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
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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