Chapter 14
14.1 Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
This Year 3 Years Ago
8.29
10.36
8.13
8.43
100
100
8.28
0
198
-5.09
0.0000
1.6526
0.0000
1.9720
t = –5.09, p-value = 0. There is overwhelming evidence to conclude that there has been a decrease over the
past three years.
14.2 a z-test of p1  p 2 (case 1)
H 0 : (p1  p 2 )  0
H1 : ( p1  p 2 )  0
z
1
2
3
4
5
6
7
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



A
B
C
D
z-Test of the Difference Between Two Proportions (Case 1)
Sample proportion
Sample size
Alpha
Sample 1 Sample 2
0.4336
0.2414
113
87
0.05
317
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
E
2.83
0.0024
1.6449
0.0047
1.9600
z = 2.83, p-value = .0024. There is enough evidence to infer that customers who see the ad are more likely
to make a purchase than those who do not see the ad.
b Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Ad
97.38
621.97
49
522.35
0
68
0.90
0.1853
1.6676
0.3705
1.9955
C
No Ad
92.01
283.26
21
t = .90, p-value = .1853. There is not enough evidence to infer that customers who see the ad and make a
purchase spend more than those who do not see the ad and make a purchase.
c z-estimator of p
p̂  z  / 2
1
2
3
4
5
6
p̂(1  p̂)
n
A
B
z-Estimate of a Proportion
Sample proportion
Sample size
Confidence level
0.4336
113
0.95
C
D
Confidence Interval Estimate
0.4336

Lower confidence limit
Upper confidence limit
E
0.0914
0.3423
0.5250
We estimate that between 34.23% and 52.50% of all customers who see the ad will make a purchase.
d t-estimator of 
x  t/2
s
n
318
1
2
3
4
5
6
7
A
B
t-Estimate: Mean
C
D
Ad
Mean
Standard Deviation
LCL
UCL
97.38
24.94
90.22
104.55
We estimate that the mean amount spent by customers who see the ad and make a purchase lies between
$90.22 and $104.55.
14.3 t-test of  D
H0 : D  0
H1 :  D  0
t
xD D
sD / n D
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Paired Two Sample for Means
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
Before
381.00
39001
25
0.96
0
24
0.70
0.2438
1.7109
0.4876
2.0639
After
373.12
40663
25
t = .70, p-value = .2438. There is not enough evidence to conclude that the equipment is effective.
14.4 a z-test of p
H 0 : p = .95
H1 : p > .95
z
p̂  p
p(1  p)
n
319
1
2
3
4
5
6
7
8
9
10
11
A
B
z-Test: Proportion
C
D
Prority
0.9714
245
0.95
1.54
0.0619
1.6449
0.1238
1.9600
Sample Proportion
Observations
Hypothesized Proportion
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
z = 1.54, p-value = .0619. There is not enough evidence to infer that the spokesperson's claim is true.
b z-test of p1  p 2 (case 1)
H 0 : (p1  p 2 )  0
H1 : ( p1  p 2 )  0
z
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



A
1 z-Test: Two Proportions
2
3
4 Sample Proportions
5 Observations
6 Hypothesized Difference
7 z Stat
8 P(Z<=z) one tail
9 z Critical one-tail
10 P(Z<=z) two-tail
11 z Critical two-tail
B
C
D
Prority
Ordinary
0.9714
0.9101
245
378
0
3.02
0.0013
1.6449
0.0026
1.9600
z = 3.02, p-value = .0013. There is enough evidence to infer that Priority Mail delivers letters within two
days more frequently than does ordinary mail.
14.5 Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
n
n
2 
 1
320
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
Discount No Discount
13.06
18.22
30.26
38.13
50
50
34.20
0
98
-4.41
0.0000
1.6606
0.0000
1.9845
t = –4.41, p-value = 0. There is enough evidence to infer that the discount plan works.
14.6 Speeds: Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
n
n
2 
 1
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
Speeds Before Speeds After
31.74
31.42
4.50
4.41
100
100
4.45
0
198
1.07
0.1424
1.6526
0.2849
1.9720
t = 1.07, p-value = .1424. There is not enough evidence to infer that speed bumps reduce speeds.
Proper stops: Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
321
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
n
n
2 
 1
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
Stops Before Stops After
7.82
7.98
1.83
1.84
100
100
1.83
0
198
-0.84
0.2021
1.6526
0.4042
1.9720
t = –.84, p-value = .2021. There is not enough evidence to infer that speed bumps increase the number of
proper stops.
14.7 t-estimator of 
x  t/2
1
2
3
4
5
6
7
s
n
A
B
t-Estimate: Mean
C
D
PSI
Mean
Standard Deviation
LCL
UCL
4.20
1.93
3.93
4.46
LCL = 3.93, UCL = 4.46. We estimate that on average tires are between 3.93 and 4.46 pounds per square
inch below the recommended amount.
Tire life: LCL = 100(3.93) = 393, UCL = 100(4.46) = 446. We estimate that the average tire life is
decreased by between 393 and 446 miles.
Gasoline consumption: LCL = .1(3.93) = .393, UCL = .1(4.46) = .446. We estimate that average gasoline
consumption increases by between .393 and .446 gallons per mile.
14.8 t-test of  D
322
H0 : D  0
H1 :  D  0
t
xD D
sD / n D
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Paired Two Sample for Means
C
Before
28.94
61.45
50
0.87
0
49
3.73
0.000
1.677
0.000
2.010
Mean
Variance
Observations
Pearson Correlation
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
After
26.22
104.30
50
t = 3.73, p-value = .0002. There is enough evidence to infer that the law discourages bicycle use.
14.9 z -test of p1  p 2 (case 1)
H 0 : (p1  p 2 )  0
H1 : ( p1  p 2 )  0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



