Download Normal Distribution and Sampling Distribution of the Mean

Normal Distribution and Sampling Distribution of the Mean
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Normal Distribution
Use of Z table
Sampling Distributions
1. Normal Distribution
Purpose: To extend the empirical rule to more than 1,2 or 3 standard deviations.
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Mound-shaped symmetric
Empirical Rule
Bell-shaped Polygon
Standard Normal -
2. Z table – Before finding probabilities of sample means, we will work with probabilities of normal
populations.
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Formulas Z = (x-)/and X =  + Z
Empirical Rule:
Range of Z
Probability
2 to 3
0.025
1 to 2
0.135
0 to 1
0.340
-1 to 0
0.340
-2 to -1
0.135
-3 to -2
0.025
Approximate Probabilities Using Empirical Rule
Z value < Z-value > Z-value Within ±Z outside ±Z
3
1
0
1.00
0.00
2
0.975
0.025
0.95
0.05
1
0.84
0.135
0.68
0.32
0
0.5
0.5
0.00
1.00
-1
0.16
0.84
-----2
0.025
0.975
-----3
0
1
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For other more accurate probabilities of Z, go to
http://wweb.uta.edu/faculty/eakin/busa3321/alternativeNormal.doc
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Examples
Pr ( Z < 1) = ?
Pr (Z < ? ) = 0.0250
Pr (X < 10 |  = 15,  = 5) = ?
Pr( 20 < X < 24 |  = 15,  = 5 ) = ?
For examples, click on the following link and press F9 to get more examples.
http://wweb.uta.edu/faculty/eakin/busa3321/NormProbOfX.xls
Exercise: Using Internet Explorer answer the questions on the following web page. The questions must
be answered in one attempt. (The page keeps track of the number of attempts.) Print the page when successful
and upload it to Blackboard.
http://wweb.uta.edu/faculty/eakin/asps/examples/NormalProb.asp
3. Sampling Distribution of Sample Mean
Purpose: To apply the normal distributions to sample means. This brings all parts of the third Building Block
together.
3.1 If repeated random samples of the same size are drawn from a very large population, the following
result:
a. The average of all the sample averages will be the same as the average of the original population since
both use the same numbers.
b. From the introduction, the typical error in the sample average is a function of two items: variability and
knowledge. The standard error is the fraction of the population standard deviation divided by the
square root of n.
The square root is used because of diminishing returns of n. As an analogy, you typically learn
more going from 1 to 2 years on the job than you learn from 28 to 29 years on the same job.
Symbol:
is the population standard error and
is the sample estimate of the standard error
c. The larger the sample size, the closer the distribution of a sample average is to a normal distribution. (If
the original data is normal, then samples of any size will result in means that are normal).
Example: Suppose you take all possible random samples of size 4 from the following population of size 6:
{1, 2, 3, 4, 5, 6}. Average of the population is 3.5
Original Population
Value
1
2
3
4
5
6
Probability
16.7%
16.7%
16.7%
16.7%
16.7%
16.7%
Possible Samples
{1, 2, 3, 4}
{1, 2, 3, 5}
{1, 2, 3, 6}
{1, 2, 4, 5}
{1, 2, 4, 6}
{1, 2, 5, 6}
{1, 3, 4, 5}
{1, 3, 4, 6}
{1, 3, 5, 6}
{1, 4, 5, 6}
{2, 3, 4, 5}
{2, 3, 4, 6}
{2, 3, 5, 6}
{2, 4, 5, 6}
{3, 4, 5, 6}
Sample
Mean
2.5
2.75
3
3
3.25
3.5
3.25
3.5
3.75
4
3.5
3.75
4
4.25
4.5
Sampling Distribution
Sample
Means
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
Probability
7%
7%
13%
13%
20%
13%
13%
7%
7%
Distribution of Original Data
Distribution of All Possible Sample Means
18.0%
25%
Probability of Sample Mean
Having this Value
16.0%
Probability
14.0%
12.0%
10.0%
8.0%
6.0%
4.0%
2.0%
0.0%
1
2
3
4
5
6
20%
15%
10%
5%
0%
2.5
2.75
Values
3
3.25
3.5
3.75
4
4.25
Possible Sam ple Means
What is the average of the original population? Average of all possible sample means?
