ECO ECO 220Y Lecture18 Sampling Distribution Part 2 Central Limit Theorem
Migiwa Tanaka Reading: 9.1,9.3,9.4
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Outline
Recap of Sampling Distribution Relationship between sample size and sampling distribution of mean Central Limit Theorem Application 1 Application 2 Sampling Distribution of difference of two means.
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Recap
Sampling distribution is distribution of a statistic calculated from samples drawn from the same population. We can derive it it
analytically remember the telework example. by simulation
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Parameters of Sampling Distribution of Mean
Mean:
X E[ X ] X
2 X
Variance:
2 X
n
Standard Deviation:
X
X
n
Homework: How do you derive these parameters? 1n X Xi Hint: n i 1 Try them first by yourself and you can find solution in CD appendix appendix I of the textbook.
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Parameters of Sampling Distribution of Mean
Standard Error : Standard Deviation of sample mean.
It measures the variation of X around its mean X which is equal equal to X . It measures the magnitude of sampling errors. Why? X High S.E. More likely to get remote value from More Low S.E. More likely to get closer value to X
Density Density .02 .02 .04 .06 .04 .06 00 .8 .008
5
XX X
Relationship between sample size and sampling distribution of mean li di
How does the sample size n relate to the sampling distribution of mean? Analysis of Parameters: Mean: E[ X ]
X X
Standard Error: X
X
n
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Relationship between sample size and n 1 sampling distribution of mean li di X Xi
Each Xi is Normal
.08 .06 .04 .02 0 -20
n
i 1
Population
mean=10 s.d. =5
.08 .06 .04 .02 0 -20
Sampling Distribution n=1
mean=? s.e.=?
0
20
40
0
20
40
.2 .15 .1 .05 0 -20
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Sampling Distribution n=5 n=5 mean=? s.e.=?
.6 .4 .2 0 40 -20
Sampling Distribution n=50 mean=? s.e.=?
0
20
0
20
40
Relationship between sample size and 1n sampling distribution of mean li di X X
Discrete Distribution: Each Xi is Bernoulli.
Population 1 1 Probability 0 .2 .4 .6 .8 Probability 0 .2 .4 .6 .8 n=1,p=.1 .4 Probability .1 .2 .3
n
i 1
i
n=10,p=.1
mean=.1 s.d.=.09
mean=? s.e.=.?
mean=? s.e.=.?
0 Probability 0 .05 .1 .15 .2 .25
1
0
1 Probability .05 .1 .15
0 0.1.2.3.4.5.6.7.8.91 n=100,p=.1 n=50,p=.1
n=30,p=.1 Probability .05 .1 .15
mean=? s.e.=.?
.2
mean=? s.e.=.?
mean=? s.e.=.?
0
8
0 .20.40.60.80 1.00
0 .20.40.60.80 1.00
0
0 .20.40.60.80 1.00
Relationship between sample size and 1n sampling sampling distribution of mean
X
Each Xi is Uniform. (mean=0.5, s.d.=0.289)
Population 0 .2 .4 .6 .8 1 0 .2 .4 .6 .8 1 n=1 2 1.5 1 .5 0 0 .25.5.75 1
n=10
X n
i 1
i
n=2
0 .25.5.75 1
n=3
0
.5
1
2.5 2 1.5 1 .5 0 0 .5 1
4 3 2 1 0 0 .5 1
What are the mean and s.e. of each case?
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Relationship between sample size and sampling distribution of mean li di 1n
Each Xi is Log normal.
Population .1 .2 .3 .4 n=1 .1 .2 .3 .4
X
X n
i 1
n=3
i
mean=1.64 s.d.=2.16
.6 .4 .2
mean=? s.e.=?
mean=? s.e.=?
0
0
0
5
10
15
0
5
10
n=30
15
0 -1 4 914
.8 .6 .4 .2 0 -1
10
n=10
1.5 1 .5 0
mean=? s.e.=?
mean=? s.e.=?
4
9
14
-1
4
9
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Central Limit Theorem
The sampling distribution of the mean of a random sample drawn from any population is approximately normal for a sufficiently sample large sample size.
How large is sufficiently large? If population is normal, n=1 is sufficient. For other population, the rule of thumb is the n=30.
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Application House prices (1)
Suppose a small town in Ontario has 2,000 houses. The The town tries to promote itself with affordable houses and published the data on house prices in the town: Mean of $234,000 with s.d. $70,000. An appraiser took 49 houses from the town as a sample and evaluated the values, and found the mean is $270,000. Should we reject the towns claim because $234,000<$270,000? How likely is such a high sample mean if the towns claim is true? P ( X 270,000 | X 234,000, X 70,000, n 49) ?
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Application House prices (2)
Assume the towns claim is correct. Population mean = $234,000 Population s.d. = $70,000 Due to the CLT, sampling distribution of mean is Normal with mean =population mean =$234,000 and s.d. =(population s.d.)/(square root of n) (population =$70000/7=$10000.
X X 270000 X P( X 270000) P X X
X 234000 270000 234000 P P ( Z 3 .6 ) 10000 10000 Z z
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Application House prices (3) P( X 270000) P( Z 3.6) 1 P( Z 3.6) 1 0.999841 0.000159
What are plausible explanations?
Sampling Error?
Non sampling error?
Something else?
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Application Tips and Friendliness (9.53)
A restaurant manager has an impression that there can be difference in amount of tips between two groups of servers servers.
Group1 Introduce themselves with their names. Group2 Do not introduce themselves.
The manager took random sample of servers from each group and compared the mean percentage of tips for two groups. Result: the mean of group1 is lower than the mean of group2! group2! Can we conclude that managers claim was rejected?
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Application Tips and Friendliness (9.53) (9
In order to conclude, we need to know if the difference we of sample means is due to sampling error or not. We need to know sampling distribution of difference di between two sample means, X 1 and X 2 . Remember a function of random variables is also random variable. In particular, X 1 X 2 is linear function (combination) of random variables.
Linear Combination: Y=a0+a1X1+a2X2++aJXJ
A Linear combination of independent normal random variables variables is a normal random variable. Why normal?
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Mean and Variance of X 1 X 2
X 1 ~ N X1 , n1
2 X1
where n1 and n2 are sample size of X 1 and X 2 , respectively.
,
X 2 ~ N X 2 , n2
2 X2
E X 1 X 2
V X 1 X 2
2
E X 1 E X 2 X 1 X 2
2
(1) V X 1 ( 1) V X 2 1 ( 1) 2COV X 1 , X 2
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2 X1
n1
2 X2
n2
Sampling Distribution of X 1 X 2 Di
2 X 2 X
s.e.
1
n1
2
n2
Which factor reduces the standard error?
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X X
1
2
Application Tips and Friendliness Ti
The manager took following samples.
Population 1(Group 1): Sample1: 44 servers
X 18, X 4.7
1 1
Population 2 (Group2): Sample2: 39 servers
X 15, X 4.7
2 2
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Application Tips and Friendliness Ti
E X 1 X 2 18 15 3
s.e.
.4
Sampling Distribution Di mean=3, S.E.=1.03
2
2 X1
n1
2 X2
n2
2
4.7
44
4.7
39
1.03
0
.1
.2
.3
-5
0
X X
1
3
5
2
10
20
0 3 P X 1 X 2 0 P Z P( Z 2.90) 1.03 0.0019