CHAPTER 9
Comparison of Paired Samples
•9.1 (a) The standard deviation of the four sample differences is given as .68. The standard error is
sd
.68
SE(y- - y- ) = SE - =
=
= .34.
d
1
2
nd
4
(b) H0: The mean yields of the two varieties are the same (µ1 = µ2)
HA: The mean yields of the two varieties are different (µ1 ≠ µ2)
ts = -1.65/.34 = -4.85. With df = 3, Table 4 gives t.01 = 4.541 and t.005 = 5.841; thus, .01 < P < .02. At
significance level α = .05, we reject H0 if P < .05. Since .01 < P < .02, we reject H0. There is sufficient
evidence (.01 < P < .02) to conclude that Variety 2 has a higher mean yield than Variety 1.
(c) H0: The mean yields of the two varieties are the same (µ1 = µ2)
HA: The mean yields of the two varieties are different (µ1 ≠ µ2)
SE(y- - y- ) =
1
2
1.762 1.722
4 + 4 = 1.230.
ts = -1.65/1.230 = -1.34. With df = 6, Table 4 gives t.20 = .906 and t.10 = 1.440. Thus, .20 < P < .40 and
we do not reject H0. There is insufficient evidence (.20 < P < .40) to conclude that the mean yields of the
two varieties are different. (By contrast, the correct test, in part (b), resulted in rejection of H0.)
9.2 (a) The standard deviation of the nine sample differences is given as 59.3. The standard error is
sd
59.3
SEd- =
=
= 19.77.
nd
9
(b) H0: The mean weight gains on the two diets are the same (µ1 = µ2)
HA: The mean weight gains on the two diets are different (µ1 ≠ µ2)
ts = 22.9/19.77 = 1.158. With df = 8, Table 4 gives t.20 = .889 and t.10 = 1.397. Thus, .20 < P < .40 and
we do not reject H0. There is insufficient evidence (.20 < P < .40) to conclude that the mean weight gains
on the two diets are different.
(c) 22.9 ± (1.860)(19.77)
(-13.9,59.7) or -13.9 < µd < 59.7 lb.
(d) We are 90% confident that the average steer gains somewhere between 59.7 pounds more and 13.9
pounds less when on Diet 1 than when on Diet 2 (in a 140-day period).
130
•9.3 Let 1 denote control and let 2 denote progesterone.
H0: Progesterone has no effect on cAMP (µ1 = µ2)
HA: Progesterone has some effect on cAMP (µ1≠µ2)
The standard error is
SE(y- - y- ) = SE - =
d
1
2
sd
.40
=
= .20.
nd
4
The test statistic is
y-1 - y-2
d
.68
ts = SE
= - = .20 = 3.4.
SEd
(y-1 - y-2)
To bracket the P-value, we consult Table 4 with df = 4 - 1 = 3. Table 4 gives t.025 = 3.182 and
t.02 = 3.482. Thus, the P-value is bracketed as
.04 < P < .05.
At significance level α = .10, we reject H0 if P < .10. Since .04 < P < .05, we reject H0. There is
sufficient evidence (.04 < P < .05) to conclude that progesterone decreases cAMP under these conditions.
•9.4 (a) Let 1 denote treated side and 2 denote control side. The standard error is
sd
1.118
SE(y- - y- ) =
=
= .2887.
1
2
nd
15
The critical value t.025 is found from Student's t distribution with df = nd - 1 =
15 - 1 = 14. From Table 4 we find that t(14).025 = 2.145.
The 95% confidence interval is
d ± t.025SEd.117 ± (2.145)(.2887)
(-.50,.74) or -.50 < µ1 - µ2 < .74 ˚C.
1.2172 1.3022
(b) SE(y- - y- ) =
15 + 15 = .460.
1
2
.117 ± (2.048)(.460)
(using df = 28)
(-.83,1.06) or -.83 < µ1 - µ2 < 1.06 ˚C.
This interval is wider than the one obtained in part (a).
9.5 Let 1 denote treated side and 2 denote control side.
H0: The electrical treatment has no effect on collagen shrinkage temperature (µ1 = µ2)
HA: The electrical treatment tends to reduce collagen shrinkage temperature (µ1 < µ2)
We note that y- 1 > y- 2, so the data do not deviate from H0 in the direction specified by HA. Thus, P > .50
and we do not reject H0. There is no evidence (P > .50) that the electrical treatment tends to reduce
collagen shrinkage temperature under these conditions.
