Download Chapter 7 Inferences Based on Two Samples: Confidence Intervals

Chapter 7
Inferences Based on Two Samples: Confidence
Intervals and Tests of Hypotheses
7.1
μ1 ± 2 σ x1  μ1 2
b.
μ2 ± 2 σ x2  μ2 ± 2
c.
μ x1  x2 = μ1 − μ2 = 150 − 150 = 0
σ x1  x2 
7.2
σ12
n1

n1
σ 22
n2
σ12
 150 ± 2

σ2
900
 150 ± 6  (144, 156)
100
 150 ± 2
n2
1600
 150 ± 8  (142, 158)
100
900 1600
2500


5
100 100
100

σ 22
900 1600

 0 ± 10  (−10, 10)
100 100
d.
(μ1 − μ2) ± 2
e.
The variability of the difference between the sample means is greater than the variability of the
individual sample means.
a.
μ x1 = μ1 = 12
σ x1 
b.
μ x2 = μ2 = 10
σ x2 
c.
μ x1  x2 = μ1 − μ2 = 12 − 10 = 2
σ x1  x2 
7.3
σ1
a.
σ12
n1
n1

σ 22
n2
n2

 (150 − 150) ± 2
42 32


64 64
σ1
n1

4
= .5
64

3
= .375
64
σ2
n2
25
= .625
64
d.
Since n1 ≥ 30 and n2 ≥ 30, the sampling distribution of x1  x2 is approximately normal by the
Central Limit Theorem.
a.
For confidence coefficient .95, α = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
The confidence interval is:
( x1  x2 )  z.025
σ12
n1

σ 22
n2
 (5,275 − 5,240) ± 1.96
1502 2002

400
400
 35 ± 24.5  (10.5, 59.5)
We are 95% confident that the difference between the population means is between 10.5 and 59.5.
376
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
b.
The test statistic is z =
( x1  x2 )  ( μ1  μ 2 )
σ12
n1

σ 22

(5275  5240)  0
150 2 200 2

400
400
n2
377
 2.8
The p-value of the test is P(z ≤ −2.8) + P(z ≥ 2.8) = 2P(z ≥ 2.8) = 2(.5 − .4974)
= 2(.0026) = .0052
Since the p-value is so small, there is evidence to reject H0. There is evidence to indicate the two
population means are different for α > .0052.
c.
The p-value would be half of the p-value in part b. The p-value = P(z ≥ 2.8) = .5 − .4974 = .0026.
Since the p-value is so small, there is evidence to reject H0. There is evidence to indicate the mean
for population 1 is larger than the mean for population 2 for
α > .0026.
d.
The test statistic is z =
( x1  x2 )  ( μ1  μ2 )
σ12
n1

σ 22

(5275  5240)  25
n2
150 2 2002

400
400
 .8
The p-value of the test is P(z ≤ −.8) + P(z ≥ .8) = 2P(z ≥ .8) = 2(.5 − .2881) = 2(.2119) = .4238
Since the p-value is so large, there is no evidence to reject H0. There is no evidence to indicate that
the difference in the 2 population means is different from 25 for α ≤ .10.
e.
7.4
We must assume that we have two independent random samples.
Assumptions about the two populations:
1.
2.
Both sampled populations have relative frequency distributions that are approximately normal.
The population variances are equal.
Assumptions about the two samples:
The samples are randomly and independently selected from the population.
7.5
7.6
a.
No. Both populations must be normal.
b.
No. Both populations variances must be equal.
c.
No. Both populations must be normal.
d.
Yes.
e.
No. Both populations must be normal.
a.
sp2 =
(n1  1) s12  (n2  1) s22 (25  1)120  (25  1)100 5280


= 110
25  25  2
48
n1  n2  2
b.
sp2 =
(20  1)12  (10  1)20 408

= 14.5714
20  10  2
28
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
378
Chapter 7
c.
sp2 =
(6  1).15  (10  1).2 2.55

= .1821
6  10  2
14
d.
sp2 =
(16  1)3000  (17  1)2500 85,000

= 2741.9355
16  17  2
31
e.
sp2 falls near the variance with the larger sample size.
7.7
Some preliminary calculations are:
x1 
x
1
n1

11.8
 2.36
5
s12 

x12
 x 

2
1
n1  1
n1

 x 

(11.8) 2
5
= .733
5 1
30.78 
2
x2 

x2
n2

14.4
 3.6
4
s22 
x
2
2
2
n2  1
n2

(14.4)2
4
= .42
4 1
53.1 
(n1  1) s12  (n2  1) s22 (5  1).773 + (4  1).42 4.192
=
=
= .5989
n1  n2  2
5+42
7
a.
sp2 
b.
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 < 0
The test statistic is t =
( x1  x2 )  D0
1 1 
sp2   
 n1 n2 

(2.36  3.6)  0
1
.5989  
5
1

4

1.24
 2.39
.5191
The rejection region requires α = .10 in the lower tail of the t-distribution with df =
n1 + n2 − 2 = 5 + 4 − 2 = 7. From Table V, Appendix B, t.10 = 1.415. The rejection region is t <
−1.415.
Since the test statistic falls in the rejection region (t = −2.39 < −1.415), H0 is rejected. There is
sufficient evidence to indicate that μ2 > μ1 at α = .10.
c.
A small sample confidence interval is needed because n1 = 5 < 30 and n2 = 4 < 30.
For confidence coefficient .90, α = .10 and α/2 = .05. From Table V, Appendix B, with df = n1 + n2 −
2 = 5 + 4 − 2 = 7, t.05 = 1.895. The 90% confidence interval for (μ1 − μ2) is:
1 1 
1
( x1  x2 )  t.05 sp2     (2.36 − 3.6) ± 1.895 .5989  
5
 n1 n2 
1

4
 −1.24 ± .98  (−2.22, −0.26)
d.
The confidence interval in part c provides more information about (μ1 − μ2) than the test of
hypothesis in part b. The test in part b only tells us that μ2 is greater than μ1. However, the
confidence interval estimates what the difference is between μ1 and μ2.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.8
σ12
σ 22
σ x1  x2 
b.
The sampling distribution of
x1  x2 is approximately normal
by the Central Limit Theorem
since n1 ≥ 30 and n2 ≥ 30.
n2

9
16

 .25 = .5
100 100
a.
n1

379
μ x1  x2 = μ1 − μ2 = 10
c.
x1 − x2 = 15.5 − 26.6 = −11.1
Yes, it appears that x1  x2 = −11.1 contradicts the null hypothesis H0: μ1 − μ2 = 10.
d.
The rejection region requires α/2 = .025 = .05/2 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
e.
H0: μ1 − μ2 = 10
Ha: μ1 − μ2 ≠ 10
The test statistic is z =
( x1  x2 )  10
σ12
n1

σ 22

(15.5  26.6)  10
= −42.2
.5
n2
The rejection region is z < −1.96 or z > 1.96. (Refer to part d.)
Since the observed value of the test statistic falls in the rejection region (z = −42.2 < −1.96), H0 is
rejected. There is sufficient evidence to indicate the difference in the population means is not equal
to 10 at α = .05.
f.
The form of the confidence interval is:
( x1  x2 )  zα / 2
σ12
n1

σ 22
n2
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z.025 = 1.96. The confidence interval is:
9
16

(15.5 − 26.6) ± 1.96
 −11.1 ± .98  (−12.08, −10.12)
100 100
We are 95% confident that the difference in the two means is between −12.08 and −10.12.
g.
The confidence interval gives more information.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
380
7.9
Chapter 7
a.
The p-value = .1150. Since the p-value is not small, there is no evidence to reject H0 for α ≤ .10.
There is insufficient evidence to indicate the two population means differ for α ≤ .10.
b.
If the alternative hypothesis had been one-tailed, the p-value would be half of the value for the twotailed test. Here, p-value = .1150/2 = .0575.
There is no evidence to reject H0 for α = .05. There is insufficient evidence to indicate the mean for
population 1 is less than the mean for population 2 at α = .05.
There is evidence to reject H0 for α > .0575. There is sufficient evidence to indicate the mean for
population 1 is less than the mean for population 2 at α > .0575.
7.10
Some preliminary calculations:
x1 
s12 
x1 
x
1
n1
x
2
1
x

654
15
 x 

n1  1
2
n2
2
6542
15  419.6 = 29.3167
15  1
14
1

n1
28934 

858
= 53.625
16
 x 

2
s22 
sp2 
a.

x22
2
n2  1
n2

8582
16  439.75 = 29.3167
16  1
15
46450 
(n1  1) s12  (n2  1) s22 (15  1)29.9714  (16  1)29.3167 859.3501


= 29.6328
n1  n2  2
15  16  2
29
H0: μ2 − μ1 = 10
Ha: μ2 − μ1 > 10
The test statistic is t =
( x1  x2 )  D0
1 1 
sp2   
 n1 n2 

(53.625  43.6)  10
1 1
29.6328   
 15 16 

.025
= .013
1.9564
The rejection region requires α = .01 in the upper tail of the t-distribution with df = n1 + n2 − 2 = 15 +
16 − 2 = 29. From Table V, Appendix B, t.01 = 2.462. The rejection region is t > 2.462.
Since the test statistic does not fall in the rejection region (t = .013  2.462), H0 is not rejected.
There is insufficient evidence to conclude μ2 − μ1 > 10 at α = .01.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
b.
381
For confidence coefficient .98, α = .02 and α/2 = .01. From Table V, Appendix B, with df = n1 + n2
− 2 = 15 + 16 − 2 = 29, t.01 = 2.462. The 98% confidence interval for (μ2 − μ1) is:
1 1 
1 1
( x1  x2 )  tα / 2 sp2     (53.625 − 43.6) ± 2.462 29.6328   
 15 16 
 n1 n2 
 10.025 ± 4.817  (5.208, 14.842)
We are 98% confident that the difference between the mean of population 2 and the mean of
population 1 is between 5.208 and 14.842.
7.11
a.
s 2p =
(n1  1) s12  (n2  1) s22
(17 -1)3.42  (12  1)4.82
= 16.237
=
17  12  2
n1  n2  2
H0: (μ1 − μ2) = 0
Ha: (μ1 − μ2) ≠ 0
The test statistic is t =
( x1  x2 )  0
sp2
1 1
 n  n 
1
2
=
(5.4  7.9)  0
1
1
16.237  + 
 17 12 
=  1.646
Since no α was given, we will use α = .05. The rejection region requires α/2 = .05/2 = .025 in each
tail of the t-distribution with df = n1 + n2 − 2 = 17 + 12 – 2 = 27. From Table V, Appendix B,
t.025 = 2.052. The rejection region is t < −2.052 or t > 2.052.
Since the observed value of the test statistic does not fall in the rejection region (t = −1.646  −2.052),
H0 is not rejected. There is insufficient evidence to indicate μ1 − μ2 is different from 0 at α = .05.
b.
For confidence coefficient .95, α = .05 and α/2 = .025. From Table V, Appendix B, with
df = n1 + n2 − 2 = 17 + 12 − 2 = 27, t.025 = 2.052. The confidence interval is:
1 1
( x1  x2 )  t.025 sp2    where t has 27 df
 n1 n2 
1
1
 (5.4 − 7.9) − 2.052 16.237     −2.50 ± 3.12  (−5.62, 0.62)
 17 12 
7.12
Let μ1 = the mean test score of students on the SAT reading test in classrooms that used educational
software and μ 2 = the mean test score of students on SAT reading test in classrooms that did not use the
technology
a.
The parameter of interest is μ1 – μ2.
b.
The null and alternative hypotheses for the test are:
H0: μ1 – μ2 = 0
Ha: μ1 – μ2 > 0
c.
Since the p-value of the test is so large (p = 0.62), we would not reject H0 for any reasonable value of
α. There is insufficient evidence to indicate that the mean test score of the students on the SAT
reading test was significantly higher in classrooms using reading software products than in
classrooms that did not use educational software. This agrees with the conclusion of the DOE.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
382
7.13
Chapter 7
a.
Let μ1 = mean number of items recalled by those in the video only group and μ2 = mean number of
items recalled by those in the audio and video group. To determine if the mean number of items
recalled by the two groups is the same, we test:
Ho: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
b.
s 2p 
n1  1 s12  n2  1 s22  20  11.982  20  1 2.132  4.22865
n1  n2  2
The test statistic is t 
 x1  x2   Do
1 1 
s 2p   
 n1 n2 

3.70  3.30  0
1 
 1
4.22865   
 20 20 

0.4
 0.62
.65028
c.
The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = n1 + n2 − 2 =
20 + 20 – 2 = 38. From Table V, Appendix B, t.05 ≈ 1.684. The rejection region is t < −1.684 or t >
1.684.
d.
Since the observed value of the test statistic does not fall in the rejection region
(t = 0.62  1.684), Ho is not rejected. There is insufficient evidence to indicate a difference in the
mean number of items recalled by the two groups at α = .10.
e.
The p-value is p = .542. This is the probability of observing our test statistic or anything more
unusual if H0 is true. Since the p-value is not less than α = .10, there is no evidence to reject H0 .
There is insufficient evidence to indicate a difference in the mean number of items recalled by the
two groups at α = .10.
f.
We must assume:
1.
2.
3.
7.14
20  20  2
a.
Both populations are normal
Random and independent samples
σ12  σ 22
Let μ1 = mean score for males and μ2 = mean score for females. For confidence
coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.025 = 1.645. The 90%
confidence interval is:
( x1  x2 )  zα / 2
σ12
n1

σ 22
n2
 (39.08  38.79)  1.645
6.732 6.942

127
114
 0.29  1.452  (1.162, 1.742)
We are 90% confident that the difference in mean service-rating scores between males and females.
b.
Because 0 falls in the 90% confidence interval, we are 90% confident that there is no difference in
the mean service-rating scores between males and females.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.15
a.
383
Let µ1 = mean forecast error of buy-side analysts and µ2 = mean forecast error of sell-side analysts.
For confidence coefficient 0.95, α = .05 and α /2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96. The 95% confidence interval is:
( x1  x2 )  z.025
σ12
n1

σ 22
n2
 (.85  (.05))  1.96
1.932
.852

 .90  .064  (.836, .964)
3,526 58,562
We are 95% confident that the difference in the mean forecast error of buy-side analysts and sell-side
analysts is between .836 and .964.
b.
Based on 95% confidence interval in part a, the buy-side analysts has the greater mean forecast error
because our interval contains positive numbers.
c.
The assumptions about the underlying populations of forecast errors that are necessary for the
validity of the inference are:
1.
2.
7.16
The samples are randomly and independently sampled.
The sample sizes are sufficiently large.
a.
No. Just looking at the sample means, as the students went from no solution to check figures, the
sample mean improvement score increased. However, as the students went from check figures to
complete solutions, the sample mean improvement score dropped to below the no solution group.
b.
The problem with using only the sample means to make inferences about the population mean
knowledge gains for the three groups of students is that we don’t know the variability or the “spread”
of the probability distributions of the populations.
c.
Let μ1 = mean knowledge gain for students in the “no solutions” group and μ2 = mean knowledge
gain for students in the “check figures” group. To determine if the test score improvement decreases
as the level of assistance increases, we test:
H0: μ1 –μ2 = 0
Ha: μ1 – μ2 > 0
d.
Since the observed significance level of the test is not less than α = .05 (p = .8248  .05), do not
reject H0. There is insufficient evidence to indicate that the mean knowledge gain of students in the
“no solutions” group is greater than the mean knowledge gain of students in the “check figures”
group at α = .05.
e.
Let μ3 = mean knowledge gain of students in the “completed solutions” group. To determine if the
test score improvement decreases as the level of assistance increases, we test:
H0: μ2 – μ3 = 0
Ha: μ2 – μ3 > 0
f.
Since the observed significance level of the test is not less than α = .05 (p = .1849  .05), do not
reject H0. There is insufficient evidence to indicate that the mean knowledge gain of students in the
“check figures” group is greater than the mean knowledge gain of students in the “complete
solutions” group at α = .05.
g.
To determine if the test score improvement decreases as the level of assistance increases, we test:
H0: μ1 – μ3 = 0
Ha: μ1 – μ3 > 0
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
384
7.17
Chapter 7
h.
Since the observed significance level of the test is not less than α = .05 (p = .2726  .05), do not
reject H0. There is insufficient evidence to indicate that the mean knowledge gain of students in the
“no solutions” group is greater than the mean knowledge gain of students in the “complete solutions”
group at α = .05.
a.
The descriptive statistics are:
Descriptive Statistics: Text-line, Witness-line, Intersection
Variable
Text-lin
WitnessIntersec
N
3
6
5
Mean
0.3830
0.3042
0.3290
Median
0.3740
0.2955
0.3190
TrMean
0.3830
0.3042
0.3290
Variable
Text-lin
WitnessIntersec
Minimum
0.3350
0.1880
0.2850
Maximum
0.4400
0.4390
0.3930
Q1
0.3350
0.2045
0.2900
Q3
0.4400
0.4075
0.3730
StDev
0.0531
0.1015
0.0443
SE Mean
0.0306
0.0415
0.0198
Let μ1 = mean zinc measurement for the text-line, μ2 = mean zinc measurement for the
line, and μ3 = mean zinc measurement for the intersection.
s 2p 
witness-
(n1  1) s12  (n3  1) s32 (3  1).05312  (5  1).04432

 .00225
35 2
n1  n3  2
For α = .05, α/2 = .05/2 = .025. Using Table V, Appendix B, with df = n1 + n2 − 2 =
3 + 5 – 2 = 6, t.025 = 2.447. The 95% confidence interval is:
1 1
1 1
( x1  x3 )  tα / 2 s 2p     (.3830  .3290)  2.447 .00225   
3 5
n
n
 1
3
 0.0540  .0848  (0.0308, 0.1388)
We are 95% confident that the difference in mean zinc level between text-line and
between −0.0308 and 0.1388.
intersection is
To determine if there is a difference in the mean zinc measurement between text-line and intersection,
we test:
H0: μ1 = μ3
Ha: μ1 ≠ μ3
The test statistic is t 
( x1  x3 )  Do
s 2p
1 1
 n  n 
1
3

