1 15.math-review statistics. 2 zlet us consider x 1, x 2,…,x n, n independent identically...
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15.Math-Review
StatisticsStatistics
15.Math-Review 2
Let us consider X1, X2,…,Xn, n independent identically distributed random variables with mean and standard deviation .
And define:
Central Limit Theorem
n
iin XS
1
15.Math-Review 3
Central Limit Theorem
The Central Limit Theorem (CLT) states: If n is large (say n30) then Sn follows approximately a
normal distribution with mean n and standard deviation
If n is large (say n30) then follows approximately a normal distribution with mean and standard deviation
n
nnS1
n
15.Math-Review 4
Central Limit Theorem
Example: sums of a Bernoulli random variable
Frequency Chart
.000
.199
.399
.598
.797
0
7974
1.00 1.25 1.50 1.75 2.00
10,000 Trials 0 Outliers
Forecast: n=1
Frequency Chart
.000
.075
.149
.224
.298
0
745.2
2981
14.50 15.88 17.25 18.63 20.00
10,000 Trials 75 Outliers
Forecast: n=10
Frequency Chart
.000
.044
.089
.133
.178
0
444.5
889
1778
48.00 51.00 54.00 57.00 60.00
10,000 Trials 28 Outliers
Forecast: n=30
Frequency Chart
.000
.036
.073
.109
.146
0
364.7
729.5
1459
82.50 86.25 90.00 93.75 97.50
10,000 Trials 70 Outliers
Forecast: n=50
15.Math-Review 5
Central Limit Theorem
Example: Averages of Bernoulli random variable
Frequency Chart
.000
.075
.149
.224
.298
0
745.2
2981
1.45 1.59 1.73 1.86 2.00
10,000 Trials 75 Outliers
Forecast: mean, n=10
Frequency Chart
.000
.044
.089
.133
.178
0
444.5
889
1778
1.60 1.70 1.80 1.90 2.00
10,000 Trials 47 Outliers
Forecast: mean, n=30
Frequency Chart
.000
.036
.073
.109
.146
0
364.7
729.5
1459
1.65 1.73 1.80 1.88 1.95
10,000 Trials 70 Outliers
Forecast: mean, n=50
Frequency Chart
.000
.199
.399
.598
.797
0
7974
1.00 1.25 1.50 1.75 2.00
10,000 Trials 0 Outliers
Forecast: n=1
15.Math-Review 6
Central Limit Theorem
Example: Compare a binomial random variable X~B(40,0.2) with its normal approximation: What is the normal approximation? Compare P(X10), P(X 20), P(X30) for the binomial and the
normal approximation.
BINOMIAL: AVERAGE: NORMAL:
X<=5 0.16133 X<5 0.07591 0.11862 0.11784X<=10 0.83923 X<10 0.73178 0.78550 0.78540X<=20 0.99999 X<20 0.99998 0.99999 1.00000X<= 30 1.00000 X<30 1.00000 1.00000 1.00000
15.Math-Review 7
Let us consider the following example. We work at a phone company and we would like to be
able to estimate the shape of the demand. We assume that monthly household telephone bills follow
a certain probability distribution (continuous) We have obtained the following data of monthly
household telephone bills by interviewing 70 randomly chosen households (or their habitants rather) for the month of October.
