 22)²/22 = 22 |

Orange | 30 | 1/6 | 132 * 1/6 = 22 | (30 - 22)²/22 = 2.9091 |

Starfruit | 14 | 1/6 | 132 * 1/6 = 22 | (14 - 22)²/22 = 2.9091 |

Total |
132 |
0.00 |
132 |
39.9091 |

Test statistic:

χ² = ∑ ((O-E)²/E) = **39.909**

df = n-1 = 5

Critical value:

χ²α = CHISQ.INV.RT(0.05, 5) = **11.070**

Decision:

Reject the null hypothesis

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