What is the range of the original population? What is the range of all possible sample means?
What shape is the distribution of the original data? The sample means?
Example: If n=64,  = 30 and  = 5, then
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Population Mean of all X’s =  X =
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Population Variance of all X’s = 2 X =
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Population Standard Deviation (Standard Error) of allX’s =
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Distributional Shape of all X’s =
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Formulas: Based on one of our basic building blocks: To evaluate an error you compare it to the
standard error:
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Z = Error / (Standard Error) = (X -) / ( X
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andX =  + Z( X
4.5
3.2 Examples
3.2.1 Pr(X < 20 |  = 19,  = 5, n = 16 )
 X =
 X
=
= Pr(Z < _____ ) = _________
3.2.2 Pr( 14.2 <X < 16 |  = 15,  = 5, n= 100 ) = ?
 X =
 X
=
= Pr(Z < _____ ) – Pr (Z < _______)
= ________ - ________ = _______
3.2.3 Pr(X < ? |  = 15,  = 5, n = 16 ) = 0.0500
 X =
 X
=
 + ( Z) X =
For examples, click on the following link and press F9 to get more examples.
http://wweb.uta.edu/faculty/eakin/busa3321/NormProbOfXbar.xls
Exercise: Using Internet Explorer answer the questions on the following web page. The questions must
be answered in one attempt. (The page keeps track of the number of attempts.) Print the page when successful
and upload it to Blackboard.
http://wweb.uta.edu/faculty/eakin/asps/examples/ProbofXbarQues.asp
4. Sampling Distribution of Sample Proportions
4.1 Background
Consider a population of size 5 where there are 3 successes and two failures. The probability of a success in
the population, p,equals 3/5= 0.60. Consider recording the five values where successes are recorded as 1’s
and failures are recorded as 0’s. Find the variance of this list of 0’s and 1’s using the rules from section 5:
Values
b. Distance to Mean
c. Squared Distance
1
1 – 0.60 = 0.40
(0.40)2= 0.16
1
1 – 0.60 = 0.40
(0.40)2= 0.16
1
1 – 0.60 = 0.40
(0.40)2= 0.16
0
0 – 0.60 = -.60
(0.60)2= 0.36
0
0 – 0.60 = -.60
(0.60)2= 0.36
a.  = 3/5 = 0.60
d. Sum = 1.20
e. 2 = 1.20/5 = 0.24 (divide by 5 since it’s a population)
Note: From a. we see the population proportion is a population mean and from e. that the population
variance is 0.60*0.40 =p(1-p)
Thus when estimating the population proportion, p, the sample proportion, p̂ , becomes a special case of a
sample mean and we can use the rules of section 3 with 2 replaced by p(1-p)and with the word “mean”
replaced with “proportion”:
4.2 If repeated random samples of the same size are drawn from a very large population and the sample
proportion, p̂ , is calculated then the following result:
a. The average of all the sample proportions will be the same as the population proportion
b. From the introduction, the typical error in the sample average is a function of two items: variability and
knowledge. The standard error is the fraction of the population standard deviation divided by the
square root of n.
The population standard deviation, , is
p(1  p)
Symbol:
p̂ 
Sp̂ 
p(1  p)
n
p̂(1  p̂)
n
is the population standard error and
is the sample estimate of the standard error
c. The larger the sample size, the closer the distribution of a sample proportion is to a normal
distribution. The general rule is: if np and n(1-p) are both greater than 5, then the distribution of the
sample proportions is approximately normal.
4.2 Example: if n=100,  = 0.4, find
Pr( p̂  0.45)
 p̂ 
 p̂ 
= Pr(Z > _____ ) = _________
For examples, click on the following link and press F9 to get more examples.
http://wweb.uta.edu/faculty/eakin/busa3321/NormProbOfP.xls
Exercise: Using Internet Explorer answer the questions on the following web page. The questions must
be answered in one attempt. (The page keeps track of the number of attempts.) Print the page when successful
and upload it to Blackboard.
http://wweb.uta.edu/faculty/eakin/asps/examples/ProbofPQues.asp