ANSWERS TO OLD EXAM 2 FROM FALL 2002

1a. Std. Error of difference = s_d/sqrt(# of pairs) = 2.87/sqrt(10) =
.9076

1b 90% CI for difference between Scale I, II means (paired data):
-2.0 +/- (1.833 * .9076) = [-3.66, -0.34]

1c. Since zero is NOT contained in the 90% confidence in 1b, zero is NOT
an acceptable value for (muI - muII). Conclude that the two scales do NOT
yield the same result, on average. Probability of a Type I error here is
alpha = 0.10.

1d. First, solve for t_beta. t_beta = sqrt(10)*(1.0/2.87) - 1.833 = -.73.
From Table B.3, obtain .10 < power < .25.

1e. Ho: Scale I - Scale II = 0
Ha: Scale I - Scale II not= 0
OR Ho: delta = 0 vs. Ha:delta not=0 where delta = general location shift
The ranks of the unsigned paired differences are, in order:
7 9 4 4 1.5 4 9 1.5 9 6,
if you assign the average of the ranks to any tied values. Or, you may
observe that all ties were in the same group, so no need to assign average
of ranks. In either case, obtain T+ = 9; T- = 46. So Tmin = 9. Compare to
critical value Tcrit = 10 from Table B.12. Since Tmin < Tcrit, we can
reject Ho and conclude that the Scales are giving different readings.
P-value: .05<P<.10.

2. Ho: Location 1 = Location 2
Ha: Location 1 not= Location 2
OR Ho: delta = 0 vs. Ha:delta not=0 where delta = general location shift

Ranks for Mann-Whitney test (in order) are:
Loc. 1: 4 9.5 9.5 2.5 6 1 8; R1=40.5
Loc. 2: 13 7 11 5 12 2.5 14; R2 = 64.5

U1 = 36.5; U2 = 12.5, so Umax = 36.5 From Table B.11, Ucrit = 41 Umax <
Ucrit so fail to reject Ho and conclude that the two locations do not
show different grain sizes: .10<P<.20.

3a. Ho: sigmasquaredA = sigmasquaredB
Ha: sigmasquaredA not= sigmasquaredB

F_obs = (2.6^2)/(2.2^2) = 1.40; compare to F(df=4,5;.10(2)) = 5.19.
F_obs < Fcrit, so fail to reject Ho; not enough evidence to declare the
variances different. P(2) > 0.50.

3b. First compute the pooled variance, sp^2.
sp^2 = [4(2.6^2) + 5(2.2^2)]/9 = 5.693333...
Std. Error of (xbarA - xbarB) = sqrt[5.693333(1/6 + 1/5)] = 1.44

3d. Ho: muA - muB < or = 0
Ha: muA - muB > 0
t_obs = (15.4 - 10.6)/1.44 = 3.32;
compare to tcrit = t(9 df, .05(1)) = 1.833
t_obs > tcrit so reject Ho and conclude that there is a reduction in
mean time to sleep in the B group as compared to the A group.
.0025<P<.005.

3d. Starting out with n=infinity for the t_alpha and t_beta values, obtain
n >= 2(5.69/4) * (1.645 + 1.282)^2 = 24.37 so try n=25 therefore df=48.
n >= 2(5.69/4) * (1.677 + 1.299)^2 = 25.20 so try n=26 therefore df=50.
n >= 2(5.69/4) * (1.676 + 1.299)^2 = 25.18 so try n=26 but this is the
same n as in the previous step, so conclude n=26 for each group, so total
sample size is 52.