A
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
B
C
D
Cardizem Placebo
0.084
0.0797
607
301
0
0.22
0.4126
1.6449
0.8252
1.9600
z = .22, p-value = .4126. There is not enough evidence to indicate that Cardizem users are more likely to
suffer headache and dizziness side effects than non-users.
323
14.10 t-test of 
H 0 :   200
H 1 :   200
t
x 
s/ n
1
2
3
4
5
6
7
8
9
10
11
12
A
t-Test: Mean
B
C
D
Pedestrians
209.13
60.01
200
39
0.96
0.1711
1.6849
0.3422
2.0227
Mean
Standard Deviation
Hypothesized Mean
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
t = .96, p-value = .1711. There is not enough evidence to infer that the franchiser should build on this site.
14.11 a z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 > 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
A
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
B
C
D
Smokers NoSmokers
0.041
0.024
1000
1000
0
2.14
0.0160
1.6449
0.0320
1.9600
z = 2.14, p-value = .0160. There is evidence to infer that children in smoke-free households are less likely
to be in fair to poor health than children in households with smokers.
b z-estimator of p
324
p̂(1  p̂)
n
p̂  z  / 2
1
2
3
4
5
6
A
z-Estimate: Proportion
Sample Proportion
Observations
LCL
UCL
B
Smokers
0.0410
1000
0.0287
0.0533
LCL = .0287, UCL = .0533. We estimate that between 2.87% and 5.33% of all children living in
households with smokers are in fair to poor health.
Number of children: LCL = 10 million(.0287) = 287,000, UCL = 10 million(.0533) = 533,000. Between
287,000 and 533,000 children living with at least one smoker are estimated to be in fair to poor health.
14.12 z-test of p
H 0 : p = .5
H1 : p > .5
z
1
2
3
4
5
6
7
8
9
10
11
p̂  p
p(1  p)
n
A
B
z-Test: Proportion
Sample Proportion
Observations
Hypothesized Proportion
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
C
D
Winner
0.5296
625
0.5
1.48
0.0694
1.6449
0.1388
1.9600
z = 1.48, p-value = .0694. There is not enough evidence to conclude that more Floridians believe that Mr.
Bush won than Floridians who believe Mr. Gore won.
14.13 t-tests of 
45 minutes:
H 0 :  = 45
H1 :  < 45
t
x 
s/ n
325
1
2
3
4
5
6
7
8
9
10
11
12
A
t-Test: Mean
B
C
45 minutes
41.75
3.63
45
19
-4.01
0.0004
1.7291
0.0008
2.0930
Mean
Standard Deviation
Hypothesized Mean
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
60 minutes:
D
H 0 :  = 60
H1 :  < 60
t
x 
s/ n
1
2
3
4
5
6
7
8
9
10
11
12
A
t-Test: Mean
B
Mean
Standard Deviation
Hypothesized Mean
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
80 minutes:
C
D
60 minutes
58.75
5.02
60
19
-1.11
0.1399
1.7291
0.2798
2.0930
H 0 :  = 80
H1 :  < 80
t
x 
s/ n
326
1
2
3
4
5
6
7
8
9
10
11
12
A
t-Test: Mean
B
C
80 minutes
69.05
6.31
80
19
-7.76
0.0000
1.7291
0.0000
2.0930
Mean
Standard Deviation
Hypothesized Mean
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
100 minutes:
D
H 0 :  = 100
H1 :  < 100
t
x 
s/ n
1
2
3
4
5
6
7
8
9
10
11
12
A
t-Test: Mean
B
Mean
Standard Deviation
Hypothesized Mean
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
125 minutes:
C
D
100 minutes
90.40
12.35
100
19
-3.48
0.0013
1.7291
0.0026
2.0930
H 0 :  = 125
H1 :  < 125
t
x 
s/ n
327
1
2
3
4
5
6
7
8
9
10
11
12
A
t-Test: Mean
B
C
Mean
Standard Deviation
Hypothesized Mean
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
D
125 minutes
110.05
17.11
125
19
-3.91
0.0005
1.7291
0.0010
2.0930
Overall Conclusion: p-values are .0004, .1399, 0, .0013, and .0005, respectively. In four of the jobs there is
overwhelming evidence to conclude that the times specified by the schedule are greater than the actual
times.
14.14 Unequal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
( x 1  x 2 )  ( 1   2 )
s 12 s 22