9.6 The data provide fairly strong evidence (P = .03) that desipramine is more effective than clomipramine in
reducing the compulsion to pull one's hair.
131
9.7 SEd- =
sd
=
nd
3
= .57. The confidence interval is 10.9 ± (2.052)(.57) or (9.7,12.1).
28
9.8 With the outliers deleted, the mean of the remaining 26 differences is 11.0 and the standard deviation is 2.1.
sd
2.1
=
SEd- =
= .41. The confidence interval is 11.0 ± (2.060)(.41) or (10.1,11.8). This interval is
nd
26
more narrow than the previous interval that was based on all of the data, including the outliers, but the
difference is not great.
9.9 There is no single correct answer. Any data set with Y1 and Y2 varying, but d not varying, is correct; for
example:
Y1
5
6
3
4
5
Y2
3
4
1
2
3
d
2
2
2
2
2
9.10 - 9.12 See Section III of this Manual.
9.13 (a)
Yield, Variety 2
34
33
32
31
30
31
32
33
Yield, Variety 1
Yes, the upward trend indicates that the pairing was effective.
132
(b)
Weight gain, Diet 2
600
570
540
510
480
440
480
520
560
Weight gain, Diet 1
The upward trend here is rather weak, which indicates that the pairing was not especially effective.
Shrinkage Temp, Control
(c)
70
69
68
67
67
68
69
70
Shrinkage Temp, Treated
Yes, the upward trend indicates that the pairing was effective.
•9.14 (a) Bs = 6. Looking under nd = 9 in Table 7, we see that there is no entry less than or equal to 6.
Therefore, P > .20.
(b) Bs = 7. Looking under nd = 9 in Table 7, we see that the only column with a critical value less than or
equal to 7 is the column headed .20 (for a nondirectional alternative), and the next column is headed .10.
Therefore, .10 < P < .20.
(c) Bs = 8. Looking under nd = 9 in Table 7, we see that the rightmost column with a critical value less
than or equal to 8 is the column headed .05 (for a nondirectional alternative), and the next column is
headed .02. Therefore, .02 < P < .05.
(d) Bs = 9. Looking under nd = 9 in Table 7, we see that the rightmost column with a critical value less
than or equal to 9 is the column headed .01 (for a nondirectional alternative), and the next column is
headed .002. Therefore, .002 < P < .01.
133
9.15 (a) P > .20
(b) .10 < P < .20
(c) .02 < P < .05
(d) .002 < P < .01
(e) P < .001
(f) P < .001
9.16 Let p denote the probability that oral conjugated estrogen will decrease PAI-1 level.
H0: Oral conjugated estrogen has no effect on PAI-1 level (p = .5)
HA: Oral conjugated estrogen has an effect on PAI-1 level (p ≠ .5)
N+ = 8, N- = 22, Bs = 22. With nd = 30, 22 falls under the .02 heading (for a nondirectional alternative) in
Table 7. Thus, .01 < P < .02 and we reject H0. There is sufficient evidence (.01 < P < .02) to conclude
that oral conjugated estrogen tends to decrease PAI-1 level.
•9.17 For the sign test, the hypotheses can be stated as
H0: p=.5
HA: p>.5
where p denotes the probability that the rat in the enriched environment will have the larger cortex. The
hypotheses may be stated informally as
H0: Weight of the cerebral cortex is not affected by environment
HA: Environmental enrichment increases cortex weight
There were 12 pairs. Of these, there were 10 pairs in which the relative cortex weight was greater for the
"enriched" rat than for his "impoverished" littermate; thus N+ = 10 and N- = 2. To check the directionality
of the data, we note that
N+ > NThus, the data so deviate from H0 in the direction specified by HA. The value of the test statistic is
Bs = larger of N+ and N= 10.
Looking in Table 7, under nd = 12 for a directional alternative, we see that the rightmost column with a
critical value less than or equal to 10 is the column headed .025 and the next column is headed .01.
Therefore, .01 < P < .025. At significance level α = .05, we reject H0 if P < .05. Since P < .025, we reject
H0. There is sufficient evidence (.01 < P < .025) to conclude that environmental enrichment increases
cortex weight.
