Question

Suppose a teacher gives her statistics class a four-question, multiple-choice quiz at the beginning of the semester to measure how well prepared they were for the class. The accompanying table shows the number of students who had 0, 1, 2, 3 and 4 questions correct. Using α = 0.05, perform a chi-square test to determine if the number of correct answers per student follows the binomial probability distribution.

Number of Correct Answers per Student |
Frequency |

0 | 5 |

1 | 6 |

2 | 8 |

3 | 10 |

4 | 11 |

Total: 40

1. Calculate the test statistic.

x^{2} = ____ (Round to two decimal places as needed.)

2. Determine the p-value for the test statistic.

p-value = ____ (Round to three decimal places as needed.)

3. State the appropriate conclusion:

**(Do not reject/reject)** H0. At the 5% significance
level, there **(is not/is)** enough video evidence to
conclude that the distribution of the number of correct answers per
student does not follow the claimed, or expected, distribution.

Please be clear and concise with your response (so I will fully understand how the problem was solved)

Answer #1

Ans:

P(x=k)=4C_{k}*0.5^{k}*(1-0.5)^{4-k}

x | fo | p(x) | fe=40*p(x) | (fo-fe)^2/fe |

0 | 5 | 0.0625 | 2.5 | 2.50 |

1 | 6 | 0.2500 | 10 | 1.60 |

2 | 8 | 0.3750 | 15 | 3.27 |

3 | 10 | 0.2500 | 10 | 0.00 |

4 | 11 | 0.0625 | 2.5 | 28.90 |

Total | 40 | 1 | 40 | 36.27 |

Test statistic:

chi square**=36.27**

p-value=CHIDIST(36.27,4)=**0.000**

**Reject** H0.At the 5% significance level, there
**is** enough video evidence to conclude that the
distribution of the number of correct answers per student
**does not follow** the claimed, or expected,
distribution.

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shows the number of students who had 0, 1, 2, 3 and 4 questions
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