(.3830  .3290)  0
1 1
.00225   
3 5
 1.56
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n1 + n2 − 2 =
3 + 5 – 2 = 6. From Table V, Appendix B, t.025 = 2.447. The rejection region is t < −2.447 or t >
2.447.
Since the observed value of the test statistic does not fall in the rejection region (t = 1.56  2.365), H0
is not rejected. There is insufficient evidence to indicate a difference in the mean zinc measurement
between text-line and intersection at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
b.
s 2p
n2  1 s22  n3  1 s32


n2  n3  2

385
(6  1).10152  (5  1).04432
 .006596
652
For α = .05, α/2 = .05/2 = .025. Using Table V, Appendix B, with
df = n1 + n2 – 2 = 6 + 5 – 2 = 9, t.025 = 2.262. The 95% confidence interval is:
1
1
1 1
( x2  x3 )  tα / 2 s 2p     .3042  .3290  2.262 .006596   
6 5
 n2 n3 
 .0248  .1112  (.1361, .0864)
We are 95% confident that the difference in mean zinc level between witness-line and intersection is
between −0.1361 and 0.0864.
To determine if the difference in mean zinc measurement between the witness-line and the
intersection, we test:
H0: μ2 = μ3
Ha: μ2 ≠ μ3
The test statistic is t 
 x2  x3   Do
s 2p
 1
1 
 n  n 
2
3

.3042  .3290  0
1 1
.006596   
6 5
 .50
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution. From Table V,
Appendix B, with df = n1 + n2 – 2 = 6 + 5 – 2 = 9, t.025 = 2.262. The rejection region is t < −2.262 or
t > 2.262.
Since the observed value of the test statistic does not fall in the rejection region (t = −.50 </ −2.262),
H0 is not rejected. There is insufficient evidence to indicate a difference in mean zinc measurement
between witness-line and intersection at
α = .05.
c.
If we order the sample means, the largest is Text-line, the next largest is intersection and the smallest
is witness-line. In parts a and b, we found that text-line is not different from the intersection and that
the witness-line is not different from the intersection. However, we cannot make any decisions about
the difference between the witness-line and the text-line.
d.
In order for the above inferences to be valid, we must assume:
1.
The three samples are randomly selected in an independent manner from the three target
populations.
2.
All three sampled populations have distributions that are approximately normal.
3.
All three population variances are equal (i.e. σ 12 = σ 22 = σ 32 )
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
386
7.18
Chapter 7
Let μ1 = the mean relational intimacy score for participants in the CMC group and μ2 = the mean relational
intimacy score for participants in the FTF group.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: CMC, FTF
Variable
CMC
FTF
N
24
24
N*
0
0
Mean
3.500
3.542
SE Mean
0.159
0.134
StDev
0.780
0.658
Minimum
2.000
2.000
Q1
3.000
3.000
Median
3.500
4.000
Q3
4.000
4.000
Maximum
5.000
5.000
Some preliminary calculations are:
s 2p 
n1  1 s12  n2  1 s22  24  1.7802  24  1.6582  0.5207
n1  n2  2
24  24  2
To determine if the mean relational intimacy score for participants in the CMC group is lower than the
mean relational intimacy score for participants in the FTF group, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 < 0
The test statistic is t 
 x1  x2   Do
1 1 
s 2p   
 n1 n2 

3.500  3.542  0  0.042  .20
1 
 1
.5207   
 24 24 
.20831
The rejection region requires α= .10 in the lower tail of the t-distribution with df = n1 + n2 – 2 =
24 + 24 – 2 = 46. From Table V, Appendix B, t.10 ≈1.303. The rejection region is t < −1.303.
Since the observed value of the test statistic does not fall in the rejection region
(t = −.20  −1.303), H0 is not rejected. There is insufficient evidence to indicate that the
mean relational intimacy score for participants in the CMC group is lower than the mean
relational intimacy score for participants in the FTF group at α = .10.
7.19
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Control, Rude
Variable
Rude
Control
N
45
53
Mean
8.511
11.81
StDev
3.992
7.38
Minimum
0.000
0.00
Q1
5.500
5.50
Median
9.000
12.00
Q3
11.000
17.50
Maximum
18.000
30.00
Let μ1 = mean performance level of students in the rudeness group and μ2 = mean performance level of
students in the control group. To determine if the true performance level for students in the rudeness
condition is lower than the true mean performance level for students in the control group, we test:
H0: µ1 − µ2 = 0
Ha: µ1 − µ2 < 0
The test statistic is z 
( x1  x2 )  0
σ12
n1

σ 22
n2

(8.511  11.81)  0
3.9222 7.382

45
53
 2.81
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
387
The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix B,
z.01 = 2.33. The rejection region is z < −2.33.
Since the observed value of the test statistic falls in the rejection region (z = −2.81 < −2.33), H0 is rejected.
There is sufficient evidence to indicate the true mean performance level for students in the rudeness
condition is lower than the true mean performance level for students in the control group at α = .01.
7.20
a.
The first population is the set of responses for all business students who have access to lecture notes
and the second population is the set of responses for all business students not having access to lecture
notes.
b.
To determine if there is a difference in the mean response of the two groups, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
The test statistic is z =
( x1  x2 )  0
s12
n1

s22
n2

(8.48  7.80)  0
= 2.19
.94 2.99

86
35
The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.58. The rejection region is z < −2.58 or z > 2.58.
Since the observed value of the test statistic does not fall in the rejection region (z = 2.19  2.58), H0
is not rejected. There is insufficient evidence to indicate a difference in the mean response of the two
groups at α = .01.
c.
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 =
2.58. The confidence interval is:
( x1  x2 )  z.005
s12 s22
.94 2.99

 (8.48 − 7.80) ± 2.58

n1 n2
86
35
 .68 ± .801  (−.121, 1.481)
We are 99% confident that the difference in the mean response between the two groups is between
−.121 and 1.481.
d.
A 95% confidence interval would be smaller than the 99% confidence interval. The z value used in
the 95% confidence interval is z.025 = 1.96 compared with the z value used in the 99% confidence
interval of z.005 = 2.58.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
388
7.21
Chapter 7
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Honey, DM
Variable
Honey
DM
N
35
33
Mean
10.714
8.333
StDev
2.855
3.256
Minimum
4.000
3.000
Q1
9.000
6.000
Median
11.000
9.000
Q3
12.000
11.500
Maximum
16.000
15.000
Let μ1 = mean improvement in total cough symptoms score for children receiving the Honey dosage and
μ2 = mean improvement in total cough symptoms score for children receiving the DM dosage. To test if
honey may be a preferable treatment for the cough and sleep difficulty associated with childhood upper
respiratory tract infection, we test:
H0: μ1 - μ2 = 0
Ha: μ1 - μ2 > 0
The test statistic is z 
( x1  x2 )  0
σ12
n1

σ 22
n2

(10.714  8.333)  0
2.8552 3.2562

35
33
 3.20
Since no α was given, we will use α = .05.The rejection region requires α = .05 in the upper tail of the
z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 3.20 > 1.645), H0 is rejected.
There is sufficient evidence to indicate that honey may be a preferable treatment for the cough and sleep
difficulty associated with childhood upper respiratory tract infection at α = .05.
7.22
a.
The bacteria counts are probably normally distributed because each count is the median of
five measurements from the same specimen.
b.
Let μ1 = mean of the bacteria count for the discharge and μ2 = mean of the bacteria count upstream.
Since we want to test if the mean of the bacteria count for the discharge exceeds the mean of the
count upstream, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Plant, Upstream
Variable
Plant
Upstream
N
6
6
Mean
32.10
29.617
Median
31.75
30.000
TrMean
32.10
29.617
Variable
Plant
Upstream
Minimum
28.20
26.400
Maximum
36.20
32.300
Q1
29.40
27.075
Q3
35.23
31.850
s 2p 
StDev
3.19
2.355
(n1  1) s12  (n2  1) s22 (6  1)3.192  (6  1)2.3552

 7.861
n1  n2  2
662
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
SE Mean
1.30
0.961
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
The test statistic is t 
( x1  x2 )  0
s 2p
1 1 
 n  n 
1
2

(32.10  29.617)  0
1 1
7.861   
6 6
389
 1.53
No α level was given, so we will use α = .05. The rejection region requires α = .05 in the upper tail of
the t-distribution with df = n1 + n2 − 2 = 6 + 6 – 2 = 10. From Table V, Appendix B, t.05 = 1.812. The
rejection region is t > 1.812.
Since the observed value of the test statistic does not fall in the rejection region (t = 1.53  1.812), H0
is not rejected. There is insufficient evidence to indicate the mean bacteria count for the discharge
exceeds the mean of the count upstream at α = .05.
d.
We must assume:
1.
2.
3.
7.23
a.
The mean counts per specimen for each location is normally distributed.
The variances of the 2 distributions are equal.
Independent and random samples were selected from each population.
Let μ1 = the mean heat rates of traditional augmented gas turbines and μ2 = the mean heat rates of
aeroderivative augmented gas turbines.
Some preliminary calculations are:
s 2p 
n1  1 s12  n2  1 s22  39  112792  7  1 26522  2,371,831.409
n1  n2  2
39  7  2
To determine if there is a difference in the mean heat rates for traditional augmented gas turbines and
the mean heat rates of aeroderivative augmented gas turbines, we test:
Ho: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
The test statistic is
t
 x1  x2   Do
1 1 
s 2p   
 n1 n2 

11,544  12,312  0
 1 1
2,371,831.409   
 39 7 

768
 1.21
632.1782
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with
df = n1 + n2 – 2 = 39 + 7 – 2 = 44. From Table V, Appendix B, t.025 ≈ 2.021. The rejection region
is t < −2.021 or t > 2.021.
Since the observed value of the test statistic does not fall in the rejection region
(t = −1.20  −2.021), H0 is not rejected. There is insufficient evidence to indicate that there is a
difference in the mean heat rates for traditional augmented gas turbines and the mean heat rates of
aeroderivative augmented gas turbines at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
390
Chapter 7
b.
Let μ3 = the mean heat rates of advanced augmented gas turbines and μ2 = the mean heat rates of
aeroderivative augmented gas turbines.
Some preliminary calculations are:
s 2p 
n3  1 s32  n2  1 s22  21  1 6392  7  1 26522
n3  n2  2
21  7  2
 1,937,117.077
To determine if there is a difference in the mean heat rates for traditional augmented gas turbines and
the mean heat rates of aeroderivative augmented gas turbines, we test:
H0: μ3 − μ2 = 0
Ha: μ3 − μ2 ≠ 0
The test statistic is
t
 x3  x2   Do
1
1
s 2p   
n
n
 3
2

9,764  12,312  0
 1 1
1,937,117.077   
 21 7 

2,548
 4.19
607.4329
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with
df = n1 + n2 – 2 = 21 + 7 – 2 = 26. From Table V, Appendix B, t.025 ≈ 2.056. The rejection region
is t < −2.056 or t > 2.056.
Since the observed value of the test statistic falls in the rejection region (t = −4.19 < −2.021), H0 is
rejected. There is sufficient evidence to indicate that there is a difference in the mean heat rates for
advanced augmented gas turbines and the mean heat rates of aeroderivative augmented gas turbines
at α = .05.
7.24
a.
We cannot make inferences about the difference between the mean salaries of male and female
accounting/finance/banking professionals because no standard deviations are provided.
b.
To determine if the mean salary for males is significantly greater than that for females, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645.
To make things easier, we will assume that the standard deviations for the 2 groups are the same.
The test statistic is
z
 x1  x2   Do  69, 484  52,012  0 
 σ12
n
 1


n2 
σ 22
1 
 1

 1400 1400 
σ2 
17,836
σ (.037796)

471,896.2038
σ
In order to reject H0 this test statistic must fall in the rejection region, or be greater than 1.645.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
391
Solving for σ we get:
471,896.2038
z
 1.645  σ 
σ
471,896.2038
 286,866.99
1.645
Thus, to reject H0 the average of the two standard deviations has to be less than $286,866.99.
7.25
7.26
c.
Yes. In fact, reasonable values for the standard deviation will be around $5,000. which is much
smaller than the required $286,866.99.
d.
These data were collected from voluntary subjects who responded to a Web-based survey. Thus, this
is not a random sample, but a self-selected sample. Generally, subjects who respond to surveys tend
to have very strong opinions, which may not be the same as the population in general. Thus, the
results from this self-selected sample may not reflect the results from the population in general.
a.
The rejection region requires α = .05 in the upper tail of the t-distribution with
df = nd − 1 = 12 − 1 = 11. From Table V, Appendix B, t.05 = 1.796. The rejection region
is t > 1.796.
b.
From Table V, with df = nd − 1 = 24 − 1 = 23, t.10 = 1.319. The rejection region is
t > 1.319.
c.
From Table V, with df = nd − 1 = 4 − 1 = 3, t.025 = 3.182. The rejection region is t > 3.182.
d.
Using Minitab, with df = nd − 1 = 80 − 1 = 79, t.01 = 2.374. The rejection region is t > 2.374.
a.
Pair
Difference
1
2
3
4
5
6
3
2
2
4
0
1
nd
d 
sd2 
b.
d
i 1
nd
i
=
12
=2
6
 nd 
 di 
nd
 i 1 
di2 
nd
i 1


nd  1
2

(12)2 
34


6 

=
=2
5
μd = μ1 − μ2
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
392
Chapter 7
c.
For confidence coefficient .95, α = .05 and α/2 = .025. From Table V, Appendix B, with
df = nD − 1 = 6 − 1 = 5, t.025 = 2.571. The confidence interval is:
d  tα / 2
d.
sd
2
 2.571
 2 ± 1.484  (.516, 3.484)
nd
6
H0: μd = 0
Ha: μd ≠ 0
The test statistic is t = t 
d
sd
nd

2
= 3.46
2/ 6
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = nD − 1 = 6
− 1 = 5. From Table V, Appendix B, t.025 = 2.571. The rejection region is t < −2.571 or t > 2.571.
Since the observed value of the test statistic falls in the rejection region (3.46 > 2.571), H0 is rejected.
There is sufficient evidence to indicate that the mean difference is different from 0 at α = .05.
7.27
Let μ1 = mean of population 1 and μ2 = mean of population 2.
a.
H0: μd = 0
Ha: μd < 0 where μd = μ1 − μ2
b.
Some preliminary calculations are:
Pair
Population 1
Population 2
1
2
3
4
5
6
7
8
9
10
19
25
31
52
49
34
59
47
17
51
24
27
36
53
55
34
66
51
20
55
nd
d 

di
i 1
nd

37
 3.7
10
The test statistic is t 
d
sd
nd
sd2 

Difference,
d
−5
−2
−5
−1
−6
0
−7
−4
−3
−4
 nd 
 di 
nd
 i 1 
2
di 
nd
i 1


nd  1
2

2
37 

181 
10
10  1
3.7
 5.29
4.9
10
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
 4.9
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
393
The rejection region requires α = .10 in the lower tail of the t-distribution with df = nd – 1 = 10 – 1 = 9.
From Table V, Appendix B, t.10 = 1.383. The rejection region is t < −1.383.
Since the observed value of the test statistic falls in the rejection region (t = −5.29
< −1.383), H0 is rejected. There is sufficient evidence to indicate the mean of population 1 is less
than the mean for population 2 at α = .10.
c.
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table V, Appendix B, with
df = nd – 1 = 10 – 1 = 9, t.05 = 1.833. The 90% confidence interval is:
d  tα / 2
sd
nd
 3.7  1.833
4.9
10
 3.7  1.28  (4.98,  2.42)
We are 90% confident that the difference in the two population means is between −4.98 and −2.42.
7.28
d.
We must assume that the population of differences is normal, and the sample of differences is
randomly selected.
a.
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 < 0
The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.10 = 1.28. The rejection region is z < −1.28.
b.
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 < 0
The test statistic is z 
d  0 3.5  0

 4.71 .
sd
21
nd
38
The rejection region is z < −1.28 (Refer to part a.)
Since the observed value of the test statistic falls in the rejection region (z = −4.71 < −1.28), H0 is
rejected. There is sufficient evidence to indicate μ1 − μ2 < 0 at α = .10.
c.
Since the sample size of the number of pairs is greater than 30, we do not need to assume that the
population of differences is normal. The sampling distribution of d is approximately normal by the
Central Limit Theorem. We must assume that the differences are randomly selected.
d.
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 =
1.645. The 90% confidence interval is:
d  z.05
e.
sd
21
 3.5  1.645
 3.5  1.223  (4.723,  2.277)
38
nd
The confidence interval provides more information since it gives an interval of possible values for
the difference between the population means.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
394
7.29
Chapter 7
a.
Some preliminary calculations:
nd
d
sd2 
d
i 1
nd
i

468
 11.7
40
 nd 
 di 
nd
 i 1 
2
di 
nd
i 1


2
nd  1

4682
40  36.0103
40  1
6,880 
To determine if (μ1 − μ2) = μd is different from 10, we test:
H0: μd = 10
Ha: μd ≠ 10
The test statistic is z 
d  D0
11.7  10