134
•9.18 We have nd = 12. The null distribution is a binomial distribution with n = 12 and p = .5. Since Bs = 10
and HA is directional, we need to calculate the probability of 10, 11, or 12 plus (+) signs. We apply the
binomial formula nCjpj(1 - p)n-j, as follows:
j = 10, n - j =
2:
j = 11, n - j =
1:
j = 12, n - j =
0:
(66)(.510)(.52) = .01611
(12)(.511)(.51) = .00293
(1)(.512)(.50) = .00024
The P-value is the sum of these probabilities:
P = .01611 + .00293 + .00024 = .01928.
9.19 Let p denote the probability that a patient will have fewer minor seizures with valproate then with placebo.
H0: Valproate is not effective against minor seizures (p = .5)
HA: Valproate is effective against minor seizures (p > .5)
N+ = 14, N- = 5, Bs = 14; the data deviate from H0 in the direction specified by HA. Eliminating the pair
with d = 0, we refer to Table 7 with nd = 19. The only entry of 14 falls under the .05 heading (for a
directional alternative). Thus, .025 < P < .05 and we reject H0. There is sufficient evidence
(.025 < P < .05) to conclude that valproate is effective against minor seizures.
9.20 We need to find the probability of 14 or more successes for a binomial distribution with n = 19 and p = .5.
For the normal approximation to the binomial, the mean is
np = (19)(.5) = 9.5
and the SD is
np(1 - p) = (19)(.5)(.5) = 2.17945.
z=
13.5 − 9.5
= 1.84; Table 3 gives .9671.
2.17945
P = 1 - .9671 = .0329.
9.21 We need to find the probability of 22 or more successes, or of 8 or fewer successes, for a binomial
distribution with n = 30 and p = .5. For the normal approximation to the binomial, the mean is
np = (30)(.5) = 15
and the SD is
np(1 - p) = (30)(.5)(.5) = 2.73861.
We find the probability of 22 or more successes and double this probability (since the normal curve is
symmetric).
21.5 - 15
z = 2.73861 = 2.37; Table 3 gives .9911.
P = 2(1 - .9911) = .0178.
9.22 Let p denote the probability that the Northern member of a pair will dominate in more episodes than the
Carolina.
135
H0: Dominance is balanced between the subspecies (p = .5)
HA: One of the subspecies tends to dominate the other (p ≠ .5)
N+ = 8, N- = 0, Bs = 8. Looking under nd = 8 in Table 7, we see that the rightmost column with a critical
value less than or equal to 8 is the column headed .01 (for a nondirectional alternative), and the next
column is headed .002. Therefore, .002 < P < .01. There is sufficient evidence (.002 < P < .01) to
conclude the Carolina subspecies tends to dominate the Northern.
9.23 P = 2(.58) = .0078125.
•9.24 (a) The null distribution is a binomial distribution with n = 7 and p = .5. Since Bs =7 and HA is
nondirectional, we need to calculate the probability of 7 successes or of 0 successes. The probability of 7
successes is .57 = .0078125. Likewise, the probability of 0 successes is .57 = .0078125. Thus,
P = 2(.0078125) = .015625.
(b) With nd = 7, the smallest possible p-value is .015625; thus P cannot be less than .01.
9.25 (a) (i) P = (2)[(105)(.513)(.52) + (15)(.514)(.5) + (1)(.515)] = .007385.
(ii) P = (2)[(15)(.514)(.5) + (1)(.515)] = .000977.
(iii) P = (2)(.515) = .000061.
(b) If Bs = 14, then P = .000977 < .002; if Bs = 13, then P = .007385 > .002. Thus, the critical value 14
corresponds to a P-value that is as close to .002 as possible without exceeding it.
(c) The entry for .005 would be 14, because .000977 < .005 but .007385 > .005.
9.26 Let p denote the probability that hunger rating is higher when taking mCPP than when taking the placebo.
H0: p = .5
HA: p ≠ .5
N+ = 3, N- = 5, Bs = 5. Looking under nd = 8 in Table 7, we see that the leftmost column has an entry of
7, so the P-value is greater than .20. There is insufficient evidence (P > .20) to conclude that hunger
ratings differ on the two treatments.
