Chapter 14
14.1 Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
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
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  p2 (case 1)
H 0 : ( p1  p2 )  0
H1 : ( p1  p2 )  0
z
( pˆ 1  pˆ 2 )
1
1 

pˆ (1  pˆ ) 
n
n
2 
 1
250
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
z Stat
P(Z<=z) one-tail
z Critical one-tail
P(Z<=z) two-tail
z Critical two-tail
2.83
0.0024
1.6449
0.0047
1.9600
z = 2.83, p-value = .0024. There is overwhelming 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
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
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
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
pˆ  z  / 2
pˆ (1  pˆ )
n
251
z-Estimate of a Proportion
Sample proportion
Sample size
Confidence level
0.4336
113
0.95
Confidence Interval Estimate

0.4336
Lower confidence limit
Upper confidence limit
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
t-Estimate: Mean
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 $194.55.
14.3 t-test of  D
H0 :D  0
H1 :  D  0
t
xD   D
s D / nD
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
Before
381.00
39001.33
25
0.9610
0
24
0.70
0.2438
1.7109
0.4876
2.0639
After
373.12
40663.28
25
t = .70, p-value = .2438. There is not enough evidence to conclude that the equipment is effective.
252
14.4 a z-test of p
H 0 : p = .95
H 1 : p > .95
z
pˆ  p
p(1  p)
n
z-Test: Proportion
Prority
0.9714
245
0.95
1.54
0.0619
1.6449
0.1238
1.96
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  p2 (case 1)
H 0 : ( p1  p2 )  0
H1 : ( p1  p2 )  0
z
( pˆ 1  pˆ 2 )
1
1
pˆ (1  pˆ ) 
 n1 n2



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
Prority
Ordinary
0.9714
0.9101
245
378
0
3.02
0.0013
1.6449
0.0026
1.96
z = 3.02, p-value = .0013. There is overwhelming evidence to infer that Priority Mail delivers letters
within two days more frequently than does ordinary mail.
253
14.5 Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
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
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 overwhelming evidence to infer that the discount plan works.
14.6 Speeds: Equal-variances t-test of 1   2
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



254
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
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
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
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
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
s
n
255
t-Estimate: Mean
Mean
Standard Deviation
LCL
UCL
PSI
4.1950
1.9328
3.9254
4.4646
LCL = 3.9254, UCL = 4.4646. We estimate that on average tires are between 3.9254 and 4.4646
pounds per square inch below the recommended amount.
Tire life: LCL = 100(3.9254) = 392.54, UCL = 100(4.4646) = 446.46. We estimate that the average
tire life is decreased by between 392.54 and 446.46 miles.
Gasoline consumption: LCL = .1(3.9254) = .39254, UCL = .1(4.4646) = .44646. We estimate that
average gasoline consumption increases by between .39254 and .44646 gallons per mile.
14.8 t-test of  D
H0 :D  0
H1 :  D  0
t
xD   D
s D / nD
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
Before
28.94
61.45
50
0.8695
0
49
3.73
0.0002
1.6766
0.0005
2.0096
After
26.22
104.30
50
t = 3.73, p-value = .0002. There is overwhelming evidence to infer that the law discourages bicycle
use.
14.9 z -test of p1  p2 (case 1)
H 0 : ( p1  p2 )  0
H1 : ( p1  p2 )  0
256
( pˆ 1  pˆ 2 )
z
1
1 

pˆ (1  pˆ ) 
n
n
2 
 1
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
Cardizem Placebo
0.084
0.0797
607
301
0
0.22
0.4126
1.6449
0.8252
1.96
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.
14.10 t-test of 
H 0 :   200
H 1 :   200
t
x 
s/ n
t-Test: Mean
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
Pedestrians
209.13
60.01
200
39
0.96
0.1711
1.6849
0.3422
2.0227
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  p2 (case 1)
H 0 : p1  p2 = 0
H 1 : p1  p2 > 0
257
z
( pˆ 1  pˆ 2 )
1
1 

