Solutions for Assignment 5 -- Hypothesis Testing |

1/ Ho: l = 1.5, Ha: l ! 1.5 alpha = 5 % reject Ho if | t | > 2.571 (2 tail)x bar = 1.6, s = 0.434 t = 0.5644 so we cannot reject Ho. |
2/ Ho: l_{ 1} = l_{ 2} , Ha: l_{ 1} ! l_{ 2} alpha = 5 % reject Ho if | t | > 2.447 (2 tail)calculated t = 1.29 cannot reject Ho |

3/ Ho: l_{ d} = 0, Ha: l_{ d} ! 0alpha = 1 % l _{d} = .02 r _{d} = .0287 reject Ho if | t | > 3.25 (2 tail)calculated t value of t = 2.2 cannot reject Ho that diff = 0 |
4/ Ho: l_{ 1} = l_{ 2} , Ha: l_{ 1} ! l_{ 2}alpha = 5 % reject Ho if | z | > 1.96 (2 tail)calculated value of z = 4.79 heave Ho. |

5/ Ho: p _{1} = p _{2} Ha: p _{1} > p_{ 2}alpha = 1 % if z > 2.33 heave Ho (top tail)p hat = .5875 z calculated = 2.89 must heave Ho. |
6/ Ho: p _{1} = p _{2} Ha: p _{1} ! p_{ 2}alpha = 10% if | z | > 1.65 heave Ho (2 tail)p hat = .77 z calculated = - 0.823 cannot heave Ho. |

7/ Ho: distribution is uniform Ha: distribution is not uniformalpha = 5% if x ^{ 2} > 7.815 heave Ho.x ^{ 2} = 14.66 so we must heave Ho. | |

8/ Ho: the distribution is uniform Ha: the distribution is not uniformalpha = 5% if x ^{ 2} > 5.991 heave Ho.x ^{ 2} = 0.34 so we cannot heave Ho. |

9/ numbers in brackets are *expected frequencies*.

northeast |
midwest |
southwest |
total | |

unemployment | 87 (90.4) | 73 (67.8) | 66 (67.8) | 226 |

inflation | 113 (109.6) | 77 (82.2) | 84 (82.2) | 274 |

total | 200 |
150 |
150 |
500 |

p
_{ 1} = 87/200 = .435p
_{ 2} = 73/150 = .487p _{ 3} = 66/150 = .44**Ho**: p_{ 1} = p_{ 2} = p_{ 3} **Ha**: the p's are not equal

alpha = 5%

if x^{ 2} > 5.991 heave **Ho**.

x^{ 2} = 1.048 cannot heave **Ho**. differences in proportions are not significant.

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