9.27 P = (2)[(56)(.55)(. 53) + (28)(.56)(. 52) + (8)(.57)(. 51) + (1)(.58)] = .7265.
9.28 (a) P > .20
(b) .10 < P < .20
(c) .02 < P < .05
(d) .01 < P < .02
9.29 (a) P > .20
(b) .05 < P < .10
(c) .002 < P < .01
(d) .002 < P < .01
9.30 H0: Hunger rating is not affected by treatment (mCPP vs. placebo)
HA: Treatment does affect hunger rating
136
The absolute values of the differences are 5, 7, 28, 47, 80, 7, 8, and 20.
The ranks of the absolute differences are 1, 2.5, 6, 7, 8, 2.5, 4, and 5.
The signed ranks are –1, 2.5, -6, -7, -8, 2.5, 4, and –5.
Thus, W+ = 2.5 + 2.5 + 4 = 9 and W- = 1 + 6 + 7 + 8 + 5 = 27.
Ws = 27 and nd = 8; reading Table 8 we find P-value > .20 and H0 is not rejected. There is insufficient
evidence (P > .20) to conclude that treatment has an effect.
9.31 H0: Weight change is not affected by treatment (mCPP vs. placebo)
HA: Treatment does affect weight change
The absolute values of the differences are 1.1, 1.6, 2.1, 0.3, 0.6, 2.2, 0.9, 0.7, and 0.4.
The ranks of the absolute differences are 6, 7, 8, 1, 3, 9, 5, 4, and 2.
The signed ranks are 6, -7, -8, -1, -3, -9, 5, 4, and -2.
Thus, W+ = 6 + 5 + 4 = 15 and W- = 7 + 8 + 1 + 3 + 9 + 2 = 30.
Ws = 30 and nd = 9; reading Table 8 we find P-value > .20 and H0 is not rejected. There is insufficient
evidence (P > .20) to conclude that treatment has an effect.
9.32 H0: HL-A compatibility has no effect on graft survival time
HA: Survival time tends to be greater when compatibility score is close
The differences tend to be positive, which is consistent with HA.
The absolute values of the differences are 12, 6, 42+, 67, 5, 5, 6, 20, 11, 18+, and 1.
The ranks of the absolute differences are 7, 4.5, 10, 11, 2.5, 2.5, 4.5, 9, 6, 8, and 1.
The signed ranks are 7, 4.5, 10, 11, 2.5, 2.5, -4.5, 9, 6, 8, and -1.
Thus, W+ = 7 + 4.5 + 10 + 11 + 2.5 + 2.5 + 9 + 6 + 8 = 60.5 and W- = 4.5 + 1 = 5.5.
Ws = 60.5 and nd = 11; reading Table 8 we find .005 < P-value < .01 and H0 is rejected. There is strong
evidence (.005 < P-value < .01) to conclude that survival time tends to be greater when compatibility
score is close.
9.33 H0: Alcoholism has no effect on brain density
HA: Alcoholism reduces brain density
The differences tend to be negative, which is consistent with HA.
The absolute values of the differences are 1.2, 1.7, .5, 4.7, 3.3, .4, 2.7, 1.8, .1, .3, and 1.4.
The ranks of the absolute differences are 5, 7, 4, 11, 10, 3, 9, 8, 1, 2, and 6.
The signed ranks are -5, -7, -4, -11, -10, 3, -9, -8, -1, 2, and -6.
Thus, W+ = 3 + 2 = 5 and W- = 5 + 7 + 4 + 11 + 10 + 9 + 8 + 1 + 6 = 61.
Ws = 61 and nd = 11; reading Table 8 we find .001 < P-value < .005 and H0 is rejected. There is strong
evidence (.001 < P-value < .005) to conclude that alcoholism is associated with reduced brain density.
This was an observational study, so drawing a cause-effect inference is risky. We should stop short of
saying that alcoholism reduces brain density.
137
9.34 Let 1 denote 5 weeks and 2 denote baseline.
(a) Let N denote no coffee.
H0: Mean cholesterol does not change in the "no coffee" condition (µN,1 = µN,2)
HA: Mean cholesterol does change in the "no coffee" condition (µN,1 ≠ µN,2)
SE = 27/ 25 = 5.40.
ts = -35/5.40 = -6.48. With df = 24, Table 4 gives t.0005 = 3.745. We reject H0. There is sufficient
evidence (P < .001) to conclude that mean cholesterol is reduced in the "no coffee" condition.