 1.79
sd
36.0103
nd
40
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.79  1.96), H0
is not rejected. There is insufficient evidence to indicate (μ1 − μ2) = μd is different from 10 at
α = .05.
7.30
b.
The p-value is p = P(z ≤ −1.79) + P(z ≥ 1.79) = (.5 − .4633) + (.5 − .4633) = .0367 + .0367 = .0734.
The probability of observing our test statistic or anything more unusual if H0 is true is .0734. Since
this p-value is not small, there is no evidence to indicate (μ1 − μ2) = μd is different from 10 at α = .05.
c.
No, we do not need to assume that the population of differences is normally distributed. Because our
sample size is 40, the Central Limit Theorem applies.
a.
The data should be analyzed using a paired-difference test because that is how the data were
collected. Evaluation scores were collected twice from each agency, once in 2007 and once in 2008.
Since the two sets of data are not independent, they cannot be analyzed using independent samples.
b.
The differences between the 2008 score and the 2007 score for each agency are:
Agency
GSA
Agriculture
Social Security
USAID
Defense
Score07
34
33
33
32
17
Score08
40
35
33
42
32
Difference
(08-07.
6
2
0
10
15
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
c.
395
The mean and standard deviation of the differences are:
d

di
nd

33
 6.6
5
sd2 
d
2
i
 d 

2
i
nd
nd  1

332
5  147.2  36.8
5 1
4
365 
sd  sd2  36.8  6.066
7.31
d  Do 6.6  0

 2.435
6.066
sd
5
nd
d.
The test statistic is t 
e.
The rejection requires α = 0.10 in the upper tail of the t-distribution with df = nd − 1 = 5 − 1 = 4.
From Table V, Appendix B, t.10 = 1.533. The rejection region is t > 1.533.
f.
Since the observed value of the test statistic falls in the rejection region (t = 2.433 > 1.533), H0 is
rejected. There is sufficient evidence to indicate that the true mean evaluation score of government
agencies in 2008 exceeds the true mean evaluation score in 2007 at α = 0.10.
a.
Let μ1 = the mean salary of technology professionals in 2003 and μ2 = the mean salary of technology
professionals in 2005. Let μd = μ1 − μ2.
To determine if the mean salary of technology professionals at all U.S. metropolitan areas has
increased between 2003 and 2005, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 < 0
H0: μd = 0
OR
Ha: μd < 0
b.
Metro Area
Silicon Valley
New York
Washington, D.C.
Los Angeles
Denver
Boston
Atlanta
Chicago
Philadelphia
San Diego
Seattle
Dallas-Ft. Worth
Detroit
2003 Salary
($ thousands)
87.7
78.6
71.4
70.8
73.0
76.3
73.6
71.1
69.5
69.0
71.0
73.0
62.3
2005 Salary
($ thousands)
85.9
80.3
77.4
77.1
77.1
80.1
73.2
73.0
69.8
77.1
66.9
71.0
64.1
Difference
(2003 – 2005)
1.8
−1.7
−6.0
−6.3
−4.1
−3.8
0.4
−1.9
−0.3
−8.1
4.1
2.0
−1.8
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
396
Chapter 7
nd
c.
d 
d
i
1
nd
nd
sd2 

1
di2


25.7
 1.98
13



nd

1

di 

nd
nd  1
2

(25.7) 2
13
 12.9819
13  1
206.59 
sd  sd2  12.9819  3.6
d  μo
The test statistic is t 
e.
The rejection region requires α = .10 in the lower tail of the t-distribution with
df = nd – 1 = 13 – 1 = 12. From Table V, Appendix B, t.10 = 1.356. The rejection region is
t < −1.356.
f.
Since the observed value of the test statistic falls in the rejection region (t = −1.98 < −1.356), H0 is
rejected. There is sufficient evidence to indicate the mean salary of technology professionals at all
U.S. metropolitan areas has increased between 2003 and 2005 at α = .10.
g.
In order for the inference to be valid, we must assume that the population of differences is normal
and that we have a random sample.
sd
nd

1.98  0
d.
3.6
13
 1.98
Using MINITAB, the histogram of the differences is:
Histogram of Diff
3.0
Fr equency
2.5
2.0
1.5
1.0
0.5
0.0
-7.5
-5.0
-2.5
0.0
2.5
5.0
Diff
The graph is fairly mound-shaped although it is somewhat skewed to the right. Since there are only
13 observations, this graph is close enough to being mound-shaped to indicate the normal assumption
is reasonable.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.32
397
a.
The data should be analyzed as a paired difference experiment because each actor who
won an Academy Award was paired with another actor with similar characteristics who did not win
the award.
b.
Let μ1 = mean life expectancy of Academy Award winners and μ2 = mean life expectancy of nonAcademy Award winners. To compare the mean life expectancies of Academy Award winners and
non-winners, we test:
H0: μ1 − μ2 = μd = 0
Ha: μd ≠ 0
7.33
c.
Since the p-value was so small, there is sufficient evidence to indicate the mean life expectancies of
the Academy Award winners and non-winners are different for any value of α > .003. Since the
sample mean life expectancy of Academy Award winners is greater than that for non-winners, we
can conclude that Academy Award winners have a longer mean life expectancy than non-winners.
a.
The data should be analyzed using a paired-difference test because that is how the data were
collected. Response rates were observed twice from each survey using the “not selling” introduction
method and the standard introduction method. Since the two sets of data are not independent, they
cannot be analyzed using independent samples.
b.
Some preliminary calculations are:
s 2p 
(n1  1) s12  (n2  1) s22 (29  1)(.12) 2  (29  1)(.11)2

 .01325
n1  n2  2
29  29  2
Let μ1 = mean response rate for those using the “not selling” introduction and μ2 = mean response
rate for those using the standard introduction. Using the independent-samples t-test to determine if
the mean response rate for “not selling” is higher than that for the standard introduction, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0
The test statistic is t 
( x1  x2 )  0
s 2p
1
1 
 n  n 
1
2

(.262  .246)  0
1 
 1
.01325   
 29 29 
 .53
The rejection region requires α = .05 in the upper tail of the t-distribution with df = n1 + n2 -2 = 29 +
29 – 2 = 56. From Table V, Appendix B, t.05 ≈ 1.671. The rejection region is t > 1. 671.
Since the observed value of the test statistic does not fall in the rejection region (t =.53  1.671), H0
is not rejected. There is insufficient evidence to indicate the mean response rate for “not selling” is
higher than that for the standard introduction at α =.05.
c.
Since p-value is less than α =.05 (p = .001 < .05), H0 is rejected. There is sufficient evidence to
indicate the mean response rate for “not selling” is higher than that for the standard introduction at
α =.05.
d.
The two inferences in parts b and c have different results because using the independent samples ttest is not appropriate for this study. The paired-difference design is better. There is much variation
in response rates from survey to survey. By using the paired difference design, we can eliminate the
survey to survey differences.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
398
7.34
Chapter 7
a.
Let μ1 = mean driver chest injury rating and μ2 = mean passenger chest injury rating. Because the
data are paired, we are interested in μ1 − μ2 = μd, the difference in mean chest injury ratings between
drivers and passengers.
b.
The data were collected as matched pairs and thus, must be analyzed as matched pairs. Two ratings
are obtained for each car – the driver’s chest injury rating and the passenger’s chest injury rating.
c.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: DrivChst, PassChst, diff
Variable
DrivChst
PassChst
diff
N
98
98
98
Mean
49.663
50.224
-0.561
Median
50.000
50.500
0.000
TrMean
49.682
50.148
-0.420
Variable
DrivChst
PassChst
diff
Minimum
34.000
35.000
-15.000
Maximum
68.000
69.000
13.000
Q1
45.000
45.000
-4.000
Q3
54.000
55.000
3.000
StDev
6.670
7.107
5.517
SE Mean
0.674
0.718
0.557
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 =
2.58. The 99% confidence interval is:
d  z.005
7.35
sd
5.517
 0.561  2.58
 0.561  1.438  (1.999, 0.877)
nd
98
d.
We are 99% confidence that the difference between the mean chest injury ratings of drivers and
front-seat passengers is between −1.999 and 0.877. Since 0 is in the confidence interval, there is no
evidence that the true mean driver chest injury rating exceeds the true mean passenger chest injury
rating.
e.
Since the sample size is large, the sampling distribution of d is approximately normal by the Central
Limit Theorem. We must assume that the differences are randomly selected.
Some preliminary calculations are:
Operator
1
2
3
4
5
6
7
8
9
10
d =
Difference
(Before - After)
5
3
9
7
2
−2
−1
11
0
5
 d = 39 = 3.9
nd
10
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
 d 

399
2
sd2 
d
2
nE
nd  1

39 2
10 = 18.5444
10  1
319 
sd =
18.5444 = 4.3063
a.
To determine if the new napping policy reduced the mean number of customer complaints, we test:
H0: μd = 0
Ha: μd > 0
The test statistic is t =
d  0 3.9  0

= 2.864
4.3063
sd
10
nd
The rejection region requires α = .05 in the upper tail of the t-distribution with df = nd − 1 = 10 − 1 =
9. From Table V, Appendix B, t.05 = 1.833. The rejection region is t > 1.833.
Since the observed value of the test statistic falls in the rejection region (t = 2.864 > 1.833), H0 is
rejected. There is sufficient evidence to indicate the new napping policy reduced the mean number
of customer complaints at α = .05.
b.
In order for the above test to be valid, we must assume that
1. The population of differences is normal
2. The differences are randomly selected
7.36
a.
Let μC1 = mean relational intimacy score for the CMC group on the first meeting and μC3 = mean
relational intimacy score for the CMC group on the third meeting. Let μCd = difference in mean
relational intimacy score between the first and third meetings for the CMC group. To determine if
the mean relational intimacy score will increase between the first and third meetings, we test:
Ho: μCd = 0
Ha: μCd < 0
b.
The researchers used the paired t-test because the same individuals participated in each of the three
meeting sessions. Thus, the samples would not be independent.
c.
Since the p-value is so small (p = .003), H0 would be rejected. There is sufficient evidence to
indicate that the mean relational intimacy score for participants in the CMC group increased from the
first to the third meeting for any value of α > .003.
d.
Let μF1 = mean relational intimacy score for the FTF group on the first meeting and μF3 = mean
relational intimacy score for the FTF group on the third meeting. Let μFd = difference in mean
relational intimacy score between the first and third meetings for the FTF group. To determine if the
mean relational intimacy score will change between the first and third meetings, we test:
H0: μFd = 0
Ha: μFd ≠ 0
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
400
Chapter 7
e.
7.37
Since the p-value is not small (p = .39), H0 would be not be rejected. There is insufficient evidence
to indicate that the mean relational intimacy score for participants in the FTF group changed from the
first to the third meeting for any value of α < .39.
Some preliminary calculations are:
Circuit
Standard Method
1
2
3
4
5
6
7
8
9
10
11
.80
.80
.83
.53
.50
.96
.99
.98
.81
.95
.99
Huffman-coding
Method
.78
.80
.86
.53
.51
.68
.82
.72
.45
.79
.77
Difference
.02
.00
-.03
.00
-.01
.28
.17
.26
.36
.16
.22
nd
d 
nd
sd2 

1
d
1
nd
di2 



i

nd

1
1.43
 0.13
11

di 

nd
nd  1
2

(1.43) 2
11  0.0194
11  1
0.3799 
sd  sd2  0.0194  0.1393
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table V, Appendix B, with
df = nd – 1 = 11 – 1 = 10, t.025 = 2.228. The 95% confidence interval is:
d  t.025
sd
.1393
 .13  2.228
 .13  .094  (0.036, 0.224)
11
n
We are 95% confident that the true difference in mean compression ratio between the standard method and
the Huffman-based coding method is between 0.036 and 0.224. Since 0 is not contained in the interval, we
can conclude there is a difference in mean compression ratios between the two methods. Since the values
of the confidence interval are positive, we can conclude that the mean compression ratio for the Huffmanbased method is smaller than the standard method.
7.38
Let μ1 = mean number of crashes caused by red light running per intersection before camera is installed and
μ2 = mean number of crashes caused by red light running per intersection after camera is installed. The data
are collected as paired data, so we will analyze the data using a paired t-test. Then, let μd = μ1 − μ2.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
401
Using MINITAB to compute the differences (Di) and the summary statistics, the results are:
Descriptive Statistics: Before, After, Di
Variable
Before
After
Di
N
13
13
13
Mean
2.513
1.506
1.007
StDev
1.976
1.448
1.209
Minimum
0.270
0.000
-0.850
Q1
0.805
0.260
0.265
Median
2.400
1.360
0.560
Q3
3.405
2.380
2.335
Maximum
7.350
4.920
2.780
To determine if the mean number of crashes caused by red light running has been reduced since the
installation of red light cameras, we test:
Ho: µd = 0
Ha: µd > 0
The test statistic is t 
d  Do 1.007  0

 3.003
1.209
sd
13
nd
The rejection region requires α = .05 in the upper tail of the t-distribution with df = nd − 1 = 13 − 1 = 12.
From Table V, Appendix B, t.05 = 1.782. The rejection region is t > 1.782.
Since the observed value of the test statistic falls in the rejection region (t = 3.003 > 1.782), H0 is rejected.
There is sufficient evidence to indicate that photo-red enforcement program is effective in reducing redlight-running crash incidents at intersections at α = 0.05.
7.39
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Male, Female, Diff
Variable
Male
Female
Diff
N
19
19
19
Mean
5.895
5.526
0.368
Median
6.000
5.000
1.000
TrMean
5.706
5.294
0.294
Variable
Male
Female
Diff
Minimum
3.000
3.000
-5.000
Maximum
12.000
12.000
7.000
Q1
4.000
4.000
-3.000
Q3
8.000
7.000
3.000
StDev
2.378
2.458
3.515
SE Mean
0.546
0.564
0.806
Let μ1 = mean number of swims by male rat pups and μ2 = mean number of swims by female rat pups.
Then μd = μ1 − μ2. To determine if there is a difference in the mean number of swims required by male and
female rat pups, we test:
H0: μd = 0
Ha: μd ≠ 0
The test statistic is t 
d  Do .368  0

 0.46
3.515
sd
19
nd
The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = nd – 1 = 19 – 1 =
18. From Table V, Appendix B, t.05 = 1.734. The rejection region is t < −1.734 or t > 1.734.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
402
Chapter 7
Since the observed value of the test statistic does not fall in the rejection region (t = 0.46  1.734), H0 is
not rejected. There is insufficient evidence to indicate that there is a difference in the mean number of
swims required by male and female rat pups at α = .10.
(using Minitab, the p-value ≈ .653.)
Since the sample size is not large, we must assume that the population of differences is normally
distributed and that the sample of differences is random. There is no indication that the sample differences
are not from a random sample. However, because the number of swims is discrete, the differences are
probably not normal.
7.40
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: HMETER, HSTATIC, Diff
Variable
HMETER
HSTATIC
Diff
N
40
40
40
N*
0
0
0
Mean
1.0405
1.0410
-0.000523
Variable
HMETER
HSTATIC
Diff
Median
1.0232
1.0237
-0.000165
Q3
1.0883
1.0908
0.000317
SE Mean
0.00638
0.00649
0.000204
StDev
0.0403
0.0410
0.001291
Minimum
0.9936
0.9930
-0.004480
Q1
1.0047
1.0043
-0.001078
Maximum
1.1026
1.1052
0.001580
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table V, Appendix B, with
df = nd – 1 = 40 – 1 = 39, t.025 ≈ 2.021. The 95% confidence interval is:
sd
0.001291
 0.000523  2.021
 0.000523  0.000413
n
40
 (0.000936,  0.000110)
d  t.025
We are 95% confident that the true difference in mean density measurements between the two methods is
between -0.000936 and -0.000110. Since the absolute value of this interval is completely less than the
desired maximum difference of .002, the winery should choose the alternative method of measuring wine
density.
7.41
a.
From the exercise, we know that x1 and x2 are binomial random variables with the number of trials
x
equal to n1 and n2. From Chapter 7, we know that for large n, the distribution of pˆ1  1 is
n1
approximately normal. Since x1 is simply p̂1 multiplied by a constant, x1 will also have an
approximate normal distribution. Similarly, the distribution of pˆ 2 
x2
is approximately normal,
n2
and thus, the distribution of x2 is approximately normal.
b.
The Central Limit Theorem is necessary to find the sampling distributions of p̂1 and p̂2 when n1 and
n2 are large. Once we have established that both p̂1 and p̂2 have normal distributions, then the
distribution of their difference will also be normal.
7.42
a.
The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.01 = 2.33. The rejection region is z < −2.33.
b.
The rejection region requires α = .025 in the lower tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.43
c.
The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z < −1.645.
d.
The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.10 = 1.28. The rejection region is z < −1.28.
From Section 5.4, it was given that the distribution of p̂ is approximately normal if npˆ  15 and nqˆ  15 .
a.
n1 pˆ1  12(.42)  5.04  15 and n1 qˆ1  12(.58)  6.96  15
n2 pˆ 2  14(.57)  7.98  15 and n2 qˆ2  14(.43)  6.02  15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1  pˆ 2 ) is
approximately normal.
b.
n1 pˆ1  12(.92)  11.04  15 and n1 qˆ1  12(.08)  0.96  15
n2 pˆ 2  14(.86)  12.04  15 and n2 qˆ2  14(.14)  1.96  15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1  pˆ 2 ) is
approximately normal.
c.
n1 pˆ1  30(.70)  21  15 and n1 qˆ1  30(.30)  9  15
n2 pˆ 2  30(.73)  21.9  15 and n2 qˆ2  30(.27)  8.1  15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1  pˆ 2 ) is
approximately normal.
d.
n1 pˆ1  100(.93)  93  15 and n1 qˆ1  100(.07)  7  15
n2 pˆ 2  250(.97)  242.5  15 and n2 qˆ2  250(.03)  7.5  15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1  pˆ 2 ) is
approximately normal.
e.
n1 pˆ1  125(.08)  10  15 and n1 qˆ1  125(.92)  115  15
n2 pˆ 2  200(.12)  24  15 and n2 qˆ2  200(.88)  176  15
Thus, the sample sizes are not large enough to conclude the sampling distribution of ( pˆ1  pˆ 2 ) is
approximately normal.
7.44
403
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96. The 95% confidence interval for p1 − p2 is approximately:
a.
( pˆ1  pˆ 2 )  zα / 2
pˆ1 qˆ1 pˆ 2 qˆ2
.65(1  .65) .58(1  .58)

 (.65 − .58) ± 1.96

400
400
n1
n2
 .07 − .067  (.003, .137)
b.
( pˆ1  pˆ 2 )  zα / 2
pˆ1 qˆ1 pˆ 2 qˆ2
 (.31 − .25) − 1.96

n1
n2
.31(1  .31) .25(1  .25)

180
250
 .06 ± .086  (−.026, .146)
c.
( pˆ1  pˆ 2 )  zα / 2
pˆ1 qˆ1 pˆ 2 qˆ2
.46(1  .46) .61(1  .61)

