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Unit 2 homework solutions
7.x a.
The sample space associated with n=3 consists of
Nn=33=27 elementary events; namely, S={(1,1,1);
(1,1,2);(1,1,3);(1,2,1)(1,2,2);(1,2,3);(1,3,1);
(1,3,2);(1,3,3);(2,1,1);(2,1,2);(2,1,3);(2,2,1);
(2,2,2);(2,2,3);(2,3,1);(2,3,2);(2,3,3);(3,1,1);
(3,1,2);(3,1,3);(3,2,1)(3,2,2);(3,2,3);(3,3,1);
(3,3,2);(3,3,3)}.
b. The sample space of the mean, then, is
_
S( x )={(1),(4/3),(5/3),(4/3),(5/3),(2),(5/3),(2),(7/3),
(4/3),(5/3),(2),(5/3),(2),(7/3),(2),(7/3),(8/3),
(5/3),(2),(7/3),(2),(7/3),(8/3),(7/3),(8/3),(3)}.
c.
Frequency
Class interval
.5-under 1.5
1.5-under 2.5
2.5-under 3.5
Frequency
4
19
4
19
4
.5
1.5 2.5 3.5
Class Endpoint
7.18 In this problem we have µ = 99.9 and σ = 30 ; and since
n=38>30, the Central Limit Theorem applies.
_
a. P( x <90) is obtained by first calculating the z
_
score associated with x =30, which is
z=(90-99.9)(30/ 38 )=-2.03. This z value corresponds
_
to area equal .4788, so P( x <90)=.5-.4788=.0212.
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d. P(93< x <96) is obtained by calculating the z scores
associated with sample means of 93 and 96. These z
scores are z=-1.42 and z=-.80, which correspond to
areas equal .4222 and .2881, respectively. So
_
P(93< x <96)=.4222-.2881=.1341.
8.1 In all of the parts (a-c) of this problem n>30, so the
Central Limit Theorem applies; hence, the sampling
distribution of the population mean is the normal
distribution. In part (a), where the population
standard deviation is known, equation (8.1) on p. 313
is used to yield the interval estimate
[24.114, 25.886]; in parts (b) and (c), where the
sample standard deviation is given, the equation at the
top of p. 319 is used to yield the two interval
estimates [113.173, 126.028] and [3.136, 3.702],
respectively.
8.4 Since the population is normally distributed with known
variance, the sampling distribution of the mean is
normal, even though the sample size n=15<30.
Accordingly, equation (8.1) on p. 313 is used to yield
the interval estimate [7.852, 9.488].
8.18 Since the sample size n=15<30 and the population
standard deviation is unknown, the sampling
distribution of the population mean is the t
distribution (even though the population is normally
distributed; cf. Problem 8.4). Accordingly, equation
(8.3) on p. 324 is used to yield the interval estimate
[1.955, 2.773].
8.34 The sampling distribution of the population variance
is the chi-square distribution, so equation (8.6) on
p. 334 is used to construct the confidence intervals
in this problem.
b. This confidence interval is
6 *1.24 2
6 *1.24 2
2
≤
σ
≤
χ.2025 , 6
χ .2975 , 6
which implies
6 *1.24 2
6 *1.24 2
≤σ 2 ≤
14.494
1.237
which implies .637 ≤σ 2 ≤7.458 .
c. This confidence interval is
19 * 32 2
19 * 32 2
19, 456
19,456
2
≤
σ
≤
which implies
≤σ 2 ≤
2
2
30.144
10.117
χ .05 ,19
χ.95 ,19
which implies 645.435 ≤σ 2 ≤1,923.100 .
9.2
This is a one-tailed test with rejection region in the
left tail of the distribution. Since the sample size
is n=96>30, the population mean has a normal
distribution. Accordingly, the test statistic is
calculated using (9.2) on p. 365; it is
z = 96 * (6.91 − 7.48) / 1.21 = 9.798*(-.57)/1.21=-4.607.
Since this one tailed hypothesis is tested with a 99%
level of confidence, the critical value is the value
of z that implies a region of rejection in the left
hand tail with area equal to .01. Inspection of the z
table shows that this value of z is the critical
z=-2.33.
Since the test statistic lies in the region of
rejection (i.e., since -4.607<-2.33), the null
hypothesis is rejected.
9.3
The test in this problem is the same as in 9.2 above
except that here it is an upper tail test.
a. The test statistic is z= 113 *(1,2151,200)/100=10.630*15/100 =1.59. The critical value
is z.1=1.28. Since the test statistic lies in the
region of rejection, the null hypothesis is
rejected.
b. The p-value is the smallest value of α at which the
null hypothesis is rejected; that is, the area of
the region of rejection at the test statistic
z=1.59. Inspection of the z table shows that this
p-value is .5-.4441=.0559.
