H* End of Chapter 6 Sampling Distributions
Random sampling
Central Limit Theorem
Let y1 , y2 , y3 , ....yn be a random sample of observation with finite
mean Μ and standard deviation Σ. For n sufficiently large the
samples HhistogramL have a normal distribution with the mean y and
standard deviation Σy of the samples are
y = Μ, and Σy = Σ ’
n *L
t = Table@8i, Random@Integer, 81, 100 000<D<, 8i, 1, 5<D
{{1, 63494}, {2, 38915}, {3, 90797}, {4, 3931}, {5, 37726}}
t = Table@Random@Integer, 81, 100 000<D, 8i, 1, 5<D
{43731, 99083, 56425, 83125, 64116}
H* Example: We want to measure the amount of sugar Hin gramsL in apples
Population : the probability of some number of grams of sugar in an
apple follows a uniform distribution *L
ud = UniformDistribution@810, 30<D
UniformDistribution@810, 30<D
StandardDeviation@udD  N
5.7735
lect15.nb
2
Random@udD
10.3893
H* n=50; We are measuring the amount of sugar in 50 apples.
are doing that 10 000 times*L
n = 50;
t = Table@Mean@Table@Random@udD, 8i, 1, n<DD, 8j, 1, 1000<D;
Max@tD
22.4795
? Frame
Frame is an option for Graphics, Grid and
other constructs that specifies whether to include a frame. ‡
We
lect15.nb
3
p1 = Histogram@t, Automatic, "ProbabilityDensity", Axes ® False,
PlotLabel -> "Mean ð of grams of sugar in samples of 50 apples",
Frame ® True, FrameLabel ® 8"Grams of sugar", "Fraction of means"<D
Mean ð of grams of sugar in samples of 50 apples
0.6
Fraction of means
0.5
0.4
0.3
0.2
0.1
0.0
18
19
20
Grams of sugar
21
22
lect15.nb
4
p2 =
Plot@PDF@NormalDistribution@Mean@udD, StandardDeviation@udD  Sqrt@nDD,
yD, 8y, 16, 24<, Axes ® False, Frame ® TrueD
0.4
0.3
0.2
0.1
0.0
16
18
20
22
24
lect15.nb
5
Show@p1, p2D
Mean ð of grams of sugar in samples of 50 apples
0.6
Fraction of means
0.5
0.4
0.3
0.2
0.1
0.0
16
18
20
Grams of sugar
Mean@tD
19.9665
Mean@udD
20
StandardDeviation@udD  [email protected]
0.816497
StandardDeviation@tD
0.787363
22
24
lect15.nb
6
StandardDeviation@udD  N
5.7735
n = 100;
t = Table@Random@udD, 8i, 1, 1000<, 8i, 1, n<D;
t@@1000DD
811.0399, 25.3798, 29.5786, 21.3661, 29.0269, 20.5107, 15.4397, 24.0242,
14.0438, 20.869, 28.9339, 18.508, 21.8394, 20.9444, 24.5356, 13.3603,
27.1961, 17.2017, 11.5997, 20.6982, 19.4829, 19.9768, 18.9535,
10.5294, 18.443, 24.5971, 19.3749, 19.1633, 19.4161, 14.0864,
13.9353, 25.1391, 15.3723, 23.2174, 15.0014, 16.6311, 23.5329,
12.273, 20.4658, 13.2708, 26.3368, 25.0713, 18.8661, 22.5726,
16.8539, 15.0945, 29.9126, 22.0432, 28.4109, 20.4974, 20.5376,
12.8799, 18.9947, 16.411, 16.6023, 17.7407, 13.6224, 23.1936,
11.6009, 11.1096, 20.0895, 20.9207, 21.1351, 27.8388, 23.7527,
25.8494, 12.269, 15.2663, 16.8988, 20.7549, 12.3565, 23.2231,
18.488, 10.2576, 21.8188, 20.3432, 29.4933, 23.8465, 15.2165,
12.6025, 25.8709, 10.6529, 13.6156, 11.4929, 15.7813, 19.7323,
22.4804, 13.654, 22.0286, 23.8829, 20.2114, 28.3878, 15.1298,
13.1279, 17.8549, 15.1647, 26.6418, 12.8704, 26.0361, 24.8215<
t1 = Table@Mean@t@@iDDD, 8i, 1, 1000<D;
lect15.nb
7
p3 = Histogram@t1, Automatic, "ProbabilityDensity", Axes ® False,
PlotLabel -> "Mean ð of grams of sugar in samples of 100 apples",
Frame ® True, FrameLabel ® 8"Grams of sugar", "Fraction of means"<D
Mean ð of grams of sugar in samples of 100 apples
0.7
Fraction of means
0.6
0.5
0.4
0.3
0.2
0.1
0.0
18.5
19.0
19.5
20.0
20.5
Grams of sugar
Mean@udD
20
n
100
StandardDeviation@udD  Sqrt@nD  N
0.57735
Mean@t1D
19.9942
21.0
21.5
22.0
lect15.nb
8
StandardDeviation@t1D
0.581893
H* Example 6.19
Let n = 5 HCentral Limit Theorem really applies to n >= 30L. Population
has an exponential distribution with mean = 1 *L
ed := ExponentialDistribution@1D
Plot@PDF@ed, yD, 8y, 0, 10<, PlotRange ® AllD
1.0
0.8
0.6
0.4
0.2
2
4
6
8
n = 5;
Mean@Table@Random@edD, 8i, 1, n<DD
0.961504
t = Table@Mean@Table@Random@edD, 8i, 1, n<DD, 8j, 1, 1000<D;
10
lect15.nb
9
p4 = Histogram@t, Automatic, "ProbabilityDensity"D
1.0
0.8
0.6
0.4
0.2
0.5
Mean@edD
1.