 (.46 − .61) ±1.96

100
120
n1
n2
 −.15 ± .131  (−.281, −.019)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
404
7.45
Chapter 7
a.
H0: (p1 − p2) = 0
Ha: (p1 − p2) ≠ 0
Will need to calculate the following:
p̂1 =
320
400
320  400
= .40 p̂2 =
= .50 p̂ =
 .45
800
800
800  800
The test statistic is z =
 pˆ1  pˆ 2   0
1
1 
ˆˆ  
pq
 n1 n2 
=
(.40  .50)  0
1 
 1

(.45)(.55) 
 800 800 
= −4.02
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic falls in the rejection region (z = −4.02 < −1.96), H0 is
rejected. There is sufficient evidence to indicate that the proportions are unequal at α = .05.
b.
The problem is identical to part a until the rejection region. The rejection region requires α/2 = .01/2
= .005 in each tail of the z-distribution. From Table IV, Appendix B, z.005 = 2.58. The rejection
region is z < −2.58 or z > 2.58.
Since the observed value of the test statistic falls in the rejection region (z = −4.02 < −2.58), H0 is
rejected. There is sufficient evidence to indicate that the proportions are unequal at α = .01.
c.
H0: p1 − p2 = 0
Ha: p1 − p2 < 0
Test statistic as above: z = −4.02
The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix
B, z.01 = 2.33. The rejection region is z < −2.33.
d.
Since the observed value of the test statistic falls in the rejection region (z = −4.02 < −2.33), H0 is
rejected. There is sufficient evidence to indicate that p1 < p2 at α = .01.
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z05 =
1.645. The confidence interval is:
pˆ qˆ
pˆ pˆ
(.4)(.6) (.5)(.5)

( pˆ1  pˆ 2 )  z.05 1 1  2 2  (.4 − .5) ± (1.645)
n1
n2
800
800
 −.10 ± .04  (−.14, −.06)
We are 90% confident that the difference between p1 and p2 is between −.14 and −.06.
7.46
p̂ =
n1 pˆ1  n2 pˆ 2 55(.7)  65(.6) 78


= .65
n1  n2
55  65
120
q̂ = 1 − p̂ = 1 − .65 = .35
H0: p1 − p2 = 0
Ha: p1 − p2 > 0
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
The test statistic is z =
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

(.7  .6)  0
1 
1
.65(.35)   
 55 65 

405
.1
= 1.14
.08739
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B,
z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.14  1.645), H0 is
not rejected. There is insufficient evidence to indicate the proportion from population 1 is greater than that
for population 2 at α = .05.
7.47
a.
pˆ1 
x1
29

 .153
n1 189
b.
pˆ 2 
x2
32

 .215
n2 149
c.
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 =
1.645. The 90% confidence interval is:
( pˆ1  pˆ 2 )  zα / 2
pˆ1 qˆ1 pˆ 2 qˆ2
.153(.847) .215(.785)

 (.153  .215)  1.645

n1
n2
189
149
 .062  .070  (.132, .008)
7.48
d.
We are 90% confident that the difference in the proportion of bidders who fall prey to the winner’s
curse between super-experienced bidders and less-experienced bidders is between −.132 and .008.
Since this interval contains 0, there is no evidence to indicate that there is a difference in the
proportion of bidders who fall prey to the winner’s curse between super-experienced bidders and
less-experienced bidders.
a.
The point estimate for the proportion of all Democrats who prefer steak as their favorite barbeque
x
662
food is pˆ1  1 
 .530 .
n1 1, 250
b.
The point estimate for the proportion of all Republicans who prefer steak as their favorite barbeque
x
586
 .630 .
food is pˆ 2  2 
n2 930
c.
The point estimate for the difference between proportions of all Democrats and all Republicans who
prefer steak as their barbeque food is pˆ1  pˆ 2  .530  .630  .100 .
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
406
Chapter 7
d.
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96. The 95% confidence interval for the difference between the proportions of all Democrats
and all Republicans who prefer steak as their barbeque food is
 pˆ1  pˆ 2   z.025
pˆ1 qˆ1 pˆ 2 qˆ2
.53(.47) .63(.37)

 (.530  .630)  1.96

n1
n2
1, 250
930
 .100  .042  (.142,  .058)
7.49
e.
We are 95 percent confident that the difference in proportions of all Democrats and all Republicans
who prefer steak as their favorite barbeque food is between -.142 and -.058. Since this interval does
not contain 0, there is a sufficient evidence to indicate that there is a significant difference between
the proportions of all Democrats and all Republicans who prefer steak as their favorite barbeque food.
f.
“95% confident” means that in repeated sampling, 95% of all confidence intervals constructed in the
same manner will contain the true population difference in proportions and 5% will not.
a.
Let p1 = proportion of employed individuals who had a routine checkup in the past year and p2 =
proportion of unemployed individuals who had a routine checkup in the past year. The researchers
are interested in whether there is a difference in these two proportions, so the parameter of interest is
p 1 – p 2.
b.
To determine if there is a difference in the proportions of employed and unemployed
individuals who had a routine checkup in the past year, we test:
H0: p1 – p2 = 0
Ha: p1 – p2 ≠ 0
c.
Some preliminary calculations are:
pˆ1 
pˆ 
x1
642

 .563
n1 1,140
pˆ 2 
x2
740

 .669
n2 1,106
x1  x2
642  740
1, 382


 .615
n1  n2 1,140  1,106 2, 246
The test statistic is z 
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

qˆ  1  pˆ  1  .615  .385
(.563  .669)  0
 1
1 
.615(.385) 

 1,140 1,106 
 5.16
d.
The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.58. The rejection region is z < −2.58 or z > 2.58.
e.
The p-value is P(z ≤ −5.16) + P(z ≥ 5.16) ≈ (.5 − .5) + (.5 − .5) = 0. This agrees with what was
reported.
f.
Since the observed value of the test statistic falls in the rejection region (z = −5.16 < −2.58), H0 is
rejected. There is sufficient evidence to indicate a difference in the proportion of employed and
unemployed individuals who had routine checkups in the past year at α = .01.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.50
a.
407
Let p1 = proportion of men who prefer to keep track of appointments in their head and
p2 = proportion of women who prefer to keep track of appointments in their head. To determine if
the proportion of men who prefer to keep track of appointments in their head is greater than that of
women, we test:
H0: p1 − p2 = 0
Ha: p1 − p2 > 0
b.
pˆ 
n1 pˆ1  n2 pˆ 2 500(.56)  500(.46)

 .51 and qˆ  1  pˆ  1  .51  .49
n1  n2
500  500
The test statistic is z 
7.51
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
n
n
 1
2 

(.56  .46)  0
1 
 1

.51(.49) 
 500 500 
 3.16
c.
The rejection region requires α = .01 in the upper tail of the z distribution. From Table IV,
Appendix B, z.01 = 2.33. The rejection region is z > 2.33.
d.
The p-value is p = P(z ≥ 3.16) ≈ .5 − .5 = 0.
e.
Since the observed value of the test statistic falls in the rejection region (z = 3.16 > 2.33), H0 is
rejected. There is sufficient evidence to indicate the proportion of men who prefer to keep track of
appointments in their head is greater than that of women at α = .01.
a.
The two populations of interest are all male cell phone users and all female cell phone users.
b.
The estimate of the proportion of men who sometimes do not drive safely while talking or texting on
a cell phone is p̂1 = .32. The estimate of the proportion of women is p̂2 = .25.
c.
For confidence coefficient .90, α = .10 and α /2 = .10/2 = .05. From Table IV, Appendix B, z.05 =
1.645. A 90% confidence interval for the difference between the proportions of men and women who
sometimes do not drive safely while talking or texting on a cell phone is:
( pˆ1  pˆ 2 )  z.05
pˆ1 qˆ1 pˆ 2 qˆ2
.32(.68) .25(.75)

 (.32  .25)  1.645

n1
n2
643
643
 .07  .041  (.029, .111)
d.
e.
Since the interval does not contain 0, then there is a sufficient evidence to indicate that there is a
difference between the proportions of men and women who sometimes do not drive safely while
talking or texting on a cell phone. Also, the interval contains all positive values so we can conclude
that men are more likely than women to sometimes not drive safely while talking or texting on a cell
phone.
The estimate of the proportion of men who used their cell phone in an emergency is p̂1 = .71. The
estimate of the proportion of women is p̂2 = .77.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
408
Chapter 7
f.
Let p1 = proportion of men who used their cell phone in an emergency and p2 = the proportion of
women who used their cell phone in an emergency.
Some preliminary calculations are:
pˆ1 
pˆ 
x1
 x1  n1 pˆ1  643(.71)  456.53
n1
x1  x2 456.53  495.11

 .74
n1  n2
643  643
pˆ 2 
x2
 x2  n2 pˆ 2  643(.77)  495.11
n2
qˆ  1  pˆ  1  .74  .26
To determine whether the proportions of men and women who used their cell phones in an
emergency differ, we test:
H0: p1 – p2 = 0
Ha: p1 – p2 ≠ 0
The test statistic is z 
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

(.71  .77)  0
1 
 1
.74(.26) 

 643 643 
 2.45 .
The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z < −1.645 or z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = −2.45 < −1.645), H0 is
rejected. There is sufficient evidence to indicate the proportions of men and women who used their
cell phone in an emergency differ at α = .10.
7.52
a.
Let p1 = proportion of customers returning the printed survey and p2 = proportion of customers
returning the electronic survey. Some preliminary calculations are:
pˆ1 
x1 261

 .414
n1 631
pˆ 2 
x2 155

 .374
n2 414
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. The 90% confidence interval is:
( pˆ1  pˆ 2 )  z.05
pˆ1 qˆ1 pˆ 2 qˆ2
.414(.586) .374(.626)

 (.414  .374)  1.645

n1
n2
631
414
 .04  .051  (.011, .091)
We are 90% confidence that the difference in the response rates for the two types of surveys is
between −.011 and .091.
b.
Since the value .05 falls in the 90% confidence interval, it is not an unusual value. Thus, there is no
evidence that the difference in response rates is different from .05. The researchers would be able to
make this inference.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.53
a.
409
Let p1 = proportion of African-American drivers searched by the LAPD and p2 = proportion of white
drivers searched by the LAPD.
Some preliminary calculations are:
pˆ1 
pˆ 
x1 12, 016

 .195
n1 61, 688
pˆ 2 
x2
5, 312

 .050
n2 106, 892
x1  x2
12, 016  5, 312
17, 328


 .103
n1  n2 61, 688  106, 892 168, 580
To determine if the proportions of African-American and white drivers searched differs, we test:
H0: p1 – p2 = 0
Ha: p1 – p2 ≠ 0
The test statistic is z 
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

.195  .050
1
 1

.103(.897) 


 61, 688 106, 892 
 94.35
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic falls in the rejection region (z = 94.35 > 1.96), H0 is
rejected. There is sufficient evidence to indicate the proportions of African-American drivers and
white drivers searched differs at α = .05.
b.
Let p1 = proportion of ‘hits’ for African-American drivers searched by the LAPD and
p2 = proportion of ‘hits’ for white drivers searched by the LAPD.
Some preliminary calculations are:
pˆ1 
pˆ 
x1
5,134

 .427
n1 12, 016
pˆ 2 
x2 3, 006

 .566
n2 5, 312
x1  x2
5,134  3, 006
8,140


 .470
n1  n2 12, 016  5, 312 17, 328
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV,
z.025 = 1.96. The 95% confidence interval is:
( pˆ1  pˆ 2 )  z.025
Appendix B,
pˆ1 qˆ1 pˆ 2 qˆ2
.427(.573) .566(.434)

 (.427  .566)  1.96

n1
n2
12, 016
5, 312
 .139  .016  (.155, -.123)
We are 95% confident that the difference in ‘hit’ rates between African-American drivers and white
drivers searched by the LAPD is between −.155 and −.123.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
410
7.54
Chapter 7
Let p1 = proportion of patients in the angioplasty group who had subsequent heart attacks and
p2 = proportion of patients in the medication only group who had subsequent attacks.
Some preliminary calculations:
x
211
pˆ1  1 
 .184
n1 1,145
qˆ1  1  pˆ1  1  .184  .816
x2
202

 .177
n2 1,142
qˆ2  1  pˆ 2  1  .177  .823
pˆ 2 
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
A 95% confidence interval for the difference in the rate of heart attacks for the two groups is
( pˆ1  pˆ 2 )  z.025
pˆ1 qˆ1 pˆ 2 qˆ2
.184(.816) .177(.823)

 (.184  .177)  1.96

n1
n2
1,145
1,142
 .007  .032  (.025, .039)
Since this interval contains 0, there is insufficient evidence to indicate that there is a difference in the rate
of heart attacks between the angioplasty group and the medication only group at α = .05. Yes, we agree
with the study’s conclusion.
7.55
Let p1 = proportion of African American MBA students who begin their career as entrepreneurs and
p2 = proportion of white MBA students who begin their career as entrepreneurs.
Some preliminary calculations:
x
209
 .16
pˆ1  1 
n1 1, 304
pˆ 2 
x2
356

 .05
n2 7,120
qˆ1  1  pˆ1  1  .160  .84
qˆ2  1  pˆ 2  1  .05  .95
We can answer the question by using either a confidence interval or a test of hypothesis. We will use the
confidence interval. Since no α was given, we will us a 95% confidence interval. For confidence
coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. A 95%
confidence interval for the difference in the proportions of two groups is:
( pˆ1  pˆ 2 )  z.025
pˆ1 qˆ1 pˆ 2 qˆ2
.16(.84) .05(.95)

 (.16  .05)  1.96

n1
n2
1, 304
7,120
 .11  .021  (.089, .131)
Since this interval does not contain 0, then there is sufficient evidence to indicate that the proportion of
African American MBA students who begin their career as entrepreneurs is significantly different than the
proportion of White MBA students who begin their career as entrepreneurs. The positive values in the
interval mean that the African American MBA students are more likely to begin their career as
entrepreneurs than the White MBA students.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.56
411
Let p1 = accuracy rate for modules with correct code and p2 = accuracy rate for modules with defective
code.
Some preliminary calculations are:
pˆ1 
x1 400

 .891
n1 449
pˆ 2 
x2 20

 .408
n2 49
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58.
The 99% confidence interval is:
( pˆ1  pˆ 2 )  z.005
pˆ1 qˆ1 pˆ 2 qˆ2
.891(.109) .408(.592)

 (.891  .408)  2.58

n1
n2
449
49
 .483  .185  (.298, .668)
We are 99% confident that the difference in accuracy rates between modules with correct code and
modules with defective code is between .298 and .668.
7.57
a.
Let p1 = proportion of women who have food cravings and p2 = proportion of men who have food
cravings.
We know that pˆ1  .97 and pˆ 2  .67 .
We know that n1p1 > 15 and n1q1 > 15 in order for the test to be valid. Thus, n1(.97) > 15  n1 >
15/.97 ≈ 16 and n1(.03) > 15  n1 > 15/.03 = 500.
Also, n2p2 > 15 and n2q2 > 15. Thus, n2(.67) > 15  n2 > 15/.67 ≈ 23 and n2(.33) > . 15  n2 >
15/.33 ≈ 46.
7.58
b.
This study involved 1,000 McMaster University students. It is very dangerous to generalize the
results of this study to the general adult population of North America. The sample of students used
may not be representative of the population of interest.
a.
Let p1 = proportion of 9th grade boys who gambled weekly or daily in 1992 and p2 = proportion of 9th
grade boys who gambled weekly or daily in 1998. The researchers are interested in whether there is
a difference in these two proportions, so the parameter of interest is p1 – p2.
Some preliminary calculations are:
pˆ1 
pˆ 
x1
4, 684

 .218
n1 21, 484
pˆ 2 
x2
5, 313

 .229
n2 23,199
x1  x2
4, 684  5, 313
9, 997


 .224
n1  n2 21, 484  23,199 44, 683
qˆ  1  pˆ  1  .224  .776
To determine if there is a difference in the proportions of 9th grade boys who gambled weekly or
daily in 1992 and 1998, we test:
H0: p1 – p2 = 0
Ha: p1 – p2 ≠ 0
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
412
Chapter 7
The test statistic is z 
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
n
n
 1
2 

(.218  .229)  0
 1
1 
.224(.776) 

 21, 484 23,199 
 2.79
The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.58. The rejection region is z < −2.58 or z > 2.58.
Since the observed value of the test statistic falls in the rejection region (z = −2.79 < −2.58), H0 is
rejected. There is sufficient evidence to indicate a difference in the proportions of 9th grade boys
who gambled weekly or daily in 1992 and 1998 at α = .01.
b.
Yes. If samples sizes are large enough, differences can almost always be found. Suppose we
compute a 99% confidence interval. For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005.
From Table IV, Appendix B, z.005 = 2.58. The 99% confidence interval is:
( pˆ1  pˆ 2 )  zα / 2
pˆ1 qˆ1 pˆ 2 qˆ2
.218(.782) .229(.771)

 (.218  .229)  2.58

n1
n2
21, 484
23,199
 .011  .010  (.021,  .001)
We are 99% confident that the difference in the proportions of 9th grade boys who gambled weekly
or daily in 1992 and 1998 is between −.021 and −.001.
7.59
a.
For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From Table IV, Appendix
B, z.005 = 2.58.
n1  n 2 
b.
( ME )
2

2.582 .4(1  .4)  .7(1  .7) 
.01
2

2.99538
 29, 953.8  29, 954
.0001
For confidence coefficient .90, α = 1 - .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. Since we have no prior information about the proportions, we use p1 = p2 = .5 to get a
conservative estimate. For a width of .05, the margin of error is .025.
n1  n 2 
c.
( zα / 2 )2 ( p1 q1  p2 q2 )
( zα / 2 )2 ( p1 q1  p2 q2 )
( ME ) 2

(1.645)2 .5(1  .5)  .5(1  .5)
.0252
= 2164.82 ≈ 2165
From part b, z.05 = 1.645.
n1  n 2 
( zα / 2 ) 2 ( p1 q1  p2 q2 )
( ME )
2