n1 n 2
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Quit
Did not quit
2.04
0.72
2.05
1.40
259
1626
0
316
14.06
0.0000
1.6497
0.0000
1.9675
t = 14.06, p-value = 0. There is enough evidence to infer that quitting smoking results in weight gains.
14.15 F-test of 12 /  22
H 0 : 12 /  22  1
H1 : 12 /  22  1
328
F
1
2
3
4
5
6
7
8
9
10
s12
s 22
A
B
C
F-Test Two-Sample for Variances
Mean
Variance
Observations
df
F
P(F<=f) one-tail
F Critical one-tail
Brand A
Brand B
145.95
144.78
16.45
4.25
100
100
99
99
3.87
0.0000
1.3941
F = 3.87, p-value = 0. There is overwhelming evidence to infer that Brand B is superior to Brand A.
14.16 a z-test of p1  p 2 (case 1) (Success = 2)
H 0 : p1  p 2 = 0
H1 : p 1  p 2  0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



A
z-Test: Two Proportions
B
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
C
D
Male most Male less
0.5962
0.8077
52
52
0
-2.36
0.0092
1.6449
0.0184
1.96
z = –2.36, p-value = .0184. There is enough evidence that men’s choices are affected by the attractiveness
of women’s pictures
b z-test of p1  p 2 (case 1) (Success = 2)
H 0 : p1  p 2 = 0
H1 : p 1  p 2  0
329
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
A
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
B
C
D
Female most Female less
0.7885
0.8269
52
52
0
-0.50
0.3094
1.6449
0.6188
1.96
z = –.50, p-value = .6188. There is not enough evidence to infer that women’s choices are affected by the
attractiveness of men’s pictures.
14.17 t-test of  D
H0 : D  0
H1 :  D  0
t
xD D
sD / n D
A
B
C
1 t-Test: Paired Two Sample for Means
2
3
Price shown Price not shown
4 Mean
56.15
60.31
5 Variance
243.68
467.71
6 Observations
100
100
7 Pearson Correlation
0.79
8 Hypothesized Mean Difference
0
9 df
99
10 t Stat
-3.12
11 P(T<=t) one-tail
0.0012
12 t Critical one-tail
1.6604
13 P(T<=t) two-tail
0.0024
14 t Critical two-tail
1.9842
t = –3.12, p-value = .0012. There is overwhelming evidence to conclude that ads with no price shown are
more effective in generating interest than ads that show the price.
14.18 t-estimator of 
330
x  t/2
1
2
3
4
5
6
7
s
n
A
B
t-Estimate: Mean
C
D
Calls
0.320
0.717
0.204
0.436
Mean
Standard Deviation
LCL
UCL
LCL = .204, UCL = .436. On average, each copier is estimated to require between .204 and .436 service
calls in the first year.
Total number of service calls: LCL = 1000(.204) = 204, UCL = 1000(.436) = 436. It is estimated that the
company's copiers will require between 204 and 436 service calls in the first year.
14.19 z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 > 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
A
z-Test: Two Proportions
B
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
C
D
Exercisers Watchers
0.4250
0.3675
400
400
0
1.66
0.0482
1.6449
0.0964
1.9600
z = 1.66, p-value = .0482. There is evidence to infer that exercisers are more likely to remember the
sponsor's brand name than those who only watch.
14.20 Unequal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
331
t
( x 1  x 2 )  ( 1   2 )
s 12 s 22

