## Answers

a)

Ho : µd= 0

Ha : µd ╪ 0

paired t test

Sample #1 | Sample #2 | difference , Di =sample1-sample2 | (Di - Dbar)² |

70 | 72 | -2 | 1.21 |

71 | 72 | -1 | 0.01 |

70 | 73 | -3 | 4.41 |

72 | 71 | 1 | 3.61 |

70 | 69 | 1 | 3.61 |

67 | 67 | 0 | 0.81 |

71 | 68 | 3 | 15.21 |

68 | 73 | -5 | 16.81 |

67 | 73 | -6 | 26.01 |

70 | 69 | 1 | 3.61 |

72 | 72 | 0 | 0.81 |

72 | 70 | 2 | 8.41 |

70 | 73 | -3 | 4.41 |

70 | 77 | -7 | 37.21 |

68 | 70 | -2 | 1.21 |

68 | 65 | 3 | 15.21 |

71 | 70 | 1 | 3.61 |

70 | 68 | 2 | 8.41 |

69 | 68 | 1 | 3.61 |

67 | 71 | -4 | 9.61 |

sample 1 | sample 2 | Di | (Di - Dbar)² | |

sum = | 1393 | 1411 | -18 | 167.800 |

mean of difference , D̅ =ΣDi / n = -0.900

std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) = 2.972

Level of Significance , α = 0.1

sample size , n = 20

std error , SE = Sd / √n = 2.9718 / √ 20 = 0.6645

t-statistic = (D̅ - µd)/SE = ( -0.9000 - 0 ) / 0.6645 = -1.354

Degree of freedom, DF= n - 1 = 19

p-value = 0.1915 [excel function: =t.dist.2t(t-stat,df) ]

Conclusion: p-value>α , Do not reject null hypothesis

Do not reject H0. There is insufficient evidence to conclude that the mean score for the first round of a golf tournament is significantly different than the mean score for the fourth and final round.

b)

point estimate of difference , D̅ =ΣDi / n = -0.900

c)

sample size , n = 20

Degree of freedom, DF= n - 1 = 19 and α = 0.1

t-critical value = t α/2,df = 1.7291 [excel function: =t.inv.2t(α/2,df) ]

std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) = 2.9718

std error , SE = Sd / √n = 2.9718 / √ 20 = 0.6645

margin of error, E = t*SE = 1.7291 * 0.6645 = 1.1490

Yes. One could check to see if the 90% confidence interval includes a difference of zero. If the interval contains 0, the difference is not statistically significant.