(1.645) 2 .2(1  .2)  .3(1  .3) 
2
.03

1.00123
= 1112.48 ≈ 1113
.0009
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.60
a.
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix
B, z.025 = 1.96.
n1 = n2 =
b.
413
 zα / 2 2 σ12  σ 22 
ME 2
(1.96) 2 (152  17 2 )

3.22
= 192.83 ≈ 193
If the range of each population is 40, we would estimate σ by:
σ ≈ 60/4 = 15
For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From Table IV,
Appendix B, z.005 = 2.58.
n1 = n2 =
7.61
(2.58) 2 (152  152 )

= 46.80 ≈ 47
ME 2
82
For confidence coefficient .9, α = 1 − .9 = .1 and α/2 = .1/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. For a width of 1, the bound is .5.
n1 = n2 =
c.
 zα / 2 2 σ12  σ 22 
n1 = n2 =
 zα / 2 2 σ12  σ 22 
ME 2
(1.645) 2 (5.82  7.52 )

.52
= 143.96 ≈ 144
( zα / 2 )2 (σ12  σ 22 )
( ME ) 2
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96.
n1 = n2 =
7.62
1.962 (14  14)
1.82
= 33.2 ≈ 34
First, find the sample sizes needed for width 5, or margin of error 2.5.
For confidence coefficient .9, α = 1 − .9 = .1 and α/2 = .1/2 = .05. From Table IV, Appendix B, z.05 = 1.645.
n1 = n2 =
 zα / 2 2 σ12  σ 22 
ME 2

(1.645) 2 (102  102 )
2.52
= 86.59 ≈ 87
Thus, the necessary sample size from each population is 87. Therefore, sufficient funds have not been
allocated to meet the specifications since n1 = n2 = 100 are large enough samples.
7.63
For confidence coefficient 0.99, α = 1 − .99 = .01 and α/2 = .01/2 = 005. From the Table IV, Appendix B,
z.005 = 2.58.
n1  n1 
 zα / 2 2 (σ12  σ 22 )
 ME 2

2.582 (12  12 )
.52
 53.25  54
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
414
7.64
Chapter 7
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
n1 = n2 =
 zα / 2 2 σ12  σ 22 
( ME )2

1.962 (3.1892  2.3552 )
1.52
= 26.8 ≈ 27
We would need to sample 27 specimens from each location.
7.65
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. If we assume that we do not know the return rates,
n1  n2 
7.66
1.6452 (.5(.5)  .5(.5))
.012
 13, 530.1  13, 531
 zα / 2 2 ( p1 q1  p2 q2 )
( ME ) 2

1.6452 .5(.5)  .5(.5) 
.022
= 3,382.5 ≈ 3,383
For confidence coefficient .95, α = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
n1  n 2 
7.68
ME 2

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. Since no information is given about the values of p1 and p2, we will be conservative and use .5
for both. A width of .04 means the bound is .04/2 = .02.
n1 = n2 =
7.67
 z.05 2 ( p1 q1  p2 q2 )
a.
( zα / 2 )2 (σ12  σ 22 )
( ME ) 2
1.962 (152  152 )
12
= 1728.72 ≈ 1729
For confidence coefficient .80, α = 1 − .80 = .20 and α/2 = .20/2 = .10. From Table IV, Appendix B,
z.10 = 1.28. Since we have no prior information about the proportions, we use p1 = p2 = .5 to get a
conservative estimate. For a width of .06, the margin of error is .03.
 zα / 2 2 ( p1 q1  p2 q2 )

(1.28)2 .5(1  .5)  .5(1  .5) 
= 910.22 ≈ 911
ME 2
.032
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. Using the formula for the sample size needed to estimate a proportion from Chapter 7,
n1 = n2 =
b.

n
 zα / 2 2
ME 2
pq

1.6452 .5(1  .5)
.022

.6765
= 1691.27 ≈ 1692
.0004
No, the sample size from part a is not large enough.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.69
a.
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96.
n1  n2 
7.70
7.72
 zα / 2 2 ( p1 q1  p2 q2 )
 ME 2

1.962 (.184(.816)  .177(.823))
.0152
 5, 050.7  5, 051
b.
The study would involve 5,051 × 2 = 10,102 patients. A study this large would be extremely time
consuming and expensive.
c.
Since a difference of .015 is so small, the practical significance detecting a 0.015 difference may not
be very worthwhile. A difference of .015 is so close to 0, that it might not make any difference.
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
n1 = n2 =
7.71
415
 zα / 2 2 σ12  σ 22 
( ME )2

1.962 (352  802 )
102
= 292.9 ≈ 293
a.
With ν1 = 9 and ν2 = 6, F.05 = 4.10.
b.
With ν1 = 18 and ν2 = 14, F.01 ≈ 3.57. (Since ν1 = 18 is not given, we estimate the value between
those for ν1 = 15 and ν1 = 20.)
c.
With ν1 = 11 and ν2 = 4, F.025 ≈ 8.81. (Since ν1 = 11 is not given, we estimate the value by averaging
those given for ν1 = 10 and ν1 = 12.)
d.
With ν1 = 20 and ν2 = 5, F.10 = 3.21.
a.
With ν1 = 2 and ν2 = 30,
P(F ≥ 5.39) = .01 (Table X, Appendix B)
b.
With ν1 = 24 and ν2 = 10,
P(F ≥ 2.74) = .05 (Table VIII, Appendix B)
Thus, P(F < 2.74) = 1 − P(F ≥ 2.74) = 1 − .05 = .95.
c.
With ν1 = 7 and ν2 = 1,
P(F ≥ 236.8) = .05 (Table VIII, Appendix B)
Thus, P(F < 236.8) = 1 − P(F ≥ 236.8) = 1 − .05 = .95.
d.
With ν1 = 40 and ν2 = 40,
P(F > 2.11) = .01 (Table X, Appendix B)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
416
7.73
7.74
7.75
Chapter 7
a.
Reject H0 if F > F.10 = 1.74. (From Table VII, Appendix B, with ν1 = 30 and ν2 = 20.)
b.
Reject H0 if F > F.05 = 2.04. (From Table VIII, Appendix B, with ν1 = 30 and ν2 = 20.)
c.
Reject H0 if F > F.025 = 2.35. (From Table IX.)
d.
Reject H0 if F > F.01 = 2.78. (From Table X.)
To test H0: σ12  σ 22 against Ha: σ12  σ 22 , the rejection region is F > Fα/2 with ν1 = 10 and ν2 = 12.
a.
α = .20, α/2 = .10
Reject H0 if F > F.10 = 2.19 (Table VII, Appendix B)
b.
α = .10, α/2 = .05
Reject H0 if F > F.05 = 2.75 (Table VIII, Appendix B)
c.
α = .05, α/2 = .025
Reject H0 if F > F.025 = 3.37 (Table IX, Appendix B)
d.
α = .02, α/2 = .01
Reject H0 if F > F.01 = 4.30 (Table X, Appendix B)
a.
The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = n1 − 1 = 25 − 1
= 24 and ν2 = n2 − 1 = 20 − 1 = 19. From Table VIII, Appendix B, F.05 = 2.11. The rejection region
is F > 2.11 (if s12 > s22 ).
b.
The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = n2 − 1 = 15 − 1
= 14 and ν2 = n1 − 1 = 10 − 1 = 9. From Table VIII, Appendix B, F.05 ≈ 3.01. The rejection region is
F > 3.01 (if s22 > s12 ).
c.
The rejection region requires α/2 = .10/2 = .05 in the upper tail of the F-distribution. If s12 > s22 , ν1
= n1 − 1 = 21 − 1 = 20 and ν2 = n2 − 1 = 31 − 1 = 30. From Table VIII, Appendix B, F.05 = 1.93. The
rejection region is F > 1.93. If s12 < s22 , ν1 = n2 − 1 = 30 and ν2 = n1 − 1 = 20. From Table VIII,
F.05 = 2.04. The rejection region is F > 2.04.
d.
The rejection region requires α = .01 in the upper tail of the F-distribution with ν1 = n2 − 1 = 41 − 1
= 40 and ν2 = n1 − 1 = 31 − 1 = 30. From Table X, Appendix B, F.01 = 2.30. The rejection region is
F > 2.30 (if s22 > s12 ).
e.
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution. If s12 > s22 , ν1
= n1 − 1 = 7 − 1 = 6 and ν2 = n2 − 1 = 16 − 1 = 15. From Table IX, Appendix B, F.025 = 3.41. The
rejection region is F > 3.41. If s12 < s22 , ν1 = n2 − 1 = 15 and ν2 = n1 − 1 = 6. From Table IX,
Appendix B, F.025 = 5.27. The rejection region is F > 5.27.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.76
a.
417
To determine if a difference exists between the population variances, we test:
H0: σ12  σ 22
Ha: σ12  σ 22
The test statistic is F =
s22
s12
8.75
= 2.26
3.87

The rejection region requires α/2 = .10/2 = .05 in the upper tail of the F-distribution with ν1 = n2 − 1
= 27 − 1 = 26 and ν2 = n1 − 1 = 12 − 1 = 11. From Table VIII, Appendix B, F.05 ≈ 2.60. The
rejection region is F > 2.60.
Since the observed value of the test statistic does not fall in the rejection region (F = 2.26  2.60),
H0 is not rejected. There is insufficient evidence to indicate a difference between the population
variances.
b.
The p-value is 2P(F ≥ 2.26). From Tables VII and VIII, with ν1 = 26 and ν2 = 11,
2(.05) < 2P(F ≥ 2.26) < 2(.10)  .10 < 2P(F ≥ 2.26) < .20
There is no evidence to reject H0 for α ≤ .10.
7.77
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Sample 1, Sample 2
Variable
Sample 1
Sample 2
N
6
5
Mean
2.417
4.36
Median
2.400
3.70
TrMean
2.417
4.36
Variable
Sample 1
Sample 2
Minimum
0.700
1.40
Maximum
4.400
8.90
Q1
1.075
1.84
Q3
3.650
7.20
StDev
1.436
2.97
SE Mean
0.586
1.33
To determine if the variance for population 2 is greater than that for population 1, we test:
H0: σ12  σ 22
Ha: σ12  σ 22
The test statistic is F =
s22
s12
=
2.972
1.4362
 4.28
The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = n2 − 1 = 5 − 1 =
4 and ν2 = n1 − 1 = 6 − 1 = 5. From Table VIII, Appendix B, F.05 = 5.19. The rejection region is
F > 5.19.
Since the observed value of the test statistic does not fall in the rejection region (F = 4.29  5.19),
H0 is not rejected. There is insufficient evidence to indicate the variance for population 2 is greater
than that for population 1 at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
418
Chapter 7
b.
The p-value is P(F ≥ 4.28). From Tables VII and VIII, with ν1 = 4 and ν2 = 5,
.05 < P(F ≥ 4.28) < .10
There is no evidence to reject H0 for α < .05 but there is evidence to reject H0 for α = .10.
7.78
Let σ12 = variance of the number of ads recalled by children in the video only group and σ 22 = variance of
the number of ads recalled by children in the A/V group
a.
To determine if the group variances are equal, we test:
H0:
σ12  σ 22
Ha: σ12  σ 22
b.
The test statistic is:
F
7.79
s 2 2.132
larger sample variance
 22 
 1.157
smaller sample variance s1 1.982
c.
The rejection region requires α/2 = 0.10/2 = .05 in the upper tail of the F-distribution with ν2 = n2 −
1 = 20 − 1 = 19 and ν1 = n1 − 1 = 20 − 1 = 19. From the Table VIII, Appendix B, F.05  2.16. The
rejection region is F > 2.16.
d.
Since the observed value of the test statistic does not fall in the rejection region (F = 1.157  2.16),
H0 is not rejected. There is insufficient evidence to indicate the variances of the number of ads
recalled by the children in the video-only group and the A/V group differ at α = 0.10.
e.
Since we could not reject H0 that the variances were equal, it indicates that the assumption of equal
variances is probably valid. The inference about the population means is probably valid.
a.
2
To determine if σ M
is less than σ F2 , we test:
2
H0: σ M
= σ F2
2
H a: σ M
< σ F2
s2
Larger sample variance
6.942
 2F 
 1.06
Smaller sample variance sM
6.732
b.
The test statistic is F 
c.
The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 =
nF – 1 = 114 – 1 = 113 and ν2 = nM – 1 = 127 – 1 = 126. From Table VII, Appendix B, F.10 ≈ 1.26.
The rejection region is F > 1.26.
d.
The p-value is P(F ≥ 1.06). From Table VII, Appendix B, with ν1 = 113 and ν2 = 126,
.10 < P(F ≥ 1.06) or p > .10
e.
Since the observed value of the test statistic does not fall in the rejection region (F = 1.063  1.26),
H0 is not rejected. There is insufficient evidence to indicate the variation in the male perception
scores is less than that for females at α = .10.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.80
419
f.
We must assume that we have two random and independent samples drawn from normal
populations.
a.
The amount of variability of GHQ scores tells us how similar or different the members of the group
are on GHQ scores. The larger the variability, the larger the differences are among the members on
the GHQ scores. The smaller the variability, the smaller the differences are among the members on
the GHQ scores.
b.
Let σ12 = variance of the mental health scores of the employed and σ 22 = variance of the mental
health scores of the unemployed. To determine if the variability in mental health scores differs for
employed and unemployed workers, we test:
H0: σ12  σ 22
Ha: σ12  σ 22
c.
The test statistic is F =
Larger sample variance s12 5.102


= 2.45
Smaller sample variance s22 3.262
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n2 − 1
= 49 − 1 = 48 and ν2 = n1 − 1 = 142 − 1 = 141. From Table IX, Appendix B, F.025 ≈ 1.61. The
rejection region is F > 1.61.
Since the observed value of the test statistic falls in the rejection region (F = 2.45 > 1.61), H0 is
rejected. There is sufficient evidence to indicate that the variability in mental health scores differs
for employed and unemployed workers for α = 05.
d.
7.81
We must assume that the 2 populations of mental health scores are normally distributed. We must
also assume that we selected 2 independent random samples.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Novice, Experienced
Variable
Novice
Experien
N
12
12
Mean
32.83
20.58
Median
32.00
19.50
TrMean
32.60
20.60
Variable
Novice
Experien
Minimum
20.00
10.00
Maximum
48.00
31.00
Q1
26.75
17.25
Q3
39.00
24.75
a.
StDev
8.64
5.74
SE Mean
2.49
1.66
Let σ12 = variance in inspection errors for novice inspectors and σ 22 = variance in inspection errors
for experienced inspectors. Since we wish to determine if the data support the belief that the
variance is lower for experienced inspectors than for novice inspectors, we test:
H0: σ12  σ 22
Ha: σ12  σ 22
The test statistic is F =
Larger sample variance s 12 8.642
 
= 2.27
Smaller sample variance s 22 5.742
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
420
Chapter 7
The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = n1 − 1 = 12 − 1
= 11 and ν2 = n2 − 1 = 12 − 1 = 11. From Table VIII, Appendix B, F.05 ≈ 2.82 (using interpolation).
The rejection region is F > 2.82.
Since the observed value of the test statistic does not fall in the rejection region (F = 2.27  2.82),
H0 is not rejected. The sample data do not support her belief at α = .05.
b.
7.82
The p-value = P(F ≥ 2.27) with ν1 = 11 and ν2 = 11. Checking Tables VII, VIII, IX, and X in
Appendix B, we find F.10 = 2.23 and F.05 = 2.82. Since the observed value of F exceeds F.10 but is
less than F.05, the observed significance level for the test is less than .10. So .05 < p-value < .10.
Let σ12 = variance zinc measurements from the text-line, σ 22 = variance zinc measurements from the
witness-line, and σ 32 = variance zinc measurements from the intersection. Using MINITAB, the descriptive
statistics are:
Descriptive Statistics: Text-line, Witness-line, Intersection
Variable
Text-lin
WitnessIntersec
N
3
6
5
Mean
0.3830
0.3042
0.3290
Median
0.3740
0.2955
0.3190
TrMean
0.3830
0.3042
0.3290
Variable
Text-lin
WitnessIntersec
Minimum
0.3350
0.1880
0.2850
Maximum
0.4400
0.4390
0.3930
Q1
0.3350
0.2045
0.2900
Q3
0.4400
0.4075
0.3730
a.
StDev
0.0531
0.1015
0.0443
SE Mean
0.0306
0.0415
0.0198
To determine if the variation in the zinc measurements for the text-line and the intersection differ, we
test:
H0: σ12 = σ 32
Ha: σ12 ≠ σ 32
The test statistic is F 
Larger sample variance s12 .05312


 1.437
Smaller sample variance s32 .04432
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n1 – 1
= 3 – 1 = 2 and ν2 = n3 – 1 = 5 – 1 = 4. From Table IX, Appendix B, F.025 = 10.65. The rejection
region is F > 10.65.
Since the observed value of the test statistic does not fall in the rejection region (F = 1.437  10.65),
H0 is not rejected. There is insufficient evidence to indicate the variation in the zinc measurements
for the text-line and the intersection differ at α = .05.
b.
To determine if the variation in the zinc measurements for the witness-line and the intersection
differ, we test:
H0: σ 22 = σ 32
Ha: σ 22 ≠ σ 32
The test statistic is F 
Larger sample variance s22 .10152


 5.250
Smaller sample variance s32 .04432
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
421
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n2 – 1
= 6 – 1 = 5 and ν2 = n3 – 1 = 5 – 1 = 4. From Table IX, Appendix B, F.025 = 9.36. The rejection
region is F > 9.36.
Since the observed value of the test statistic does not fall in the rejection region (F = 5.250  9.36),
H0 is not rejected. There is insufficient evidence to indicate the variation in the zinc measurements
for the witness-line and the intersection differ at α = .05.
7.83
c.
There is no indication that the variances of the zinc measurements for three locations differ.
d.
With only 3, 6, and 5 measurements, it is very difficult to check the assumptions.
a.
Let σ12 = variance of the order-to-delivery times for the Persian Gulf War and σ 22 = variance of the
order-to-delivery times for Bosnia.
Descriptive Statistics: Gulf, Bosnia
Variable
Gulf
Bosnia
N
9
9
Mean
25.24
7.38
Median
27.50
6.50
TrMean
25.24
7.38
Variable
Gulf
Bosnia
Minimum
9.10
3.00
Maximum
41.20
15.10
Q1
15.30
5.25
Q3
32.15
9.20
StDev
10.5204
3.6537
SE Mean
3.51
1.22
To determine if the variances of the order-to-delivery times for the Persian Gulf and Bosnia
shipments are equal, we test:
H0:
Ha:
σ12
σ 22
σ12
σ 22
=1
≠1
The test statistic is F 
Larger sample variance s12 10.52042