n1 n 2
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Leftover Returned
61.71
70.57
48.99
203.98
14
53
0
44
-3.27
0.0011
1.6802
0.0021
2.0154
t = –3.27, p-value = .0011. There is enough evidence to support the professor's theory.
14.21 Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
( x 1  x 2 )  ( 1   2 )
s 12 s 22

n1 n 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
Echinacea Placebo
7.02
7.06
2.51
2.42
262
262
2.47
0
522
-0.31
0.3799
1.6478
0.7598
1.9645
t = –.31, p-value = .3799. There is not enough evidence to conclude that Echinacea is effective.
14.22 a z-test of p
332
H 0 : p = 104,320/425,000 = .245
H1 : p > .245
z
1
2
3
4
5
6
7
8
9
10
11
p̂  p
p(1  p)
n
A
B
z-Test: Proportion
C
D
Deliver
0.2825
400
0.245
1.74
0.0406
1.6449
0.0812
1.9600
Sample Proportion
Observations
Hypothesized Proportion
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
z = 1.74, p-value = .0406. There is enough evidence to indicate that the campaign will increase home
delivery sales.
b z-test of p
H 0 : p = 110,000/425,000 = .259
H1 : p > .259
z
1
2
3
4
5
6
7
8
9
10
11
p̂  p
p(1  p)
n
A
B
z-Test: Proportion
C
Sample Proportion
Observations
Hypothesized Proportion
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
D
Deliver
0.2825
400
0.259
1.07
0.1417
1.6449
0.2834
1.9600
z = 1.07, p-value = .1417. There is not enough evidence to conclude that the campaign will be successful.
14.23 Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
333
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
n
n
2 
 1
A
B
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
C
ABS speed No ABS speed
34.72
33.94
25.27
25.63
100
100
25.45
0
198
1.09
0.1394
1.6526
0.2788
1.9720
t = 1.09, p-value = .2788. There is not enough evidence that operating an ABS-equipped car changes a
driver's behavior.
14.24a Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
( x 1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



A
B
C
1 t-Test: Two-Sample Assuming Equal Variances
2
3
Expenses MSA Expenses Regular
4 Mean
347.24
479.25
5 Variance
21043
21128
6 Observations
63
141
7 Pooled Variance
21102
8 Hypothesized Mean Difference
0
9 df
202
10 t Stat
-6.00
11 P(T<=t) one-tail
0.0000
12 t Critical one-tail
1.6524
13 P(T<=t) two-tail
0.0000
14 t Critical two-tail
1.9718
t = –6.00, p-value = 0. There is enough evidence to infer that medical expenses for those under the MSA
plan are lower than those who are not.
334
b z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 < 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
A
z-Test: Two Proportions
B
C
D
Health MSA Health Regular
0.7619
0.7801
63
141
0
-0.29
0.3867
1.6449
0.7734
1.9600
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
z = –.29, p-value = .3867. There is not enough evidence to support the critics of MSA.
14.25 a z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 > 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
A
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
B
C
D
Topiramate Placebo
0.2364
0.1042
55
48
0
1.76
0.0390
1.6449
0.0780
1.96
z = 1.76, p-value = .0390. There is enough evidence to conclude that topiramate is effective in causing
abstinence for the first month.
335
b z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 > 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
A
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
B
C
D
Topiramate Placebo
0.5091
0.1667
55
48
0
3.64
0.0001
1.6449
0.0002
1.96
z = 3.64, p-value = .0001. There is enough evidence to conclude that topiramate is effective in causing
alcoholics to refrain from binge drinking in the final month.
14.26 Time to solve the 48 problems: Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



A
B
C
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
Diet
581.95
2716.6
20
2469.1
0
38
1.94
0.0300
1.6860
0.0601
2.0244
336
Not
551.5
2221.5
20
t = 1.94, p-value = .0300. There is enough evidence to conclude that dieters take longer to solve the 48
problems than do nondieters.
Successfully repeat string of five letters: z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 < 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1
p̂(1  p̂) 
 n1 n 2



