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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