 8.29
Smaller sample variance s22
3.6537 2
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n1 – 1
= 9 – 1 = 8 and ν2 = n2 – 1 = 9 – 1 = 8. From Table IX, Appendix B, F.025 = 4.43. The rejection
region is F > 4.43.
Since the observed value of the test statistic falls in the rejection region (F = 8.29 > 4.43), H0 is
rejected. There is sufficient evidence to indicate the variances of the order-to-delivery times for the
Persian Gulf and Bosnia shipments differ at α = .05.
b.
No. One assumption necessary for the small sample confidence interval for (μ1 − μ2) is that σ12  σ 22 .
For this problem, there is evidence to indicate that σ12  σ 22 .
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
422
7.84
Chapter 7
Using MINITAB, some preliminary calculations are:
Descriptive Statistics: HEATRATE
Variable
HEATRATE
ENGINE
Advanced
Aeroderiv
Traditional
N
21
7
39
Variable
HEATRATE
ENGINE
Advanced
Aeroderiv
Traditional
Q3
10060
14628
11964
a.
N*
0
0
0
Mean
9764
12312
11544
SE Mean
139
1002
205
StDev
639
2652
1279
Minimum
9105
8714
10086
Q1
9252
9469
10592
Median
9669
12414
11183
Maximum
11588
16243
14796
To determine if the heat rate variances for traditional and aeroderivative augmented gas turbines
differ, we test:
H0: σ 22 = σ 32
Ha: σ 22 ≠ σ 32
The test statistic is F 
Larger sample variance s22 26522


 4.299
Smaller sample variance s32 12792
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F distribution with numerator
df = ν2 = n2 – 1 = 7 – 1 = 6 and denominator df = ν3 = n3 – 1 = 39 – 1 = 38. From Table IX,
Appendix B, F.025 ≈ 2.74. The rejection region is F > 2.74.
Since the observed value of the test statistic falls in the rejection region (F = 4.299 > 2.74), H0 is
rejected. There is sufficient evidence to indicate the heat rate variances for traditional and
aeroderivative augmented gas turbines differ at α = .05.
Since the test in Exercise 7.23 a assumes that the population variances are the same, the validity of
the test is suspect since we just found the variances are different.
b.
To determine if the heat rate variances for advanced and aeroderivative augmented gas turbines
differ, we test:
H0: σ12 = σ 22
Ha: σ12 ≠ σ 22
The test statistic is F 
Larger sample variance s12 26522


 17.224
Smaller sample variance s12
6392
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F distribution with numerator
df = ν1 = n1 – 1 = 7 – 1 = 6 and denominator df = ν2 = n2 – 1 = 21 – 1 = 20. From Table IX,
Appendix B, F.025 = 3.13. The rejection region is F > 3.13.
Since the observed value of the test statistic falls in the rejection region
(F = 17.224 > 3.13), H0 is rejected. There is sufficient evidence to indicate the heat rate variances for
advanced and aeroderivative augmented gas turbines differ at α = .05.
Since the test in Exercise 7.23 b assumes that the population variances are the same, the validity of
the test is suspect since we just found the variances are different.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.85
423
Let σ12 = variance of improvement scores in the honey dosage group and σ 22 = variance of improvement
scores in the DM dosage group. From Exercise 7.21, s1 = 2.855 and s2 = 3.256.
To determine if the variability in coughing improvement scores differs for the two groups, we test:
H0: σ 12 = σ 22
Ha: σ 12 ≠ σ 22
The test statistic is F =
s 2 3.256 2
larger sample variance
= 22 =
= 1.30
smaller sample variance s1
2.855 2
The rejection region requires α/2 = .10/2 = .05 in the upper tail of the F-distribution with
ν2 = n2 − 1 = 33 − 1 = 32 and ν1 = n1 − 1 = 35 − 1 = 34. From Table VIII, Appendix B, F.05 ≈ 1.84. The
rejection region is F > 1.84.
Since the observed value of the test statistic does not fall in the rejection region (F = 1.3  1.84), H0 is not
rejected. There is insufficient evidence to indicate the variability in the coughing improvement scores
differs for the two groups at α = .10.
7.86
a.
The 2 samples are randomly selected in an independent manner from the two populations. The
sample sizes, n1 and n2, are large enough so that x1 and x2 each have approximately normal
sampling distributions and so that s12 and s22 provide good approximations to σ12 and σ 22 . This will
be true if n1 ≥ 30 and n2 ≥ 30.
7.87
b.
1.
2.
3.
Both sampled populations have relative frequency distributions that are approximately normal.
The population variances are equal.
The samples are randomly and independently selected from the populations.
c.
1.
2.
The relative frequency distribution of the population of differences is normal.
The sample of differences are randomly selected from the population of differences.
d.
The two samples are independent random samples from binomial distributions. Both samples should
be large enough so that the normal distribution provides an adequate approximation to the sampling
distributions of p̂1 and p̂2 .
e.
The two samples are independent random samples from populations which are normally distributed.
a.
sp2 
(n1  1) s12  (n1  1) s22 11(74.2)  13(60.5)

= 66.7792
n1  n2  2
12  14  2
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0
The test statistic is t =
( x1  x2 )  0
sp2
1
1 
 n  n 
1
2
=
(17.8 - 15.3)  0
1
1
66.7792  + 
 12 14 
= .78
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
424
Chapter 7
The rejection region requires α = .05 in the upper tail of the t-distribution with df = n1 + n2 − 2 = 12 +
14 − 2 = 24. From Table V, Appendix B, for df = 24, t.05 = 1.711. The rejection region is t > 1.711.
Since the observed value of the test statistic does not fall in the rejection region (0.78  1.711), H0 is
not rejected. There is insufficient evidence to indicate that μ1 > μ2 at α = .05.
b.
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table V, Appendix B, with
df = n1 + n2 − 2 = 12 + 14 − 2 = 24, t.005 = 2.797. The confidence interval is:
1
1 
1
1
( x1  x2 )  t.005 s 2p     (17.8  15.3)  2.797 66.7792   
 12 14 
 n1 n2 
 2.50  8.99  (6.49, 11.49)
c.
For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B,
z.005 = 2.58.
n1  n2 
7.88
a.

( zα / 2 ) σ12  σ 22
( ME ) 2
  (2.58) (74.2  60.5)  224.15  225
2
22
H0: σ12  σ 22
Ha: σ12  σ 22
The test statistic is F =
s2
Larger sample variance
120.1
 22 
 3.84
Smaller sample variance
31.3
s1
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with numerator
df ν1 = n2 − 1 = 15 − 1 = 14 and denominator df ν2 = n1 − 1 = 20 − 1 = 19. From Table IX, Appendix
B, F.025 ≈ 2.66. The rejection region is F > 2.66.
b.
7.89
a.
Since the observed value of the test statistic falls in the rejection region (F = 3.84 > 2.66), H0 is
rejected. There is sufficient evidence to conclude σ12  σ 22 at α = .05.
No, we should not use a small sample t test to test H0: (μ1 − μ2) = 0 against Ha: (μ1 − μ2) ≠ 0 because
the assumption of equal variances does not seem to hold since we concluded σ12  σ 22 in part b.
For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The
confidence interval is:
( x1  x2 )  z.05
s12 s22
2.1 3.0

 (12.2  8.3)  1.645

n1 n2
135 148
 3.90  .31  (3.59, 4.21)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
b.
425
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
The test statistic is z =
( x1  x2 )
s12
n1

s22

(12.2  8.3)  0
2.1 3.0

135 148
n2
 20.60
The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV,
Appendix B, z.005 = 2.58. The rejection region is z < −2.58 or z > 2.58.
Since the observed value of the test statistic falls in the rejection region (20.60 > 2.58), H0 is rejected.
There is sufficient evidence to indicate that μ1 ≠ μ2 at α = .01.
c.
For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B, z.05 = 1.645.
n1  n2 
7.90

( zα / 2 ) σ12  σ 22
( ME ) 2
  (1.645) (2.1  3.0)  345.02  346
2
.2 2
Some preliminary calculations are:
p̂1 =
a.
x1 110
x
x x
130
110  130 240

= .55; p̂2 = 2 
= .65; pˆ  1 2 

n1 200
n2 200
n1  n2 200  200 400
H0: (p1 − p2) = 0
Ha: (p1 − p2) < 0
The test statistic is z =
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

(.55  .65)  0
1 
 1
.6(1  .6) 

 200 200 

.10
= −2.04
.049
The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV,
Appendix B, z.10 = 1.28. The rejection region is z < −1.28.
Since the observed value of the test statistic falls in the rejection region (z = −2.04 < −1.28), H0 is
rejected. There is sufficient evidence to conclude (p1 − p2 < 0) at α = .10.
b.
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B,
z.025 = 1.96. The 95% confidence interval for (p1 − p2) is approximately:
pˆ1 qˆ2 pˆ 2 qˆ2

( pˆ1  pˆ 2 )  zα / 2
n1
n2
.55(1  .55) .65(1  .65)

200
200
 −.10 ± .096  (−.196, −.004)
 (.55 − .65) ± 1.96
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
426
Chapter 7
c.
From part b, z.025 = 1.96. Using the information from our samples, we can use p1 = .55 and p2 = .65.
For a width of .01, the margin of error is .005.
n1 = n2 =
7.91
a.
 zα / 2 2 ( p1 q1  p2 q2 )
ME
2
(1.96) 2 .55(1  .55)  .65(1  .65) 

2
.005
= 72990.4 ≈ 72,991

1.82476
.000025
This is a paired difference experiment.
Pair
Difference
(Pop. 1 - Pop. 2)
1
2
3
4
5
6
4
4
3
2
nd
nd
d 
d
i 1
nd
i

19
 3.8
5
sd2 

di2

 nd

di 



i 1

2
nd
i 1
nd  1

19 2
5  2.2
5 1
81 
sd  2.2  1.4832
H0: μd = 0
Ha: μd ≠ 0
The test statistic is t =
d 0
sd
nd

3.8  0
1.4832 / 5
= 5.73
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n − 1 = 5 −
1 = 4. From Table V, Appendix B, t.025 = 2.776. The rejection region is t < −2.776 or t > 2.776.
Since the observed value of the test statistic falls in the rejection region (5.73 > 2.776), H0 is rejected.
There is sufficient evidence to indicate that the population means are different at α = .05.
b.
For confidence coefficient .95, α = .05 and α/2 = .025. Therefore, we would use the same t value as
above, t.025 = 2.776. The confidence interval is:
xd  tα / 2
c.
sd
nd
 3.8  3.8  2.776
1.4832
5
 3.8  1.84  (1.96, 5.64)
The sample of differences must be randomly selected from a population of differences which has a
normal distribution.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.92
a.
427
Let p1 = proportion of Opening Doors students enrolled full time and p2 = proportion of traditional
students enrolled full time.
The target parameter for this comparison is p1 – p2.
b.
Let μ1 = mean GPA of Opening Doors students and μ2 = mean GPA of traditional students.
The target parameter for this comparison is μ1 – μ2.
7.93
a.
Let μ1 = average size of the right temporal lobe of the brain for the short-recovery group and
μ2 = average size of the right temporal lobe of the brain for the long-recovery group.
The target parameter is μ1 – μ2. We must assume that the two samples are random and independent,
the two populations being sampled from are approximately normal, and the two population variances
are equal.
b.
Let p1 = proportion of athletes who have a good self-image of their body and
p2 = proportion of non-athletes who have a good self-image of their body.
The target parameter for this comparison is p1 – p2. We must assume that the two samples are
random and independent and that the sample sizes are sufficiently large.
c.
Let μ1 = average weight of eggs produced by a sample of chickens on regular feed and
μ2 = average weight of eggs produced by a sample of chickens fed a diet supplemented by corn oil.
Let μd = average difference in weight between eggs produced by the chickens on regular feed and
then on a diet supplemented with corn oil.
The target parameter is μd. We must assume that we have a random sample of differences and that
the population of differences is approximately normal.
7.94
Using MINITAB, some preliminary calculations are:
Descriptive Statistics: Spillage
Variable
Spillage
Cause
Collision
Fire
Grounding
HullFail
Unknown
N
10
12
14
12
2
N*
0
0
0
0
0
Mean
76.6
70.9
47.79
54.4
26.00
SE Mean
22.3
17.5
7.61
16.3
1.00
Variable
Spillage
Cause
Q3 Maximum
Collision 102.0
257.0
Fire
80.5
239.0
Grounding 59.00
124.00
HullFail
46.0
221.0
Unknown
*
27.00
StDev
70.4
60.7
28.47
56.4
1.41
Minimum
31.0
26.0
21.00
24.0
25.00
Q1
35.0
32.3
30.25
29.3
*
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Median
41.5
49.0
37.50
31.5
26.00
428
Chapter 7
a.
Let μ1 = mean spillage for accidents caused by collision and μ2 = mean spillage for accidents caused
by fire/explosion.
s 2p 
n1  1 s12  n2  1 s22 10  1 70.42  12  1 60.72
n1  n2  2

10  12  2
 4, 256.7415
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table V, Appendix B,
with df = n1 + n2 − 2 = 10 + 12 – 2 = 20, t.05 = 1.725. The confidence interval is:
1
1 
1 1
( x1  x2 )  t.05 s 2p     (76.6  70.9)  1.725 4256.7415   
 10 12 
 n1 n2 
 5.7  48.19  ( 42.59, 53.89)
b.
Let μ3 = mean spillage for accidents caused by grounding and μ4 = mean spillage for accidents
caused by hull failure.
s 2p

n1  1 s12  n2  1 s22 14  1 28.472  12  1 56.42
n1  n2  2

14  12  2
 1, 896.9830
To determine if the mean spillage amount for accidents caused by grounding is different from the
mean spillage amount caused by hull failure, we test:
H0: μ3 − μ4 = 0
Ha: μ3 − μ4 ≠ 0
The test statistic is t 
 x1  x2   Do
1
1 
s 2p   
 n1 n2 

47.79  54.4  0
 1 1
1896.983   
 14 12 

6.61
 .39
17.1342
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with
df = n1 + n2 – 2 = 14 +12 – 2 = 24. From Table V, Appendix B, t.025 = 2.064. The rejection region
is t < −2.064 or t > 2.064.
Since the observed value of the test statistic does not fall in the rejection region
(t = −.39  −2.064), H0 is not rejected. There is insufficient evidence to indicate the mean spillage
amount for accidents caused by grounding is different from the mean spillage amount caused by hull
failure at α = .05.
c.
The necessary assumptions are:
We must assume that the distributions from which the samples were selected are approximately
normal, the samples are independent, and the variances of the two populations are equal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
429
Below are the stem-and-leaf plots for each of the samples:
Stem-and-leaf of Spillage Cause = Collision
Leaf Unit = 10
(6)
4
2
1
1
1
0
0
1
1
2
2
0
0
0
0
1
1
1
1
1
2
2
5
N
= 12
2333
4455
7
8
3
3
Stem-and-leaf of Spillage Cause = Grounding
Leaf Unit = 1.0
3
(5)
6
4
3
2
2
2
1
1
1
2
3
4
5
6
7
8
9
10
11
12
N
0
0
0
0
1
1
1
1
1
2
2
= 14
168
11678
15
8
2
1
4
Stem-and-leaf of Spillage Cause = Hull Failure
Leaf Unit = 10
(8)
4
2
2
2
1
1
1
1
1
1
= 10
333444
69
2
Stem-and-leaf of Spillage Cause = Fire
Leaf Unit = 10
4
(4)
4
3
2
2
1
1
1
1
1
N
N
= 12
22233333
44
0
2
Based on the shapes of the stem-and-leaf plots, it does not appear that the data are normally
distributed.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
430
Chapter 7
Also, we know that if the data are normally distributed, then the Interquartile Range, IQR, divided by
the standard deviation should be approximately 1.3. We will compute IQR/s for each of the samples:
Collision:
Fire:
Grounding:
Hull Failure:
IQR/s = (102.0 – 35.0) / 70.4 = .95
IQR/s = (80.5 – 32.3) / 60.7 = .79
IQR/s = (59.0 – 30.25) / 28.47 = 1.01
IQR/s = (46.0 – 29.3) / 56.4 = .29
Since all of these ratios are quite a bit smaller than 1.3, it indicates that none of the samples come
from normal distributions.
Thus, it appears that the assumption of normal distributions is violated.
The sample standard deviations are:
Collision:
s = 70.4
Fire:
s = 60.7
Grounding: s = 28.47
Hull Failure:
s = 56.4
Without doing formal tests, it appears that the variances of the groups Collision, Fire, and Hull
Failure are probably not significantly different. However, it appears that the variance for the
grounding group is smaller than the others.
d.
Let σ12 = variance of spillage for accidents caused by collision and σ 22 = variance of spillage for
accidents caused by grounding.
To determine if the variances of the amounts of spillage due to collision and grounding differ, we test:
H0: σ12 = σ 22
Ha: σ12 ≠ σ 22
The test statistic is F 
Larger sample variance s12
70.42
 2 
 6.11
Smaller sample variance s2 28.472
The rejection region requires α/2 = .02/2 = .01 in the upper tail of the F distribution with numerator
df = ν1 = n1 – 1 = 10 – 1 = 9 and denominator df = ν2 = n2 – 1 = 14 – 1 = 13. From Table X,
Appendix B, F.01 = 4.19. The rejection region is F > 4.19.
Since the observed value of the test statistic falls in the rejection region (F = 6.11 > 4.19), H0 is
rejected. There is sufficient evidence to indicate the variances of the amounts of spillage due to
collision and grounding differ at α = .02.
7.95
a.
Yes. The sample mean of the virtual-reality group is 10.67 points higher than the sample mean of the
simple user interface group.
b.
Let μ1 = mean improvement score for the virtual-reality group and μ2 = mean improvement score for
the simple user interface group.
To determine if the mean improvement scores for the virtual-reality group is higher than that for the
simple user interface group, we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
The test statistic is z =
( x1 − x2 ) − Do
σ
σ 
+


n2 
 n1
2
1
2
2
=
( 43.15 − 32.48) − 0
 12.57
9.26 
+


45 
 45
2
2
=
431
10.67
= 4.58
2.3274
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 4.58 > 1.645), H0 is
rejected. There is sufficient evidence to indicate the mean improvement scores for the virtual-reality
group is higher than that for the simple user interface group α = .05.
7.96
a.
Since there is much variability among cars, by using matched pairs, we can block out the variability
among the cars and compare the means of the 2 types of shocks.
b.
Let µ1 = mean strength of manufacturer’s shock and µ2 = mean strength of competitor’s shock. Also,
let µd = µ1 − µ2.
Using MINITAB the descriptive statistics are:
Descriptive Statistics: Manufacturer, Competitor, Di
Variable
Manufacturer
Competitor
Di
N
6
6
6
Mean
10.717
10.300
0.4167
StDev
1.752
1.818
0.1329
Minimum
8.800
8.400
0.2000
Q1
9.400
8.850
0.3500
Median
10.100
9.700
0.4000
Q3
12.675
12.250
0.5250
Maximum
13.200
13.000
0.6000
To determine if there is a difference in the mean strength of the two types of shocks after 20,000
miles, we test:
Ho: µd = 0
Ha: µd  0
The test statistic is t 
d  0 .4167  0