(b) Let U denote usual coffee.
H0: Mean cholesterol does not change in the "usual coffee" condition (µU,1 = µU,2)
HA: Mean cholesterol does change in the "usual coffee" condition (µU,1 ≠ µU,2)
SE = 56/ 8 = 19.8.
ts = 26/19.8 = 1.31. With df = 7, Table 4 gives t.20 = 0.896 and t.10 = 1.415. We do not reject H0. There
is insufficient evidence (.20 < P < .40) to conclude that mean cholesterol is changed in the "usual coffee"
condition.
(c) Let N denote no coffee, U denote usual coffee, and d denote the change from baseline.
H0: Mean cholesterol is not affected by discontinuing coffee (µN,d = µU,d)
HA: Mean cholesterol is affected by discontinuing coffee (µN,d ≠ µU,d)
SE =
272 562
25 + 8 = 20.52.
ts = (-35 - 26)/20.52 = -2.97. Formal (7.1) gives df = 8.1 and the conservative df value is min{24,7} = 7,
whereas the liberal df value is n1 + n2 - 2 = 31. Using df = 8, Table 4 gives t.01 = 2.896 and t.005 = 3.355,
which implies that .01 < P < .02. (Using df = 30, the closest value to 31 in Table 4, we get t.005 = 2.750
and t.0005 = 3.646, which implies that .001 < P < .01.) We reject H0.
(d) There is sufficient evidence (.01 < P < .02) to conclude that mean cholesterol is reduced by
discontinuing coffee.
9.35
Food Intake (cal)
2400
2000
1600
1200
Premenstrual
Postmenstrual
138
9.36 No. This result suggests that the population mean difference between the right eye and the left eye is less
than 1.1 mg/dl, but positive and negative differences cancel each other out when such a mean is
calculated. The result says nothing at all about the typical or average magnitude of the difference between
the two eyes.
9.37 No. "Accurate" prediction would mean that the individual differences (d's) are small. To judge whether
this is the case, one would need the individual values of the d's; using these, one could see whether most
of the magnitudes (|d|'s) are small.
9.38 (a) y- 1 - y- 2 = d = -1, sd = 1.2.
SE(y- - y- ) = 1.2/ 15 = .3098.
1
2
-1 ± (2.145)(.3098)
(df = 14)
(-1.66,-.34) or -1.66 < µ1 - µ2 < -.34.
(b) SE(y- - y- ) =
1
2
1.82 2.82
= .8595.
+
15
15
-1 ± (2.145)(.8595)
(using df = 14)
(-2.84,0.84) or -2.84 < µ1 - µ2 < 0.84. This interval is much wider than the one constructed in part (a).
9.39 H0: The before and after means are the same (µ1 = µ2)
HA: The before and after means are different (µ1 ≠ µ2)
SE(y- - y- ) = 1.2/ 15 = .3098. ts = -1/.3098 = -3.23. With df = 14, Table 4 gives t.005 = 2.977 and
1
2
t.0005 = 4.140; thus, .001 < P < .01. We reject H0; there is strong evidence (.001 < P < .01) of a before and
after difference.
9.40 (a) Let p denote the probability that a before count is higher than the corresponding after count.
H0: p = .5
HA: p ≠ .5
N+ = 2, N- = 10, Bs = 10. Looking under nd = 12 in Table 7, we see that .02 < P < .05. There is sufficient
evidence (.02 < P < .05) to conclude that the after count tends to be higher than the before count.
(b) P = (2)[(66)(.510)( 52) + (12)(.511)( 51) + (1)(.512)] = .0386.
139
9.41 The scatterplot shows a positive relationship between before and after counts. The pairing removes the
variability between cats from the analysis and is, therefore, effective.
After (Y )
2
6.0
4.5
3.0
1.5
0.0
0
1
2
3
4
Before (Y )
1
9.42 Let 1 denote central and 2 denote top.
y- 1 - y- 2 = d = 2.533, sd = .41312.
SE(y- - y- ) = .41312/ 6 = .1687.
