Chapter 8
Sampling Distributions and Estimation
8.1
a.
n
σ
=
32
4
= 16
b.
n
σ
=
32
16
= 8
c.
n
σ
=
32
64
= 4
8.2
a.
1.96
n
σ
μ ±
=
12
200
1.960
36
±
or (196.08, 203.92).
b.
1.96
n
σ
μ ±
=
15
1000
1.960
9
±
or (990.2, 1009.80).
c.
1.96
n
σ
μ ±
=
1
50
1.960
25
±
or (49.608, 50.392).
8.3
a.
1.96
μ
σ
±
= 4.035
1.96*0.005
±
or (4.0252, 4.0448).
b.
1.96
n
σ
μ ±
=
0.005
4.035
1.96
25
±
or (4.03304, 4.03696).
c. In either case, we would conclude that our sample came from a population that did
not
have a population
mean equal to 4.035.
8.4
a.
1.
No, for
n
= 1 the 100 samples don’t represent a normal distribution 2.The distribution of the sample
means becomes more normally distributed as n increases.
3. The standard error becomes closer to that
predicted by the CLT the larger the sample becomes.
4.
This demonstration reveals that if numerous
samples were taken and analyzed we can confirm the CLT.
In the real word, based on our notion of the
true mean, we can assess this.
We can generate the 95% range and determine if our values are within this
range or not. Also, recognize that there is a low probability of this single range not being representative.
b.
1.
No, for
n
= 1 the 100 samples don’t represent a normal distribution 2.The distribution of the sample
means becomes more normally distributed as n increases.
The standard error becomes closer to that
predicted by the CLT the larger the sample becomes.
4.
This demonstration reveals that if numerous
samples taken and analyzed we can confirm the CLT.
In the real word, based on our notion of the true
mean, we can assess this.
We can generate the 95% range and determine if our values are within this
range or not. Also, recognize that there is a low probability of this single range not being representative.
8.5
a.
1.645
x
n
σ
±
=
4
14
1.645
5
±
or (11.057, 16.943).
b.
2.576
x
n
σ
±
=
5
37
2.576
15
±
or (33.675, 40.325).
68
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c.
1.96
x
n
σ
±
=
15
121
1.96
25
±
or (115.12, 126.88).
8.6
Exam 1:
1.96
x
n
σ
±
=
7
75
1.96
10
±
or
(70.661, 79.339).
Exam 2:
1.96
x
n
σ
±
=
7
79
1.96
10
±
or
(74.661, 83.339).
Exam 3:
1.96
x
n
σ
±
=
7
65
1.96
10
±
or
(60.661, 69.339)
The first two confidence intervals overlap.
This suggests that the first two exams had the same population
mean.
8.7
1.96
x
n
σ
±
=
0.005
2.475
1.96
15
±
or (2.4725, 2.4775).
8.8
a.
s
x
t
n
±
=
3
24
1.9432
7
±
or (21.797, 26.203).
b.
s
x
t
n
±
=
6
42
2.8982
18
±
or (37.901, 46.099).
c.
s
x
t
n
±
=
14
119
2.0518
28
±
or (113.571, 124.429).
Note: t values are found using the Excel
formula =tinv( (1

cc),n

1)
where cc is the confidence coefficient.
For a. this, this would be = tinv((1.90), 6)
8.9
a. Appendix D = 2.262, Excel = tinv(.05, 9) = 2.2622
b. Appendix D = 2.602, Excel = tinv(.02, 15) = 2.6025
c. Appendix D = 1.678 ,Excel = tinv(.10, 47) =1.6779
8.10
a. Appendix D = 2.021, Excel = tinv(.05, 40) = 2.0211
b. Appendix D = 1.990, Excel = tinv(.05, 80) = 1.9901
c. Appendix D = 1.984, Excel = tinv(.05, 100) = 1.984
All are fairly close to 1.96.
8.11
a.
s
x
t
n
±
=
27.79
45.66
2.0860
21
±
or (33.01, 58.31).
b. The confidence interval could be narrower increasing the size of the sample or decreasing the confidence
level.
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 Spring '08
 HEATHERADAMS
 Normal Distribution, producer, 1.96, representative., Megastat

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