 7.68
sd
.1329
nd
6
The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with
df = nd – 1 = 6 – 1 = 5. From Table V, Appendix B, t.025 = 2.571. The rejection region is t < −2.571
or t > 2.571.
Since the observed value of the test statistic falls in the rejection region (t = 7.68 > 2.571), H0 is
rejected. There is sufficient evidence to indicate a difference in the mean strength of the two types of
shocks after 20,000 miles at α = .05.
c.
The observed significance level is P(t ≥ 7.68) + P(t ≤ -7.68) = 2P(t ≥ 7.68) < 2(.0005) < .0010 (using
Table V, Appendix B).
d.
We must assume that the population of differences is normally distributed and that the sample is
random.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
432
Chapter 7
e.
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table V, Appendix B, with
df = nd – 1 = 6 – 1 = 5, t.025 = 2.571. The 95% confidence interval is:
d  z.025
sd
nd
 .4167  2.571
.1329
6
 .4167  .1395  (.2772, .5562)
We are 95% confident that the difference in mean strength between the two types of shocks after
20,000 miles is between .2772 and .5562.
f.
Some preliminary calculations are:
s 2p 
(n1  1) s12  (n2  1) s22 (6  1)1.7522  (6  1)1.8182

 3.1873
n1  n2  2
662
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table V, Appendix B, with
df = n1 + n2 – 2 = 6 + 6 – 2 = 10, t.025 = 2.228. The 95% confidence interval is:
1
1 
1 1
( μ1  μ 2 )  t.025 s 2p     (10.717  10.300)  2.228 3.1873   
6 6
 n1 n2 
 .417  2.2965  (1.8795, 2.7135)
We are 95% confident that the difference in mean strength between the two types of shocks after
20,000 miles is between -1.8795 and 2.7135.
7.97
g.
The interval assuming independent sample in part f is (−1.8795, 2.7135) while the interval assuming
paired differences in part e is (.2772, .5562). The interval assuming independent samples is wider.
The interval from part e gives more information because the interval is narrower.
h.
No. If the data were collected using a paired experiment, then the data must be analyzed as a paired
experiment.
a.
The data should be analyzed using a paired-difference analysis because that is how the data
were collected. Reaction times were collected twice from each subject, once under the
random condition and once under the static condition. Since the two sets of data are not
independent, they cannot be analyzed using independent samples analyses.
b.
Let μ1 = mean reaction time under the random condition and μ2 = mean reaction time under
the static condition. Let μd = μ1 − μ2. To determine if there is a difference in mean reaction
time between the two conditions, we test:
H0: μd = 0
Ha: μd ≠ 0
c.
The test statistic is t = 1.52 with a p-value of .15. Since the p-value is not small, there is no
evidence to reject H0 for any reasonable value of α. There is insufficient evidence to
indicate a difference in the mean reaction times between the two conditions. This supports
the researchers' claim that visual search has no memory.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.98
a.
433
Let μ1 = mean carat size of diamonds certified by GIA and μ2 = mean carat size of
diamonds certified by HRD. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From
Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is:
σ12
( x1  x2 )  zα / 2
n1

σ 22
n2
 (.6723  .8129)  1.96
.24562 .18312

151
79
 .1406  .0563  (.1969,  .0843)
b.
We are 95% confident that the difference in mean carat size between diamonds certified by GIA and
those certified by HRD is between -.1969 and -.0843.
c.
Let μ3 = mean carat size of diamonds certified by IGI.
( x1  x3 )  zα / 2
σ12
n1

σ 32
n3
 (.6723  .3665)  1.96
.24562 .21632

151
78
 .3058  .0620  (.2438, .3678)
d.
We are 95% confident that the difference in mean carat size between diamonds certified by GIA and
those certified by IGI is between .2438 and .3678.
e.
( x2  x3 )  zα / 2
σ 22
n2

σ 32
n3
 (.8129  .3665)  1.96
.18312 .21632

79
78
 .4464  .0627  (.3837, .5091)
f.
We are 95% confident that the difference in mean carat size between diamonds certified by HRD and
those certified by IGI is between .3837 and .5091.
Let σ12 = variance of carat size for diamonds certified by GIA, σ 22 = variance of carat size for diamonds
certified by HRD, and σ 32 = variance of carat size for diamonds certified by IGI.
g.
To determine if the variation in carat size differs for diamonds certified by GIA and diamonds
certified by HRD, we test:
H0: σ12 = σ 22
Ha: σ12 ≠ σ 22
The test statistic is F 
Larger sample variance s12 .24562


 1.799
Smaller sample variance s22 .18312
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with
ν1 = n1 – 1 = 151 – 1 = 150 and ν2 = n2 – 1 = 79 – 1 = 78. From Table IX, Appendix B, F.025 ≈ 1.43.
The rejection region is F > 1.43.
Since the observed value of the test statistic falls in the rejection region (F = 1.799 > 1.43), H0 is
rejected. There is sufficient evidence to indicate the variation in carat size differs for diamonds
certified by GIA and those certified by HRD at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
434
Chapter 7
h.
To determine if the variation in carat size differs for diamonds certified by GIA and
diamonds certified by IGI, we test:
H0: σ12 = σ 32
Ha: σ12 ≠ σ 32
The test statistic is F 
Larger sample variance s12 .24562


 1.289
Smaller sample variance s32 .21632
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with
ν1 = n1 – 1 = 151 – 1 = 150 and ν2 = n3 – 1 = 78 – 1 = 77. From Table IX, Appendix B, F.025 ≈ 1.43.
The rejection region is F > 1.43.
Since the observed value of the test statistic does not fall in the rejection region (F = 1.289  1.43),
H0 is not rejected. There is insufficient evidence to indicate the variation in carat size differs for
diamonds certified by GIA and those certified by IGI at α = .05.
i.
To determine if the variation in carat size differs for diamonds certified by HRD and
diamonds certified by IGI, we test:
H0: σ 22 = σ 32
Ha: σ 22 ≠ σ 32
The test statistic is F 
2
Larger sample variance s3 .21632
 2 
 1.396
Smaller sample variance s2 .18312
The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with
ν1 = n3 – 1 = 78 – 1 = 77 and ν2 = n2 – 1 = 79 – 1 = 78. From Table IX, Appendix B, F.025 ≈ 1.67.
The rejection region is F > 1.67.
Since the observed value of the test statistic does not fall in the rejection region (F = 1.396  1.67),
H0 is not rejected. There is insufficient evidence to indicate the variation in carat size differs for
diamonds certified by HRD and those certified by IGI at α = .05.
We will look at the 4 methods for determining if the data are normal. First, we will look at
histograms of the data. Using MINITAB, the histograms of the carat sizes for the 3 certification
bodies are:
40
40
30
30
20
Percent
Percent
j.
20
10
10
0
0
0.0
0.5
GIA
1.0
0.0
0.5
HRD
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
1.0
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
435
40
Percent
30
20
10
0
0.0
0.5
1.0
IGI
From the histograms, none of the data appear to be mound-shaped. It appears that none of the data
sets are normal.
Next, we look at the intervals x  s, x  2s, x  3s . If the proportions of observations falling in each
interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using
MINITAB, the summary statistics are:
Descriptive Statistics: GIA, IGI, HRD
Variable
GIA
IGI
HRD
N
151
78
79
Mean
0.6723
0.3665
0.8129
Median
0.7000
0.2900
0.8100
TrMean
0.6713
0.3406
0.8169
Variable
GIA
IGI
HRD
Minimum
0.3000
0.1800
0.5000
Maximum
1.1000
1.0100
1.0900
Q1
0.5000
0.2100
0.6500
Q3
0.9000
0.4850
1.0000
StDev
0.2456
0.2163
0.1831
SE Mean
0.0200
0.0245
0.0206
For GIA:
x  s  .6723  .2456  (.4267, .9179) 84 of the 151 values fall in this interval. The proportion is
.56. This is much smaller than the .68 we would expect if the data were normal.
x  2s  .6723  2(.2456)  .6723  .4912  (.1811, 1.1635) 151 of the 151 values fall in this
interval. The proportion is 1.00. This is much larger than the .95 we would expect if the data were
normal.
x  3s  .6723  3(.2456)  .6723  .7368  ( .0645, 1.4091) 151 of the 151 values fall in this
interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
For IGI:
x  s  .3665  .2163  (.1502, .5828) 69 of the 78 values fall in this interval. The proportion is
.88. This is much larger than the .68 we would expect if the data were normal.
x  2s  .3665  2(.2163)  .3665  .4326  ( .0661, .7991) 74 of the 78 values fall in this
interval. The proportion is .95. This is the same as the .95 we would expect if the data were normal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
436
Chapter 7
x  3s  .3665  3(.2163)  .3665  .6489  (.2824, 1.0154) 78 of the 78 values fall in this
interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
For HRD:
x  s  .8129  .1831  (.6298, .9960) 30 of the 79 values fall in this interval. The proportion is
.38. This is much smaller than the .68 we would expect if the data were normal.
x  2s  .8129  2(.1831)  .8129  .3662  (.4467, 1.1791) 79 of the 79 values fall in this
interval. The proportion is 1.00. This is much larger than the .95 we would expect if the data were
normal.
x  3s  .8129  3(.1831)  .8129  .5493  (.2636, 1.3622) 79 of the 79 values fall in this
interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were
normal.
From this method, it appears that the data are not normal.
Next, we look at the ratio of the IQR to s.
For GIA:
IQR = QU – QL = 1.1 − .3 = .8.
IQR
.8

 3.26 This is much larger than the 1.3 we would expect if the data were normal.
s
.2456
This method indicates the data are not normal.
For IGI:
IQR = QU – QL = 1.01 - .18 = .83.
IQR
.83

 3.84 This is much larger than the 1.3 we would expect if the data were normal.
s
.2163
This method indicates the data are not normal.
For HRD:
IQR = QU – QL = 1.09 - .5 = .59.
IQR
.59

 3.22 This is much larger than the 1.3 we would expect if the data were normal.
s
.1831
This method indicates the data are not normal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
437
Finally, using MINITAB, the normal probability plot for GIA is:
Normal Probability Plot for GIA
ML Estimates - 95% CI
ML Estimates
99
Percent
95
90
Mean
0.672252
StDev
0.244757
Goodness of Fit
80
70
60
50
40
30
20
AD*
3.332
10
5
1
0.0
0.5
1.0
1.5
Data
Since the data do not form a straight line, the data are not normal.
Using MINITAB, the normal probability plot for IGI is:
Normal Probability Plot for IGI
ML Estimates - 95% CI
ML Estimates
99
Mean
0.366538
StDev
0.214863
Percent
95
90
Goodness of Fit
80
70
60
50
40
30
20
AD*
10
5
1
0.0
0.5
1.0
Data
Since the data do not form a straight line, the data are not normal.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
5.622
438
Chapter 7
Using MINITAB, the normal probability plot for HRD is:
Normal Probability Plot for HRD
ML Estimates - 95% CI
ML Estimates
99
Mean
0.812911
StDev
0.181890
Percent
95
90
Goodness of Fit
80
70
60
50
40
30
20
AD*
3.539
10
5
1
0.5
1.0
1.5
Data
Since the data do not form a straight line, the data are not normal.
7.99
a.
From the 4 different methods, all indications are that the carat size data are not normal for any of the
certification bodies.
Let μ1 = mean response by noontime watchers and μ2 = mean response by non-noontime watchers.
To determine if the mean response differs for noontime and non-noontime watchers, we test:
H0: μ1 = μ2
Ha: μ1 ≠ μ2
7.100
b.
Since the p-value (p = .02) is less than α = .05, H0 is rejected. There is sufficient evidence to indicate
the mean response differs for noontime and non-noontime watchers at α = .05.
c.
Since the p-value (p = .02) is greater than α = .01, H0 is not rejected. There is insufficient evidence
to indicate the mean response differs for noontime and non-noontime watchers at α = .01.
d.
Since the two sample means are so close together, there appears to be no “practical” difference
between the two means. Even if there is a statistically significant difference between the two means,
there is no practical difference.
a.
Let p1 = proportion of managers and professionals who are male and p2 = proportion of part-time
MBA students who are male. To see if the samples are sufficiently large:
n1 pˆ1  162(.95)  153.9 and n1 qˆ1  162(.05)  8.1
n2 pˆ 2  109(.689)  75.101 and n2 qˆ2  109(.311)  33.899
Since n1 qˆ1  8.1  15, the normal approximation may not be adequate. We will go ahead and
perform the test
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
439
First, we calculate the overall estimate of the common proportion under H0.
pˆ 
n1 pˆ1  n2 pˆ 2 162(.95)  109(.689)

= .845
n1  n2
162  109
To determine if the population of managers and professionals consists of more males than the
part-time MBA population, we test:
H0: p1 − p2 = 0
Ha: p 1 > p 2 > 0
The test statistic is z =
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

(.95  .689)  0
1 
 1

.845(.155) 
 162 109 
= 5.82
The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.05 = 1.645. The rejection region is z > 1.645.
b.
Since the observed value of the test statistic falls in the rejection (z = 5.82 > 1.645), H0 is rejected.
There is sufficient evidence to indicate that population of managers and professionals consists of
more males than the part-time MBA population at α = .05.
We had to assume:
c.
1. Both samples were randomly selected
2. Both sample sizes are sufficiently large.
First, we calculate the overall estimate of the common proportion under H0.
pˆ 
n1 pˆ1  n2 pˆ 2 162(.912)  109(.534)

= .760
n1  n2
162  109
To determine if the population of managers and professionals consists of more married individuals
than the part-time MBA population, we test:
H0: p1 = p2
Ha: p 1 > p 2
The test statistic is z =
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

(.912  .534)  0
1 
 1
.760(.240) 

 162 109 
= 7.14
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix
B, z.01 = 2.33. The rejection region is z > 2.33.
Since the observed value of the test statistic falls in the rejection (z = 7.14 > 2.33), H0 is rejected.
There is sufficient evidence to indicate that population of managers and professionals consists of
more married individuals than the part-time MBA population at α = .01.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
440
Chapter 7
d.
We had to assume:
1. Both samples were randomly selected
2. Both sample sizes are sufficiently large.
7.101
To determine if there is a difference in the proportions of consumer/commercial and industrial product
managers who are at least 40 years old, we could use either a test of hypothesis or a confidence interval.
Since we are asked only to determine if there is a difference in the proportions, we will use a test of
hypothesis.
Let p1 = proportion of consumer/commercial product managers at least 40 years old and p2 = proportion of
industrial product managers at least 40 years old.
p̂1 = .40
q̂1 = 1 − p̂1 = 1 − .40 = .60
p̂2 = .54
q̂2 = 1 − p̂1 = 1 − .54 = .46
p̂ =
n1 pˆ1  n2 pˆ 2
93(.40)  212(.54)
=
= .497
n1  n2
93  212
q̂ = 1 − p̂ = 1 − .497 = .503
To see if the samples are sufficiently large:
n1 pˆ1  93(.4)  37.2 and n1 qˆ1  93.6(.6)  55.8
n2 pˆ 2  212(.54)  114.48 and n2 qˆ2  212(.46)  97.52
Since all values are greater than 15, the normal approximation will be adequate.
To determine if there is a difference in the proportions of consumer/commercial and industrial product
managers who are at least 40 years old, we test:
H0: p1 − p2 = 0
Ha: p 1 − p 2 ≠ 0
The test statistic is z =
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 
=
(.40  .54)  0
1 
1
.497(.503)  
 93 212 
= −2.25
We will use α = .05. The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution.
From Table IV, Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic falls in the rejection region (z = −2.25 < −1.96), H0 is rejected.
There is sufficient evidence to indicate that there is a difference in the proportions of consumer/commercial
and industrial product managers who are at least 40 years old at α = .05.
Since the test statistic is negative, there is evidence to indicate that the industrial product managers tend to
be older than the consumer/commercial product managers.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.102
a.
441
The descriptive statistics are:
Descriptive Statistics: US, Japan
Variable
US
Japan
N
5
5
Mean
6.562
3.118
Median
6.870
3.220
TrMean
6.562
3.118
Variable
US
Japan
Minimum
4.770
1.920
Maximum
8.000
4.910
Q1
5.415
1.970
Q3
7.555
4.215
s 2p 
StDev
1.217
1.227
SE Mean
0.544
0.549
(n1  1) s12  (n2  1) s22 (5  1)1.217 2  (5  1)1.227 2