1
2
2.533 ± (2.015)(.1687)
(df = 5)
(2.19,2.87) or 2.19 < µ1 - µ2 < 2.87 percent.
9.43 The standard error is SE(y- - y- ) = 1.86/ 15 = .48.
1
2
2.2 ± (1.761)(.48)
(df = 14)
(1.35,3.05) or 1.35 < µ1 - µ2 < 3.05 species.
9.44 It must be reasonable to regard the 15 differences as a random sample from a normal population. We must
trust the researchers that their sampling method was random. The normality condition can be verified
with a normal probability plot. The plot below is fairly linear (although the plateaus show that there are
several differences that have the same value), which supports the normality condition.
5.00
Difference
3.75
2.50
1.25
-0.00
-1
0
nscores
1
140
•9.45 The null and alternative hypotheses are
H0: The average number of species is the same in pools as in riffles (µ1=µ2)
HA: The average numbers of species in pools and in riffles differ (µ1≠µ2)
The standard error is
SE(y- - y- ) = SE - =
d
1
2
sd
1.86
=
= .48.
nd
15
The test statistic is
y-1 - y-2
d
2.2
ts = SE
= - = .48 = 4.58.
SEd
(y-1 - y-2)
To bracket the P-value, we consult Table 4 with df = 15 - 1 = 14. Table 4 gives t.0005 = 4.140. Thus, the
P-value for the nondirectional test is bracketed as
P < .001.
At significance level α = .10, we reject H0 if P < .10. Since P < .001, we reject H0. There is sufficient
evidence (P < .001) to conclude that the average number of species in pools is greater than in riffles.
9.46 (a) Let p denote the probability that there are more species in a pool than in its adjacent riffle.
H0: The two habitats support equal levels of diversity (p = .5)
HA: The two habitats do not support equal levels of diversity (p ≠ .5)
N+ = 12, N- = 1, Bs = 12. Eliminating the two pairs with d = 0, we refer to Table 7 with nd = 13. The
rightmost column with a critical value of 12 is the column headed .01 for a nondirectional alternative (i.e.,
for a two-tailed test), and the next column is headed .002. Therefore, .002 < P < .01. There is sufficient
evidence (.002 < P < .01) to conclude that species diversity is greater in pools than in riffles.
(b) P = (2)[(13)(.512)(.51) + .513] = .0034.
9.47 H0: Pools and riffles support equal levels of diversity
HA: Pools and riffles support different levels of diversity
The absolute values of the differences are 3, 3, 4, 3, 4, 5, -1, 1, 1, 4, 1, 4, and 1.
The ranks of the absolute differences are 7, 7, 10.5, 7, 10.5, 13, 3, 3, 3, 10.5, 3, 10.5 and 3.
The signed ranks are 7, 7, 10.5, 7, 10.5, 13, -3, 3, 3, 10.5, 3, 10.5 and 3.
Thus, W+ = 7 + 7 + 10.5 + 7 + 10.5 + 13 + 3 + 3 + 3 + 10.5 + 3 + 10.5 + 3 = 88 and W- = 3.
Ws = 88 and nd = 13; reading Table 8 we find .001 < P-value < .002 and H0 is rejected. There is strong
evidence (.001 < P-value < .002) to conclude that the diversity levels differ between pools and riffles.
9.48 There are several ties in the data, which means that the P-value from the Wilcoxon test is only
approximate.
141
•9.49 The null and alternative hypotheses are
H0: Caffeine has no effect on RER (µ1=µ2)
HA: Caffeine has some effect on RER (µ1≠µ2)
We proceed to calculate the differences, the standard error of the mean difference, and the test statistic.
Subject
1
2
3
4
5
6
7
8
9
Mean
SD
Placebo
105
119
92
97
96
101
94
95
98
Caffeine
96
99
89
95
88
95
88
93
88
Difference
9
20
3
2
8
6
6
2
10
7.33
5.59
The standard error is
SE(y- - y- ) = SE - =
d
1
2
sd
5.59
=
= 1.86.
nd
9
The test statistic is
y-1 - y-2
d
7.33
ts = SE
= - = 1.86 = 3.94.
SE
(y1 - y2)
d
To bracket the P-value, we consult Table 4 with df = 9 - 1 = 8. Table 4 gives t.005 = 3.355 and
t.0005 = 5.041. Thus, the P-value for the nondirectional test is bracketed as
.001 < P < .01.