pˆ (1  pˆ ) 
n
n
2 
 1
z-Test: Two Proportions
Smokers NoSmokers
0.041
0.024
1000
1000
0
2.14
0.0160
1.6449
0.032
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 = 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
pˆ  z  / 2
pˆ (1  pˆ )
n
z-Estimate: Proportion
Sample Proportion
Observations
LCL
UCL
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
H 1 : p > .5
z
pˆ  p
p(1  p)
n
258
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
Winner
0.5296
625
0.5
1.48
0.0694
1.6449
0.1388
1.96
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
H 1 :  < 45
t
x 
s/ n
t-Test: Mean
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:
45 minutes
41.75
3.63
45
19
-4.01
0.0004
1.7291
0.0008
2.093
H 0 :  = 60
H 1 :  < 60
t
x 
s/ n
259
t-Test: Mean
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:
60 minutes
58.75
5.02
60
19
-1.11
0.1399
1.7291
0.2798
2.093
H 0 :  = 80
H 1 :  < 80
t
x 
s/ n
t-Test: Mean
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:
80 minutes
69.05
6.31
80
19
-7.76
0
1.7291
0
2.093
H 0 :  = 100
H 1 :  < 100
t
x 
s/ n
260
t-Test: Mean
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
90.4
12.35
100
19
-3.48
0.0013
1.7291
0.0026
2.093
H 0 :  = 125
125 minutes:
H 1 :  < 125
t
x 
s/ n
t-Test: Mean
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
110.05
17.11
125
19
-3.91
0.0005
1.7291
0.001
2.093
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
H 1 : (1   2 )  0
t
( x1  x 2 )  ( 1   2 )
s12 s 22

n1 n 2
261
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
2.038
0.721
2.052
1.398
259
1626
0
316
14.06
0.0000
1.6497
0.0000
1.9675
t = 14.06, p-value = 0. There is overwhelming evidence to infer that quitting smoking results in
weight gains.
14.15 F-test of 12 /  22
H 0 : 12 /  22  1
H 1 : 12 /  22  1
F
s12
s 22
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 Equal-variances t-tests of 1   2
Memory:
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
262
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
Memory-Player Memory-Non
73.12
67.34
69.99
86.60
50
50
78.29
0
98
3.27
0.0008
1.6606
0.0015
1.9845
Reasoning:
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



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
Reasoning-Player Reasoning-Non
78.22
72.16
41.15
67.40
50
50
54.28
0
98
4.11
0.0000
1.6606
0.0001
1.9845
Reaction time:
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



263
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
Reaction-Player Reaction-Non
70.60
71.22
31.18
26.09
50
50
28.64
0
98
-0.58
0.2819
1.6606
0.5637
1.9845
Vocabulary:
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



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
Vocabulary-Player Vocabulary-Non
82.22
80.12
60.66
81.70
50
50
71.18
0
98
1.24
0.1081
1.6606
0.2163
1.9845
Overall Conclusion: p-values are .0008, 0, .5637, and .2163. There is overwhelming evidence to
indicate that bridge-players score higher on memory and reasoning tests. There is not enough
evidence of a difference in reaction time and vocabulary between players and nonplayers.
14.17 t-test of  D
H0 :D  0
H1 :  D  0
264
t
xD   D
s D / nD
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
Price shown Price not shown
56.15
60.31
243.68
467.71
100
100
0.7903
0
99
-3.12
0.0012
1.6604
0.0024
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 
x  t / 2
s
n
t-Estimate: Mean
Mean
Standard Deviation
LCL
UCL
Calls
0.320
0.717
0.204
0.436
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  p2 (case 1)
H 0 : p1  p2 = 0
H 1 : p1  p2 > 0
265
z
( pˆ 1  pˆ 2 )
1
1 

pˆ (1  pˆ ) 
n
n
2 
 1
z-Test: Two Proportions
Exercisers Watchers
0.4250
0.3675
400
400
0
1.66
0.0482
1.6449
0.0964
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 = 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
H 1 : (1   2 )  0
t
( x1  x 2 )  ( 1   2 )
s12 s 22