 1.4933
n1  n2  2
55 2
To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants,
we test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0
The test statistic is t =
( x1  x2 )  D0
1
1 
sp2   
 n1 n2 

(6.562  3.118)  0
1 1
1.4933   
5 5
= 4.456
The rejection region requires α = .05 in the upper tail of the t-distribution with df = n1 + n2 − 2 = 5 +
5 − 2 = 8. From Table V, Appendix B, t.05 = 1.860. The rejection region is t > 1.860.
Since the observed value of the test statistic falls in the rejection region (t = 4.46 > 1.860), H0 is
rejected. There is sufficient evidence to indicate the mean annual percentage turnover for U.S. plants
exceeds that for Japanese plants at α = .05.
b.
The p-value = P(t ≥ 4.456). Using Table V, Appendix B, with df = n1 + n2 − 2
= 5 + 5 – 2 = 8, .005 < P(t ≥ 4.456) < .001. Since the p-value is so small, there is evidence to reject
H0 for α > .005.
c.
The necessary assumptions are:
1.
2.
3.
Both sampled populations are approximately normal.
The population variances are equal.
The samples are randomly and independently sampled.
There is no indication that the populations are not normal. Both sample variances are similar, so
there is no evidence the population variances are unequal. There is no indication the assumptions are
not valid.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
442
7.103
Chapter 7
Let p1 = unemployment rate for the urban industrial community and p2 = unemployment rate for the
university community.
Some preliminary calculations are:
p̂1 =
x1 47

= .0895
n1 525
p̂2 =
x2
22

= .0587
n2 375
For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025
= 1.96. The confidence interval is:
( pˆ1  pˆ 2 ) ± z.025
pˆ1 qˆ1 pˆ 2 qˆ2

n1
n2
.0895(.9105) .0587(.9413)

525
375
 .0308 ± .0341  (−.0033, .0649)
 (.0895 − .0587) ± 1.96
We are 95% confident the difference in unemployment rates in the two communities is between −.0033 and
.0649.
7.104
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B,
z.05 = 1.645. We estimate p1 = p2 = .5.
n1 − n2 =
7.105
a.
 zα / 2 2 ( p1 q1  p2 q2 )
( ME )2

(1.645)2 .5(.5)  .5(.5) 
.052
= 541.205 ≈ 542
Let μ1 = mean ingratiatory score for managers and μ2 = mean ingratiatory score for clerical
personnel. To determine if there is a difference in ingratiatory behavior between managers and
clerical personnel, we test:
H0: μ1 = μ2
Ha: μ1 ≠ μ2
b.
The test statistic is z =
( x1  x2 )  D0
s12
n1

s22
n2

(2.41  1.90)  0
(.74) 2 (.59) 2

288
110
= 7.17
The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Since the observed value of the test statistic falls in the rejection region (z = 7.17 > 1.96), H0 is
rejected. There is sufficient evidence to indicate a difference in ingratiatory behavior between
managers and clerical personnel at α = .05.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
c.
443
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 =
1.96. The 95% confidence interval is:
( x1  x2 )  z.025
s12 s22
.74 2 .59 2

 (2.41  1.90)  1.96

n1 n2
288 110
 .51 ± .14  (.37, .65)
We are 95% confident that the difference in mean ingratiatory scores between managers and clerical
personnel is between .37 and .65. Since this interval does not contain 0, it is consistent with the test
of hypothesis which rejected the hypothesis that there was no difference in mean scores for the two
groups.
7.106
Let p1 = proportion of larvae that died in containers containing high carbon dioxide levels and p2 =
proportion of larvae that died in containers containing normal carbon dioxide levels. The parameter of
interest for this problem is p1 − p2, or the difference in the death rates for the two groups.
Some preliminary calculations are:
pˆ 
x1  x2 .10(80)  .05(80)

= .075
n1  n2
80  80
q̂ = 1 − p̂ = 1 − .075 = .925
To determine if an increased level of carbon dioxide is effective in killing a higher percentage of leafeating larvae, we test:
H0: p1 − p2 = 0
Ha: p 1 − p 2 > 0
The test statistic is z =
( pˆ1  pˆ 2 )  0
1 
 1
ˆˆ  
pq
 80 80 

(.10  .05)  0
1 
1
.075(.925)   
 80 80 
= 1.201
The rejection region requires α = .01 in the upper tail of the z distribution. From Table IV, Appendix B,
z.01 = 2.33. The rejection region is z > 2.33.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.201  2.33), H0 is
not rejected. There is insufficient evidence to indicate that an increased level of carbon dioxide is effective
in killing a higher percentage of leaf-eating larvae at α = .01.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
444
7.107
Chapter 7
a.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Purchasers, Nonpurchasers
Variable
Purchase
Nonpurch
N
20
20
Mean
39.80
47.20
Median
38.00
52.00
TrMean
39.67
47.56
Variable
Purchase
Nonpurch
Minimum
23.00
22.00
Maximum
59.00
66.00
Q1
32.25
33.50
Q3
48.75
58.75
s 2p 
StDev
10.04
13.62
SE Mean
2.24
3.05
(n1  1) s12  (n2  1) s22 (20  1)10.042  (20  1)13.622

 143.153
n1  n2  2
20  20  2
Let μ1 = mean age of nonpurchasers and μ2 = mean age of purchasers.
To determine if there is a difference in the mean age of purchasers and nonpurchasers, we
test:
H0: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
The test statistic is t 
( x1  x2 )  0
s 2p
1
1 
 n  n 
1
2

(47.20  39.80)  0
1 
 1
143.153   
 20 20 
 1.956
The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with
df = n1 + n2 − 2 = 20 + 20 − 2 = 38. From Table V, Appendix B, t.05 ≈ 1.684. The rejection
region is t < −1.684 or t > 1.684.
Since the observed value of the test statistic falls in the rejection region (t = 1.956 > 1.684),
H0 is rejected. There is sufficient evidence to indicate the mean age of purchasers and
nonpurchasers differ at α = .10.
b.
The necessary assumptions are:
1.
2.
3.
c.
Both sampled populations are approximately normal.
The population variances are equal.
The samples are randomly and independently sampled.
The p-value is P(z ≤ −1.956) + P(z ≥ 1.956) = (.5 − .4748) + (.5 − .4748) = .0504. The probability of
observing a test statistic of this value or more unusual if H0 is true is .0504. Since this value is less
than α = .10, H0 is rejected. There is sufficient evidence to indicate there is a difference in the mean
age of purchasers and nonpurchasers.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
d.
445
For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table V, Appendix B,
with df = 38, t.05 ≈ 1.684. The confidence interval is:
1
1 
1 
 1
( x2  x1 )  t.05 sp2     (39.8 − 47.2) ± 1.684 143.1684   

20 20 
 n1 n2 
 −7.4 ± 6.38  (−13.77, −1.03)
We are 90% confident that the difference in mean ages between purchasers and nonpurchasers is
between −13.77 and −1.03.
7.108
a.
Let p1 = proportion of female students who switched due to loss of interest in SME and
p2 = proportion of male students who switched due to lack of interest in SME.
Some preliminary calculations are:
p̂1 =
x1
x
x  x2
74
72
74  72

= .43; p̂2 = 2 
= .44; p̂ = 1

= .436
n1 172
n2 163
n1  n2 172  163
To determine if the proportion of female students who switch due to lack of interest in SME differs
from the proportion of males who switch due to a lack of interest, we test:
H0: p1 − p2 = 0
Ha: p 1 − p 2 ≠ 0
The test statistic is z =
( pˆ1  pˆ 2 )  0
1
1 
ˆˆ  
pq
 n1 n2 

(.43  .44)  0
1 
 1

.436(.564) 
 172 163 
= −0.18
The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z < −1.645 or z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region (z = −0.18  1.645), H0 is not rejected. There is insufficient evidence to indicate the proportion of female students
who switch due to lack of interest in SME differs from the proportion of males who switch due to a
lack of interest in SME at α = .10.
b.
Let p1 = proportion of female students who switched due to low grades in SME and p2 = proportion
of male students who switched due to low grades in SME.
Some preliminary calculations are:
p̂1 =
x1
33

= .19;
n1 172
p̂2 =
x2
44

= .27
n2 163
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
446
Chapter 7
For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 =
1.645. The confidence interval is:
( pˆ1  pˆ 2 )  z.05
pˆ1 qˆ1 pˆ 2 qˆ2
.19(.81) .27(.73)
 (.19 − .27) ± 1.645


172
163
n1
n2
 −.08 ± .075  (−.155, −.005)
We are 90% confident that the difference between the proportions of female and male switchers who
lost confidence due to low grades in SME is between −.155 and −.005. Since the interval does not
include 0, there is evidence to indicate the proportion of female switchers due to low grades is less
than the proportion of male switchers due to low grades.
51
7.109
d
i 1
i
μd 
b.
We do not need to estimate anything – we know the parameter’s value.
c.
Using MINITAB, the descriptive statistics are:
51

335
 6.57
51
a.
Descriptive Statistics: SAT2007, SAT2000, Difference
Variable
SAT2007
SAT2000
Difference
N
Mean StDev
51 1072.7
78.1
51 1066.1
65.9
51
6.57 22.41
Minimum
Q1 Median
Q3
931.0 1004.0 1048.0 1143.0
966.0 1007.0 1054.0 1120.0
-73.00
-5.00
5.00
19.00
Maximum
1221.0
1197.0
54.00
Let μ1 = mean SAT score in 2007 and μ2 = mean SAT score in 2000. Then μd = μ1 − μ2.
determine if the true mean SAT score in 2007 differs from that in 2000, we test:
To
H0: μd = 0
Ha: μd ≠ 0
The test statistic is z 
d  Do 6.57  0

 2.09
sd
22.41
nd
51
The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV,
Appendix B, z.05 = 1.645. The rejection region is z < −1.645 or z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 2.09 > 1.645), H0 is
rejected. There is sufficient evidence to indicate the true mean SAT score in 2007 is different than
that in 2000 at α = .10.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
7.110
For confidence level .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The
standard deviation can be estimated by dividing the range by 4:
Range
4
=
=1
4
4
σ≈
n1 = n2 =
7.111
 zα / 2 2 σ12  σ 22 

( ME )2
1.962 (12  12 )
.22
= 192.08 ≈ 193
For probability .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.
Since we have no prior information about the proportions, we use p1 = p2 = .5 to get a conservative
estimate.
n1  n2 
7.112
447
( zα / 2 )2 ( p1 q1  p2 q2 )
( ME )
2

(1.96) 2 .5(1  .5)  .5(1  .5)
.02
2

1.9208
 4, 802
.0004
Some preliminary calculations are:
x
2
1
s12 
s22 
 x 

2
1
n1

n1  1

x22
 x 

2252
5  126 = 31.5
5 1
4
10, 251 
2
2
n2
n2  1

227 2
5  45.2 = 11.3
5 1
4
10, 351 
Let σ12 = variance for instrument A and σ 22 = variance for instrument B. Since we wish to determine if
there is a difference in the precision of the two machines, we test:
H0: σ12  σ 22
Ha: σ12  σ 22
The test statistic is F =
Larger sample variance s12 31.5
=

= 2.79
Smaller sample variance s22 11.3
The rejection region requires α/2 = .10/2 = .05 in the upper tail of the F-distribution with ν1 = n1 − 1 = 5 −
1 = 4 and ν2 = n2 − 1 = 5 − 1 = 4. From Table VIII, Appendix B, F.05 = 6.39. The rejection region is F >
6.39.
Since the observed value of the test statistic does not fall in the rejection region (F = 2.79  6.39), H0 is
not rejected. There is insufficient evidence of a difference in the precision of the two instruments at
α = .10.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
448
7.113
Chapter 7
a.
Let μd = mean difference in pupil dilation between pattern 1 and pattern 2.
To determine if the pupil dilation differs for the two patterns, we test:
H0: μd = 0
Ha: μd ≠ 0
b.
Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Pattern1, Pattern2, Diff
Variable
Pattern1
Pattern2
Diff
N
15
15
15
Mean
1.122
0.883
0.2393
Median
1.000
0.910
0.2100
TrMean
1.141
0.892
0.2338
Variable
Pattern1
Pattern2
Diff
Minimum
0.150
0.050
-0.0400
Maximum
1.850
1.600
0.5900
Q1
0.850
0.650
0.1200
Q3
1.460
1.220
0.3100
The test statistic is t 
StDev
0.505
0.453
0.1608
SE Mean
0.130
0.117
0.0415
d  D0 .2393  0

 5.76
sd
0.1608
nd
15
Since no α is given, we will use α = .05. The rejection region requires α/2 = .05/2 = .025 in each tail
of the t-distribution with df = nd – 1 = 15 – 1 = 14. From Table V, Appendix B, t.025 = 2.145. The
rejection region is t < −2.145 or t > 2.145.
Since the observed value of the test statistic falls in the rejection region (t = 5.76 > 2.145), H0 is
rejected. There is sufficient evidence to indicate that there is a difference in the mean pupil dilation
between pattern 1 and pattern 2 at α = .05.
7.114
c.
The paired difference design is better. There is much variation in pupil dilation from person to
person. By using the paired difference design, we can eliminate the person to person differences.
a.
Let μ1 = mean scale score for employees who report positive spillover of work skills and μ2 = mean
scale score for employees who did not report positive work spillover. To determine if the mean scale
score for employees who report positive spillover of work skills differs from the mean scale score for
employees who did not report positive work spillover, we test:
Ho: μ1 − μ2 = 0
Ha: μ1 − μ2 ≠ 0
b.
It is appropriate to apply the large sample z-test because there are 114 workers that have been studied
and divided into two groups.
c.
From the printout, the test statistics is t = 8.847 (equal variances not assumed). Since the sample
sizes are so large, we can assume that this test statistic is essentially a z. Thus, the test statistic is
z = 8.847.
The rejection region requires α/2 = .05/2 = .025 in each tail of the z distribution. From Table IV,
Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses
449
Since the observed value of the test statistic falls in the rejection region (z = 8.847 > 1.96), Ho is
rejected. There is sufficient evidence to indicate the mean scale score for employees who report
positive spillover of work skills is different from the mean scale score for employees who did not
report positive work spillover at α = .05.
d.
We are 95% confident that the difference between the mean use of creative ideas scale scores for the
two groups in between .627 and .988. Since interval does not contain 0, then we can say that there is
a significant difference on the mean scale scores between the two groups. Yes, the inference derived
from the confidence interval agrees with that from the hypothesis test.
e.
Let p1 = proportion of male workers who reported positive work spillover and p2 = proportion of
male workers who did not report positive spillover of work skills. To determine if the proportions of
male workers in the two groups are significantly different, we test:
H0: p1 − p2 = 0
Ha: p1 − p2  0
From the printout, the test statistic is z = .75 and the p-value is .453. Since the p-value is not small,
there is no evidence to reject H0. There is insufficient evidence to indicate the proportions of male
workers in the two groups are significantly different at any value of α < .453.
7.115
Attitude towards the Advertisement:
The p-value = .091. There is no evidence to reject H0 for α = .05. There is no evidence to indicate the first
ad will be more effective when shown to males for α = .05. There is evidence to reject H0 for α = .10.
There is evidence to indicate the first ad will be more effective when shown to males for α = .10.
Attitude toward Brand of Soft Drink:
The p-value = .032. There is evidence to reject H0 for α > .032. There is evidence to indicate the first ad
will be more effective when shown to males for α > .032.
Intention to Purchase the Soft Drink:
The p-value = .050. There is no evidence to reject H0 for α = .05. There is no evidence to indicate the first
ad will be more effective when shown to males for α = .05. There is evidence to reject H0 for α > .050.
There is evidence to indicate the first ad will be more effective when shown to males for α > .050.
No, I do not agree with the author’s hypothesis. The results agree with the author’s hypothesis for only the
attitude toward the advertisement id using α = .05. If we want to use α = .10, then the author’s hypotheses
are all supported.
7.116
a.
To determine if the mean salary of all males with post-graduate degrees exceeds the mean salary of
all females with post-graduate degrees, we test:
H0: μM = μF
Ha: μM > μF
( xM  xF )  0
The test statistic is z 
c.
The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV,
Appendix B, z.01 = 2.33. The rejection region is z > 2.33.
s x2M
 s x2F

(61, 340  32, 227)
b.
2,1852  9322
 12.26
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.
450
Chapter 7
d.
7.117
Since the observed value of the test statistic falls in the rejection region (z = 12.26 > 2.33), H0 is
rejected. There is sufficient evidence to indicate the mean salary of all males with post-graduate
degrees exceeds the mean salary of all females with post-graduate degrees at α = .01.
Some preliminary calculations are:
Difference
(Design 1 - Design 2)
−53
−271
−206
−266
−213
−183
−118
−87
Working Days
8/16
8/17
8/18
8/19
8/20
8/23
8/24
8/25
d =
sd2 =
sd =
d
=
nd
d
2
1, 397
= −174.625
8
 d 

2
nd
nd  1
s d2 =
(1, 397) 2
8
= 6,548.839
8 1
289, 793 
=
6, 548.839 = 80.925
To determine if Design 2 is superior to Design 1, we test:
H0: μd = 0
Ha: μd < 0
The test statistic is t 
d  μo
sd
nd

174.625  0
80.925
8
 6.103
Since no α value was given, we will use α = .05. The rejection region requires α = .05 in the lower tail of
the t-distribution with df = nd – 1 = 8 – 1 = 7. From Table V, Appendix B, t.05 = 1.895. The rejection
region is t < −1.895.
Since the observed value of the test statistic falls in the rejection region (t = −6.103 < −1.895), H0 is
rejected. There is sufficient evidence to indicate Design 2 is superior to Design 1 at α = .05.
For confidence coefficient .95, α = .05 and α/2 = .025. From Table V, Appendix B, with df = nd − 1 =
8 − 1 = 7, t.025 = 2.365. A 95% confidence interval for μd is:
d  t.025
sd
nd
 174.625  2.365
80.925
8
 174.625  67.666  (242.29,  106.96)
Since this interval does not contain 0, there is evidence to indicate Design 2 is superior to Design 1.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.