Basic Statistics
Professor Sungsue Rheem (임성수 교수) 고려대학교 행정전문대학원/경제통계학부
Course Material 6 (강의 자료 6) * 1)
Contents
Example 2-10. A Die Rolling Experiment with an Interest in the Mean of a Multi-Value Variable (여러 값을 갖는 변수의 평균에 관심을 둔 주사위 굴리기 실험)
* Copyright © 2016 임성수 (Sungsue Rheem, [email protected]). All rights reserved.
Example 2-10. A Die Rolling Experiment with Interest in the Mean of a Multi-Value Variable (여러 값을 갖는 변수의 평균에 관심을 둔 주사위 굴리기 실험)
[1] The original distribution of an outcome from rolling a die
Let us determine the outcome space of a random trial of rolling a die. Here, an
outcome is the number on the face of a die. Also, determine the original distribution of an outcome from rolling a die.
Answer:
Table 1. Outcome space from rolling a die and the probability for each point Point in the outcome space Probability
1 1/6 (1 over six, one sixth)
2 1/6 (1 over six, one sixth)
3 1/6 (1 over six, one sixth)
4 1/6 (1 over six, one sixth)
5 1/6 (1 over six, one sixth)
6 1/6 (1 over six, one sixth)
Let X denote an outcome from rolling a die. Then, the possible values of X are 1, 2, 3, 4, 5, 6. This X is a random variable. The population of X is {1, 2, 3, 4, 5, 6}, and since each of the points in the population are equally likely, the original distribution of X is (1/6, 1/6, 1/6, 1/6, 1/6, 1/6).
Table 2. Table of the original distribution of X, an outcome from rolling a die
X Probability
1 1/6
2 1/6
3 1/6
4 1/6
5 1/6
6 1/6
1/6
1 2 3 4 5 6 X
Figure 1. Graph of the original distribution of X, which is an outcome from rolling a die
[2] The Expectation (the Original Mean) of a Random Variable
The expectation (expected value) of a random variable is defined as
E(X) =
=
[mu] (1)Let X denote an outcome from rolling a die. Find E(X), the expectation of X. The probability distribution of X is given in Table 2. Use formula (1) to calculate this expectation. When you do calculations, use the table like the below to avoid possible mistakes.
Answer: E(X) =
=
x P(x) x P(x)
1 1/6 1/6
2 1/6 2/6
3 1/6 3/6
4 1/6 4/6
5 1/6 5/6
6 1/6 6/6
sum 21/6 = 7/2 = 3.5
E(X) = sum of the products of x and its probability = (1+2+3+4+5+6) / 6 = 3.5 As you see in Example 2, the expectation of X is the original mean (평균) of X, a
measure of location (위치 측도). A Greek letter
[mu] is used to denote the original mean.
is a parameter (모수).[3] The Original Variance and Standard Deviation of a Random Variable
Another parameter is the original variance.The expectation of a function g of a random variable X, say g(X), is defined as
E[g(X)] =
The original variance (모분산) of X is a measure of dispersion (산포의 측도) of X. It is defined as
=
[sigma squared] (2)
is read as the variance of X.
, which is an individual value minus the mean, is called a deviation (편차). The variance (분산) is the expectation of the squared deviation (편차 제곱의 평균). A shortcoming of the variance is that its unit is the square of the unit of the variable. For example, when the unit of X is cm (length, 길이), the unit of V(X) is cm2(area, 넓이).The original standard deviation (모 표준편차), also a measure of dispersion, is the square root of the original variance.
x P(x) mu x-mu (x-mu)^2 (x-mu)^2*P(x)
1 0.1667 3.5 -2.5 6.25 1.041667
2 0.1667 3.5 -1.5 2.25 0.375000
3 0.1667 3.5 -0.5 0.25 0.041667
4 0.1667 3.5 0.5 0.25 0.041667
5 0.1667 3.5 1.5 2.25 0.375000
6 0.1667 3.5 2.5 6.25 1.041667
sum
1.0000 sum 2.916667 = V(X) =
1.707825 = sqrt[V(X)] =
(3)The standard deviation,
[sigma] is a better measure of dispersion than the variance, because its unit is the same as the unit of the variable.Now, let us find the original variance and original standard deviation of X, an outcome from rolling a die.
Let X denote an outcome from rolling a die. Find V(X) =
, the population variance of X, and
, the population standard deviation of X. The probability distribution of X is given in Table 2. Use formulas (2) and (3) to calculate
and
. When you do calculations, use the table like the below to avoid possible mistakes.Answer:
[4] The sampling distribution of the sample mean from a sample of size 2
Suppose we roll a die twice. This can be considered as taking a sample of size 2 from the population of an outcome from rolling a die. Let X1 denote the first outcome as a random variable and X2 the second outcome as a random variable. Find the sampling distribution of
, the sample mean from this sample of size 2.Answer: Let us construct the sample space of this random experiment. Since the sample space consists of all possible (x1, x2)s, this sample space is two-dimensional.