Mean@tD
1.00116
1.0
1.5
2.0
2.5
lect15.nb
10
p5 =
Plot@PDF@NormalDistribution@Mean@edD, StandardDeviation@edD  Sqrt@nDD,
yD, 8y, 0, 3<, Axes ® False, Frame ® TrueD
0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
2.0
2.5
3.0
Show@p4, p5D
1.0
0.8
0.6
0.4
0.2
0.5
1.0
1.5
lect15.nb
11
StandardDeviation@edD  [email protected]
0.447214
StandardDeviation@tD
0.456752
n = 30;
t = Table@Mean@Table@Random@edD, 8i, 1, n<DD, 8j, 1, 1000<D;
p6 = Histogram@t, Automatic, "ProbabilityDensity"D
2.5
2.0
1.5
1.0
0.5
0.6
Mean@tD
0.987712
Mean@edD
1
0.8
1.0
1.2
1.4
1.6
lect15.nb
12
StandardDeviation@edD  [email protected]
0.182574
StandardDeviation@tD
0.182736
p7 =
Plot@PDF@NormalDistribution@Mean@edD, StandardDeviation@edD  Sqrt@nDD,
yD, 8y, 0, 2<, Axes ® False, Frame ® TrueD
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
lect15.nb
13
Show@p6, p7D
2.5
2.0
1.5
1.0
0.5
0.0
1.0
Example 6.21
nd = NormalDistribution@60, 2D
NormalDistribution@60, 2D
1 - CDF@nd, 65D  N
0.00620967
1.5
2.0
lect15.nb
14
Plot@PDF@nd, yD, 8y, 50, 70<D
0.20
0.15
0.10
0.05
55
60
65
70
lect15.nb
15
1 - CDF@nd, 73D  N
4.016 ´ 10-11
H* 6.22
Why is the proportion of the number of heads HsuccessesL subtracted by p,
then that quantity is divided by Sqrt@ p q  nD a standard normal
variable for large values of n? *L
z = Hphat - pL  Sqrt@ p q  nD
H*
The expected value of a coin toss is p.
deviation of a coin toss Sqrt@ p qD.
The standard
The proportion of the number of heads is a mean Hð of headsLn
and when we compute the proportion,
we are doing a central lmit type of experiment. So the standard
deviation, from the central limit theorem is Sqrt@ p qnD *L
H* Central Limit Theorem applied to sums
Sy *L
mean = n Μ; Σsum = Σ Sqrt@nD
H* The binomial distribution is Hð of headsL nn = Hð of headsL*L
bd = BinomialDistribution@10, .5D
BinomialDistribution@10, 0.5D
H* For large n,
the binomial distribution looks like the normal distribution
with mean = n p, and with standard deivation= Sqrt@n p qD *L
lect15.nb
16
nd = NormalDistribution@5, Sqrt@10 ´ .5 ´ .5DD
NormalDistribution@5, 1.58114D
t1 = Table@8y - .5, PDF@bd, yD<, 8y, 0, 10<D
88-0.5, 0.000976563<, 80.5, 0.00976562<, 81.5, 0.0439453<,
82.5, 0.117187<, 83.5, 0.205078<, 84.5, 0.246094<, 85.5, 0.205078<,
86.5, 0.117187<, 87.5, 0.0439453<, 88.5, 0.00976562<, 89.5, 0.000976563<<
p1 = ListPlot@t1, PlotLabel ® "Binomial Distribution: n=10, p=.5",
AxesLabel ® 8"y", "pHyL"< , Joined ® True, InterpolationOrder ® 0,
Mesh ® FullD
Binomial Distribution: n=10, p=.5
pHyL
0.25
0.20
0.15
0.10
0.05
y
2
4
6
8
lect15.nb
17
p2 = Plot@PDF@nd, yD, 8y, 0, 10<, PlotRange -> AllD
0.25
0.20
0.15
0.10
0.05
2
4
6
8
10
Show@p1, p2, PlotRange -> All,
PlotLabel ® "Binomial and Normal Distributions"D
Binomial and Normal Distributions
pHyL
0.25
0.20
0.15
0.10
0.05
y
2
4
6
8
10
lect15.nb
18
CDF@bd, 4D
0.376953
CDF@nd, 4D
0.263545
CDF@nd, 4.5D
0.375915
H* Thm 6.10
Let y1 ,y2 ,
y3 ,....yn be normally distributed random variables with Ei =Μi ,
VHyi L=Σi2 ,
CovIyi ,yj M=Σij
If
l = â ai yi , then
n
i=1
E HlL = Μ = â ai Μi ,
n
i=1
V HlL = Σ2 = â a2i Σi2 + 2
n
Σij = Cov Iyi yj M
i=1
â
n
â ai aj Σij ,
n
i=1,i¹j, j=1
1
6.21
Μ = 60, n = 25 Σ = 10 therefore y = 60, Σy = Σ ’
25 = 2 *L
lect15.nb
19
nd = NormalDistribution@60, 2.D
NormalDistribution@60, 2.D
Plot@PDF@nd, yD, 8y, 52, 68<D
0.20
0.15
0.10
0.05
60
65
1 - CDF@nd, 65D
0.00620967
H* If we got a result of an LCN>65,
it is unlikely that the mean is 60. Therefore we conclude that
the new concrete has an LCN>60. *L