n1 n 2
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 overwhelming evidence to support the professor's theory.
14.21 Unequal-variances t-test of 1   2
H 0 : (1   2 )  0
H 1 : (1   2 )  0
266
t
( x1  x 2 )  ( 1   2 )
s12 s 22

n1 n 2
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
Coupon used Coupon not used
49.42
46.29
200.89
317.90
143
357
0
327
2.07
0.0197
1.6495
0.0395
1.9672
t = 2.07, p-value = .0197. There is evidence to infer that shoppers with coupons spend more money on
groceries than do non-coupon users.
14.22 a z-test of p
H 0 : p = 104,320/425,000 = .245
H 1 : p > .245
z
pˆ  p
p(1  p)
n
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
Deliver
0.2825
400
0.245
1.74
0.0406
1.6449
0.0812
1.96
z = 1.74, p-value = .0406. There is evidence to indicate that the campaign will increase home delivery
sales.
b z-test of p
H 0 : p = 110,000/425,000 = .259
H 1 : p > .259
267
z
pˆ  p
p(1  p)
n
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
Deliver
0.2825
400
0.259
1.07
0.1417
1.6449
0.2834
1.96
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
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
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
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
268
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1
s 2p  
 n1 n 2



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
Expenses MSA Expenses Regular
347.24
479.25
21042.80
21127.51
63
141
21101.51
0
202
-6.00
0.0000
1.6524
0.0000
1.9718
t = -6.00, p-value = 0. There is overwhelming evidence to infer that medical expenses for those under
the MSA plan are lower than those who are not.
b z-test of p1  p2 (case 1)
H 0 : p1  p2 = 0
H 1 : p1  p2 < 0
z
( pˆ 1  pˆ 2 )
1
1 

pˆ (1  pˆ ) 
n
n
2 
 1
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
Health MSA Health Regular
0.7619
0.7801
63
141
0
-0.29
0.3867
1.6449
0.7734
1.96
z = -.29, p-value = .3867. There is not enough evidence to support the critics of MSA.
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Case 14.1 Memory test scores: Equal-variances t-tests of 1   2
H 0 : (1   2 )  0
H 1 : (1   2 )  0
t
( x1  x 2 )  (1   2 )
 1
1 

s 2p  
 n1 n 2 
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
Host Mark Announcer Mark
7.76
6.97
4.52
6.37
121
121
5.44
0
240
2.65
0.0043
1.6512
0.0087
1.9699
t = 2.65, p-value = .0087. There is overwhelming evidence to infer that the memory test scores of the
two groups of children differ.
Cereal choices: z-test of
p1  p2 (case 1) (success = 4, Canary Crunch)
H 0 : p1  p2 = 0
H 1 : p1  p2
z
0
( pˆ 1  pˆ 2 )
1
1 

pˆ (1  pˆ ) 
n
n
2 
 1
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
Host Cereal Announcer Cereal
0.3636
0.3058
121
121
0
0.95
0.1702
1.6449
0.3404
1.96
270
z = .95, p-value = .3404. There is not enough evidence to infer that there is a difference in proportions
between the two groups of children in their choice of the Canary Crunch.
Overall Conclusion: Children who watch the host commercials remember more about the details of
the commercial but are no more (or less) likely to choose the advertised cereal than children who
watch announcer commercials.
Case 14.2 a z-test of p (success = 1, vote "No")
H 0 : p = .5
H 1 : p > .5
z
pˆ  p
p(1  p)
n
z-Test: Proportion
Planned vote
0.5382
641
0.5
1.94
0.0265
1.6449
0.053
1.96
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  p2 (success = 2, vote "Yes")
( pˆ 1  pˆ 2 )  z  / 2
pˆ 1 (1  pˆ 1 ) pˆ 2 (1  pˆ 2 )

n1
n2
z-Estimate: Two Proportions
Sample Proportions
Observations
LCL
UCL
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%.
271