At significance level α = .05, we reject H0 if P < .05. Since P < .01, we reject H0. To determine the
directionality of departure from H0, we note that
d > 0; that is, y- 1 > y- 2.
There is sufficient evidence (.001 < P < .01) to conclude that caffeine tends to decrease RER under these
conditions.
142
9.50
RER (%)
112.5
105.0
97.5
90.0
Placebo
Caffeine
9.51 Let p denote the probability that RER for a subject is higher after taking placebo than after taking caffeine.
H0: RER is not affected by caffeine (p = .5)
HA: RER is affected by caffeine (p ≠ .5)
N+ = 9, N- = 0, Bs = 9. Looking under nd = 9 in Table 7, we see that the rightmost column with a critical
value less than or equal to 9 is the column headed .01 (for a nondirectional alternative), and the next
column is headed .002. Therefore, .002 < P < .01. There is sufficient evidence (.002 < P < .01) to
conclude that caffeine tends to decrease RER under these conditions.
9.52 H0: Mean CP is the same in regenerating and in normal tissue (µ1 = µ2)
HA: Mean CP is different in regenerating and in normal tissue (µ1 ≠ µ2)
SE(y- - y- ) = 4.89/ 8 = 1.727.
1
2
ts = 4.64/1.727 = 2.69. With df = 7, Table 4 gives t.02 = 2.517 and t.01 = 2.998. Thus, .02 < P < .04 and
we reject H0. There is sufficient evidence (.02 < P < .04) to conclude that mean CP is different in
regenerating and in normal tissue.
9.53 (a) Let 1 denote control and 2 denote benzamil.
H0: Benzamil does not impair healing (µ1 = µ2)
HA: Benzamil impairs healing (µ1 > µ2)
y- 1 - y- 2 = d = .09706; sd = .14768.
SE(y- - y- ) = .14768/ 17 = .03582.
1
2
ts = .09706/.03582 = 2.71. P = .0077, so we reject H0. There is sufficient evidence (P = .0077) to
conclude that benzamil impairs healing.
143
(b) Let p denote the probability that the control limb heals more than the benzamil limb.
H0: Benzamil does not impair healing (p = .5)
HA: Benzamil impairs healing (p > .5)
N+ = 11, N- = 4, Bs = 1. The two animals with d = 0 are eliminated. P = .059, so we do not reject H0.
There is insufficient evidence (P = .059) to conclude that benzamil impairs healing.
[Remark: Unlike the t test in part (a), the sign test does not take account of the fact that the negative d's
are smaller in magnitude than the positive d's. This illustrates the inferior power of the sign test.]
(c) (.021,.173) or .021 < µ1 - µ2 < .173 mm2.
(d)
0.375
Benzamil
0.300
0.225
0.150
0.075
0.000
0.000
0.125
0.250
0.375
0.500
Control
Yes, the upward trend indicates that the pairing was effective.
9.54 Summary statistics are as follows:
Mean
SD
Rest
(1)
6.169
.621
Experimental group
Work
Difference
(2)
(3)
9.301
-3.133
3.323
2.862
Rest
(4)
5.291
.652
Control group
Work
Difference
(5)
(6)
4.973
.319
.703
.544
(a) Column (1) versus column (4)
H0: Mean ventilation at rest is the same in the two conditions (µ1 = µ4)
HA: Mean ventilation at rest is different in the two conditions (µ1 ≠ µ4)
ts = 2.757, df = 13.97, P = .015. We reject H0. There is sufficient evidence (P = .015) to conclude that
mean ventilation at rest is higher in the "to be hypnotized" condition than in the "control" condition.
(b) (i) Column (1) versus column (2):
H0: Hypnotic suggestion does not change mean ventilation (µ1 = µ2)
HA: Hypnotic suggestion increases mean ventilation (µ1 < µ2)
ts = -3.096, df = 7, P = .0087. We reject H0. There is sufficient evidence (P = .0087) to conclude that
hypnotic suggestion increases mean ventilation.
144
(ii) Column (4) versus column (5):
H0: Waking suggestion does not change mean ventilation (µ4 = µ5)
HA: Waking suggestion increases mean ventilation (µ4 < µ5)
Because y- 4 > y- 5, the data do not deviate from H0 in the direction specified by HA. Thus, P > .50 and we
do not reject H0. There is no evidence that waking suggestion increases mean ventilation.