A
z-Test: Two Proportions
B
C
Diet
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
D
Not
0.50
20
0
-1.99
0.0234
1.6449
0.0468
1.96
0.80
20
z = –1.99, p-value = .0234. There is enough evidence to conclude that dieters are less successful at
repeating string of five letters.
Successfully repeat string of five words: z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 > 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
A
z-Test: Two Proportions
B
C
Diet
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
0.35
20
0
-1.58
0.0567
1.6449
0.1134
1.96
D
Not
0.60
20
z = –1.58, p-value = .0567. There is not enough evidence to conclude that dieters are less successful at
repeating string of five words.
337
Case 14.1 California Verbal Learning Test (CVLT) scores
Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
1
2
3
4
5
6
7
8
9
10
11
12
13
14
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
A
B
C
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
User
Nonuser
71.53
69.66
82.64
49.71
58
47
67.93
0
103
1.16
0.1246
1.6598
0.2491
1.9833
t = 1.16, p-value = .1246.
Logical Memory Test (LMT) score
H 0 : (1   2 )  0
H1 : (1   2 )  0
t
( x 1  x 2 )  (1   2 )
 1
1 

s 2p  
n
n
2 
 1
338
1
2
3
4
5
6
7
8
9
10
11
12
13
14
A
B
C
t-Test: Two-Sample Assuming Equal Variances
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
User
Nonuser
57.67
56.64
27.24
28.98
58
47
28.02
0
103
1.00
0.1609
1.6598
0.3218
1.9833
t = 1.00, p-value = .1609.
Conclusion of both tests: There is not enough evidence to conclude that HRT helps improve cognitive
performance among women over 75.
Case 14.2 a z-estimator of p
1
2
3
4
5
6
A
z-Estimate: Proportion
Sample Proportion
Observations
LCL
UCL
B
Aware
0.9717
283
0.9524
0.9910
Total: LCL = 836,200(.9524) = 796,397, UCL = 836,200(.9910) = 828,674
b z-estimator of p
A
1 z-Estimate: Proportion
2
3 Sample Proportion
4 Observations
5 LCL
6 UCL
B
Modify
0.6466
283
0.5910
0.7023
Total: LCL = 836,200(.5910) = 494,194, UCL = 836,200(.7023) = 587,263
c z-test of p1  p 2 (case 1)
H 0 : p1  p 2 = 0
H1 : p 1  p 2 < 0
339
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
n
n
2 
 1
A
z-Test: Two Proportions
B
C
D
LSES
HSES
0.1521
0.6732
447
153
0
-12.32
0
1.6449
0
1.96
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
z = –12.32, p-value = 0. There is enough evidence to conclude that women in the HSES group are more
aware than are women in the LSES group.
z-test of p1  p 2 (case 1)
d
H 0 : p1  p 2 = 0
H1 : p 1  p 2 < 0
z
1
2
3
4
5
6
7
8
9
10
11
(p̂1  p̂ 2 )
 1
1 

p̂(1  p̂) 
 n1 n 2 
A
z-Test: Two Proportions
Sample Proportions
Observations
Hypothesized Difference
z Stat
P(Z<=z) one tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
B
C
D
LSES
HSES
0.1588
0.4641
447
153
0
-7.67
0
1.6449
0
1.96
z = –7.67, p-value = 0. There is enough evidence to conclude that women in the HSES group are more
likely to use HRT than are women in the LSES group.
340
Case 14.3 a z-test of p (success = 1, vote "No")
H 0 : p = .5
H1 : p > .5
z
1
2
3
4
5
6
7
8
9
10
11
p̂  p
p(1  p)
n
A
B
z-Test: Proportion
C
D
Planned vote
0.5382
641
0.5
1.94
0.0265
1.6449
0.0530
1.9600
Sample Proportion
Observations
Hypothesized Proportion
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
z = 1.94, p-value = .0265. There is evidence to infer that if the referendum were held on the day of the poll,
the majority of Quebec would vote to remain in Canada.
b z-estimator p1  p 2 (success = 2, vote "Yes")
(p̂1  p̂ 2 )  z  / 2
1
2
3
4
5
6
7
8
p̂1 (1  p̂1 ) p̂ 2 (1  p̂ 2 )

n1
n2
A
B
C
z-Estimate: Two Proportions
Sample Proportions
Observations
LCL
UCL
D
Francophone Anglophone
0.5553
0.0794
515
126
0.4123
0.5397
LCL = .4123, UCL = .5397. We estimate that the difference between French-speaking and Englishspeaking Quebecers in their support for separation lies between 41.23% and 53.97%.
341
342