9.11 This is a two tailed test. Since the sample size is
n=20<30 and the population standard deviation is
unknown, the sampling distribution is the t
distribution. The test statistic is
t= 20 (16.45-16)/3.59=4.472*.45/3.59=.561. The
critical values are ±t.025,19=-2.093 and 2.093. Since
the test statistic lies between these two critical
values (i.e., in the region of non-rejection), the
null hypothesis is not rejected.
9.12
This is a one tailed test with rejection region in
the lower tail. Since the sample size is n=8 and the
population variance is unknown, the sampling
distribution is the t distribution. The test
statistic is t= 8 *(58.42-60)/ 25.68 =-.882. The
critical value is -t.01,7=-2.998. Since the test
statistic lies above the region of rejection, the
null hypothesis cannot be rejected.
9.28
These tests are about the population variances of
normally distributed variables, so the sampling
distribution is the chi-square distribution.
a. This is a one tailed test with rejection region in
the upper tail. The test statistic is
χ 2 = (n − 1)s 2 / σ 2 =14*32/20=22.4. The critical value is
χ.205 ,14 =23.6848. Since the test statistic lies below
the region of rejection, the null hypothesis cannot
be rejected.
b. This is a two tailed test. The test statistic is
21*17/8.5=42. The critical values are χ.205 , 21 =32.671
and χ.295 , 21 =11.591. Since the test statistic lies in
the upper region of rejection, the null hypothesis
is rejected.
c. This is a one tailed test with rejection region in
the lower tail. The test statistic is
7*4.122/45=2.640. The critical value is χ.299 , 7 =1.239.
Since the test statistic lies above the region of
rejection, the null hypothesis cannot be rejected.
10.1 a.
The null and alternative hypotheses are:
Ho: µ1 = µ 2
Ha: µ1 ≠ µ 2
Since both n1>30 and n2>30, the sampling
distribution of the difference in population
means is normal. The test statistic is calculated
using equation (10.2) on p. 413, to yield
(51.3 − 53.2) − 0
Z=
=-1.9/1.871=-1.015.
52 60
+
32 32
The critical values are -z.05=-1.645 and
z.05=1.645. Since the test statistic lies in
between these critical values, the null
hypothesis cannot be rejected.
10.2 Since both sample sizes are greater than 30 the
sampling distribution of the difference in population
means is normal. Accordingly, formula (10.4) on
p. 416 yields the 90% confidence interval:
(70.4-68.7)-z.05
5.76 2 6.12
5.76 2 6.12
+
≤ µ1 − µ 2 ≤(70.4 − 68.7 ) + z .05
+
32
31
32
31
Substituting z.05=1.645 yields − .76 ≤ µ1 − µ 2 ≤4.16 .
10.15 This is a one tailed test with rejection region in
the lower tail. Since both sample sizes are less
than 30 and the population variances are unknown, the
sampling distribution of the difference in population
means has a t distribution. The test statistic is
calculated using equation (10.5) on p. 422; it is
t=
(24.56 − 26.42) − 0
7 *12.4 + 10 *15.8 1 1
+
8 + 11 − 2
8 11
=-1.05.
The critical value is -t.01,17=-2.567. Since the test
statistic lies above the rejection region, the null
hypothesis cannot be rejected.
10.63 This is a two tailed test. The sampling distribution
for the difference in population variances is the F
distribution. The test statistic is calculated using
equation (10.12) on page 454; it is
F=4.682/2.782=2.83. The critical values are found in
Table A.7 on p. A-20; they are F.025,4,18=3.61 and
1/F.025,4,18=1/3.61=.277. Since the test statistic lies
in between these critical values, the null hypothesis
cannot be rejected.
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11.5
_
_
The class means are x1 =2, x 2 =4, and x3 =4.1667; the
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grand mean is x =3.5294. The between class variation
is SSC=6*((2-3.5294)2+(4.1667-3.5294)2)+5(4-3.5294)2=
22.202 and the within class variation is
SSE=(2-2)2+(1-2)2+…+(1-2)2+(5-4)2+…+(5-4)2+(3-4.1667)2+…
+(5-4.667)2=14.033. The test statistic is
=[(22.202/2)]/[(14.033/14)]=11.075 and the critical
value is F.05,2,14=3.74, as shown on page A-18. Since
the test statistic is larger than the critical value
(i.e., the former lies in the rejection region), the
null hypothesis is rejected.