(iii) Column (3) versus column (6):
H0: Hypnotic and waking suggestion produce the same mean change in ventilation
(µ3 = µ6)
HA: Hypnotic suggestion increases mean ventilation more than does waking suggestion
(µ3 < µ6)
ts = -3.351, df = 7.5, P = .0055. We reject H0. There is sufficient evidence (P = .0055) to conclude that
hypnotic suggestion increases mean ventilation more than does waking suggestion.
(c) (i) Sign test for column (1) versus column (2). Let p1 denote the probability that a person's ventilation
after hypnotic suggestion will be higher than that at rest.
H0: Hypnotic suggestion does not change mean ventilation (p1 = .5)
HA: Hypnotic suggestion increases mean ventilation (p1 > .5)
Bs = 8, P = .0039. We reject H0. There is sufficient evidence (P = .0039) to conclude that hypnotic
suggestion increases mean ventilation.
(ii) Sign test for column (4) versus column (5). Let p2 denote the probability that a person's ventilation after
waking suggestion will be higher than that at rest.
H0: Waking suggestion does not change mean ventilation (p2 = .5)
HA: Waking suggestion increases mean ventilation (p2 > .5)
N+ = 6, N- = 2. Thus, the data do not deviate from H0 in the direction specified by HA, so P > .50 and we
do not reject H0. There is no evidence that waking suggestion increases mean ventilation.
(iii) Wilcoxon-Mann-Whitney test for column (3) versus column (6):
H0: Hypnotic and waking suggestion produce the same mean change in ventilation
HA: Hypnotic suggestion increases mean ventilation more than does waking suggestion
Us = 63, P = .000155. We reject H0. There is sufficient evidence (P = .000155) to conclude that hypnotic
suggestion increases mean ventilation more than does waking suggestion.
145
(d) A normal probability plot of column (3) shows that the data are quite skewed. This could account for
two discrepancies: First, to compare column (1) to column (2), we used the differences in column (3); the
t test gave P = .0087 whereas the sign test gave P = .0039. Second, to compare column (3) to column (6),
the t test gave P = .0055 whereas the Wilcoxon-Mann-Whitney test gave P = .000155. Both of the t tests
rest on the questionable condition that the population distribution corresponding to column (3) is normal.
The failure of this condition inflates the standard deviation and robs the t test of power, so that the
nonparametric tests give stronger conclusions (smaller P-values).
Ventillation
15.0
12.5
10.0
7.5
-0.75
0.00
0.75
nscores
A normal probability plot of column (6) shows that the normality condition appears to be met for these
data.
Ventillation
5.5
5.0
4.5
4.0
-0.75
0.00
0.75
nscores
9.55 (a) By using matched pairs we eliminate the variability that is associated with the variables used to create
the pairs (age, sex, etc.). This provides for greater precision and more power in the test.
(b) It may be that the pairing variables (age, sex, etc.) are unrelated to blood pressure. If this is the case,
then the pairing accomplishes nothing, but it reduces the number of degrees of freedom, and therefore the
power, of the test.
9.56 N+ = 10, N- = 10, Bs = 10. In this case, the data are as evenly balanced as possible, so P = 1. (Table 7
indicates that P > .20.) Thus, we do not reject H0. There is no evidence that transdermal estradiol has an
effect on PAI-1 level.
146
9.57 A normal probability plot of the data shows that the normality condition is not met. However, a sign test
can be conducted. Let p denote the probability that urinary protein excretion will go down after
plasmapheresis.
H0: Plasmapheresis affects urinary protein excretion (p = .5)
HA: Plasmapheresis does not affect urinary protein excretion (p ≠ .5)
N+ = 6, N- = 0, Bs = 6. From Table 7, .02 < P < .05 (for a two-sided test). The exact P-value is
(2)(.56) = .03125. Thus, there is evidence (P = .03125) to conclude that urinary protein excretion tends to
go down after plasmapheresis.
Note: Another approach would be to transform the data and then conduct a t test in the transformed scale.
For example, taking the reciprocal of each difference yields a fairly symmetric distribution; a t test then
gives ts = 5.4 and P = .003.