Sampling
15.Math-Review 8
Table:
Sampling
Respondent October Respondent October Respondent OctoberNumber Phone Bill Number Phone Bill Number Phone Bill
1 95.67$ 25 79.32$ 49 90.02$ 2 82.69$ 26 89.12$ 50 61.06$ 3 75.27$ 27 63.12$ 51 51.00$ 4 145.20$ 28 145.62$ 52 97.71$ 5 155.20$ 29 37.53$ 53 95.44$ 6 80.53$ 30 97.06$ 54 31.89$ 7 80.81$ 31 86.33$ 55 82.35$ 8 60.93$ 32 69.83$ 56 60.20$ 9 86.67$ 33 77.26$ 57 92.28$
10 56.31$ 34 64.99$ 58 120.89$ 11 151.27$ 35 57.78$ 59 35.09$ 12 96.93$ 36 61.82$ 60 69.53$ 13 65.60$ 37 74.07$ 61 49.85$ 14 53.43$ 38 141.17$ 62 42.33$ 15 63.03$ 39 48.57$ 63 50.09$ 16 139.45$ 40 76.77$ 64 62.69$ 17 58.51$ 41 78.78$ 65 58.69$ 18 81.22$ 42 62.20$ 66 127.82$ 19 98.14$ 43 80.78$ 67 62.47$ 20 79.75$ 44 84.51$ 68 79.25$ 21 72.74$ 45 93.38$ 69 76.53$ 22 75.99$ 46 139.23$ 70 74.13$ 23 80.35$ 47 48.06$ 24 49.42$ 48 44.51$
15.Math-Review 9
From this information we would like to be able to estimate, for example: What is an estimate of the shape of the distribution of October household telephone
bills? What is an estimate of the percentage of households whose October telephone bill
is bellow $45.00 What is an estimate of the percentage of households whose October telephone bill
is between $60.00 and $100.00? What is an estimate of the mean of the distribution of October household telephone
bills? What is an estimate of the standard deviation of the distribution of October
household telephone bills?
Sampling
15.Math-Review 10
A population (or “universe”) is the set of all units of interest. A sample is a subset of the units of a population. A random sample is a sample collected in such a way that
every unit in the population is equally likely to be selected. It is hard to ensure that a sample will be random.
Sampling
15.Math-Review 11
In our example the population corresponds to all the households in our area of coverage.
The random sample selected were the 70 households (or their inhabitants) interviewed.
And for the random variables X1,X2,… ,Xn corresponding to households 1, 2,… , n we observed x1=$95.67, x2=$82.69,… , xn=$74.13.
Note that if we had chosen a different random set of households we would have observed a different collection of values.
Sampling
15.Math-Review 12
To fix notation: n will be our random sample size. X1,X2,… ,Xn correspond to the random variables of unknown
distribution f(x), which is common to our population, and what we want to study.
x1,x2,… ,xn are the observations obtained by observing the outcome of our random sample. These are numbers!!
We try to use these numbers to estimate the characteristics of f(x), for example what is the distribution, what is its mean, variance, etc.
Sampling
15.Math-Review 13
To “look” at the shape of the distribution of X it is useful to create a frequency table and histogram of the sample values x1,x2,… ,xn.
Sampling
Interval Limit Frequency % Cumulative %-30 0 0.00% 0.00%
30-40 3 4.29% 4.29%40-50 6 8.57% 12.86%50-60 7 10.00% 22.86%60-70 13 18.57% 41.43%70-80 12 17.14% 58.57%80-90 11 15.71% 74.29%
90-100 9 12.86% 87.14%100-110 0 0.00% 87.14%110-120 0 0.00% 87.14%120-130 2 2.86% 90.00%130-140 2 2.86% 92.86%140-150 3 4.29% 97.14%150-160 2 2.86% 100.00%
160- 0 0.00% 100.00%
Histogram of Sample of October Telephone Bills
0
2
4
6
8
10
12
14
-30
30
-40
40
-50
50
-60
60
-70
70
-80
80
-90
90
-10
0
10
0-1
10
110
-12
0
12
0-1
30
13
0-1
40
14
0-1
50
15
0-1
60
16
0-
Range for Oct. Bill
Nu
mb
er
of
ho
us
eh
old
s
15.Math-Review 14
A histogram can be obtained from excel, the output looks something like this:
Sampling
Bin FrequencyCumulative % Bin FrequencyCumulative %30 0 .00% 70 13 18.57%40 3 4.29% 80 12 35.71%50 6 12.86% 90 11 51.43%60 7 22.86% 100 9 64.29%70 13 41.43% 60 7 74.29%80 12 58.57% 50 6 82.86%90 11 74.29% 40 3 87.14%
100 9 87.14% 150 3 91.43%110 0 87.14% 130 2 94.29%120 0 87.14% 140 2 97.14%130 2 90.00% 160 2 100.00%140 2 92.86% 30 0 100.00%150 3 97.14% 110 0 100.00%160 2 100.00% 120 0 100.00%
More 0 100.00% More 0 100.00%
15.Math-Review 15
From this analysis we can give the following description of the shape of this distribution (qualitative): An estimate of the shape of the distribution of October telephone bills
in the site area is that it is shaped like a Normal distribution, with a peak near $65.00, except for a small but significant group in the range between $125.00 and $155.00.
