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 ! 0 alpha = 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 uniform alpha = 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 uniform alpha = 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 = .435 | p 2 = 73/150 = .487 | p 3 = 66/150 = .44 |
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