The two-dimensional coordinate system (2차원 좌표 체계) can be used to describe this sample space. This sample space consists of the 36 dot(●)s on the (X1, X2) plane shown below.
X2 |
6 + ● ● ● ● ● ● |
5 + ● ● ● ● ● ● |
4 + ● ● ● ● ● ● |
3 + ● ● ● ● ● ● |
2 + ● ● ● ● ● ● |
1 + ● ● ● ● ● ● |
+---+---+---+---+---+---+ X1
1 2 3 4 5 6
Figure 2. The sample space of an experiment of rolling a die twice
Figure 2 says that this sample space is { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }
which is a set that consists of 36 two-dimensional points.
Then, first, let us calculate all the sample means of x1 and x2 in the following table.
Point in the
sample space x1 x2
1 1 1 1.0
2 1 2 1.5
3 1 3 2.0
4 1 4 2.5
5 1 5 3.0
6 1 6 3.5
7 2 1 1.5
8 2 2 2.0
9 2 3 2.5
10 2 4 3.0
11 2 5 3.5
12 2 6 4.0
13 3 1 2.0
14 3 2 2.5
15 3 3 3.0
16 3 4 3.5
17 3 5 4.0
18 3 6 4.5
19 4 1 2.5
20 4 2 3.0
21 4 3 3.5
22 4 4 4.0
23 4 5 4.5
24 4 6 5.0
25 5 1 3.0
26 5 2 3.5
27 5 3 4.0
28 5 4 4.5
29 5 5 5.0
30 5 6 5.5
31 6 1 3.5
32 6 2 4.0
33 6 3 4.5
34 6 4 5.0
35 6 5 5.5
36 6 6 6.0
Now, let us construct the table of
and its frquency and probability according to the possible values of
. The following are the table and graph of the sampling distribution of
.
Frequency(
) P(
)1.0 1 1/36 = 0.0278
1.5 2 2/36 = 0.0556
2.0 3 3/36 = 0.0833
2.5 4 4/36 = 0.1111
3.0 5 5/36 = 0.1389
3.5 6 6/36 = 0.1667
4.0 5 5/36 = 0.1389
4.5 4 4/36 = 0.1111
5.0 3 3/36 = 0.0833
5.5 2 2/36 = 0.0556
6.0 1 1/36 = 0.0278
6/36 5/36 4/36 3/36 2/36 1/36
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
Notice that the shape of this sampling distribution of
is closer to a bell (종) than that of the original distribution of X in Figure 1, even though the sample has its minimum size, which is 2.[5] The sampling expectation of the sample mean from a sample of size 2
The expectation from the sampling distribution is called the sampling expectation.
Find the sampling expectation of the sample mean
in Example 4.Answer: The table below is to calculate E(
) =
.x bar P(x) mu x bar-mu (x bar-mu)^2 (x bar-mu)^2*P(x)
1.0 0.0278 3.5 -2.5 6.25 0.173611
1.5 0.0556 3.5 -2.0 4.00 0.222222
2.0 0.0833 3.5 -1.5 2.25 0.187500
2.5 0.1111 3.5 -1.0 1.00 0.111111
3.0 0.1389 3.5 -0.5 0.25 0.034722
3.5 0.1667 3.5 0.0 0.00 0.000000
4.0 0.1389 3.5 0.5 0.25 0.034722
4.5 0.1111 3.5 1.0 1.00 0.111111
5.0 0.0833 3.5 1.5 2.25 0.187500
5.5 0.0556 3.5 2.0 4.00 0.222222
6.0 0.0278 3.5 2.5 6.25 0.173611
sum 1.0000 sum 1.458333 = V(
) =
P( ) P( )
1.0 0.0278 0.0278
1.5 0.0556 0.0833
2.0 0.0833 0.1667
2.5 0.1111 0.2778
3.0 0.1389 0.4167
3.5 0.1667 0.5833
4.0 0.1389 0.5556
4.5 0.1111 0.5000
5.0 0.0833 0.4167
5.5 0.0556 0.3056
6.0 0.0278 0.1667
sum 1.0000 3.5000
The calculation in the above table yields the answer: E(
) =
= 3.5.This sampling expectation 3.5 is the same as the original expectation 3.5, which was found in Example 2. In general,
= sampling E(
) = original E(
) =
[6] The sampling variance of the sample mean from a sample of size 2
Find the sampling variance of the sample mean
in Example 4.Answer: The table below is to calculate
This sampling variance is
1.458333 = 2.916667 / 2 = 2.916667 / n (n is the sample size),
where 2.916667 = V(X) =
. In general,
= sampling V(
) = [original V(
)] / n =
[7] The standard error of the sample mean from a sample of size 2
The standard error (표준 오차, SE) of a statistic is the square root of the sampling variance of it, which is the sampling standard deviation of it.
=
=
Find the standard error of the sample mean
.Answer:
=