Sampling
15.Math-Review 16
In order to answer the other relevant questions we can use the original data, and count favorable outcomes and divide by total possible outcomes (70): P(X 45.00) = 5/70 = 0.07 P (60.00 X 100.00) = 45/70 = 0.64
Here we are approximating the continuous unknown distribution by the discrete distribution given by the outcomes of the sample
Sampling
15.Math-Review 17
Sample mean, variance and standard deviation: From our observed values x1,x2,… ,xn, we can compute:
Sampling
n
ii
n
ii
n
ii
n
xxn
s
xxn
s
xnn
xxx
1
2
1
22
1
1
1
1
deviation, stardard sample observed The
1
1
variance,sample observed The
1
mean, sample observed The
15.Math-Review 18
In our example we have:
Sampling
79.28$69
)40.7913.74()40.7967.95(
1
1
deviation, stardard sample observed theand
40.79$70
13.7469.8267.95
mean, sample observed The
22
1
2
1
n
ii
n
xxn
s
n
xxx
15.Math-Review 19
We will use these observed values to estimate the unknown mean , and standard deviation , of our unknown underlying distribution.
In other words:
Sampling
estimate will
estimate will
s
x
Also note that if we pick a different sample of the population, our observed values will be different.
We can define the random variables: sample mean, sample standard deviation, of which x and s are observations.
15.Math-Review 20
Before the sample is collected, the random variables X1,X2,… ,Xn, can be used to define:
Sampling
n
ii
n
ii
n
ii
n
XXn
S
XXn
Xnn
XXX
1
2
1
22
1
1
1
1
deviation, stardard sample observed The
1
1S
variance,sample observed The
1
mean, sample The
15.Math-Review 21
X and S are random variables
We distinguish between the sample mean X, which is a random variable, and the observed sample mean x, which is a number.
Similarly, the sample standard deviation S is a random variable, and the observed sample standard deviation s is a number.
Sampling
15.Math-Review 22
Distribution of X From the formula that defines the sample mean we see that according
to CLT it should follow approximately a normal distribution (if n30)
The mean is E(X) = The standard deviation is E(X) =
In summary:
Sampling
n
" ," ,~ 22 nsxNnNX
15.Math-Review 23
Example: At two different branches of the G-Mart department store, they randomly sampled 100 customers on August 13. At Store 1, the average amount purchased was $41.25 per customer, with a sample standard deviation of $24.00. At Store 2, the average amount purchased was $45.75 with a sample standard deviation of $34.00 Let X denote the amount of a random purchase by a single customer at Store 1 and let Y denote the amount of a random purchase by a single customer at Store 2. Assuming that X and Y satisfy a joint normal distribution, what is the distribution of X-Y? What is the probability that the mean of X exceeds the mean of Y?
Sampling
15.Math-Review 24
Example: In the quality control department of our company, knobs are inspected to make sure that they meet quality standards. Since it is not practical to test every knob, we draw a random sample to test. It is extremely necessary that our knobs weigh at least 0.45 pounds. If we know that the average weight is less than 0.45 pounds, we stop the production line and reset all the machines. In a day we produce 300,000 knobs, and draw a random sample of 1,000 knobs to test. If yesterday (Wednesday) the observed sample mean was 0.42 pounds, and observed sample standard deviation was 0.2, how confident are you that the average weigh of knobs is less than 0.45 pounds? If the average weight of knobs produced is 0.45 pounds, with standard deviation of 0.2,
what is the probability that the average weight of the sample will be 0.42 or lower? Are these questions the same?
Sampling