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Statistics
Sample
Population:





Sample function:
1
Sample: independent identically distributed (iid)


Examples:
• production
• marketing research
a function of the observed values in
the sample used for making general
conclusion about the entire population.
lecture 6
Sample
Mean, mode & median
Sample mean / average:
1 n
X  Xi
n i 1
Mode:
Example: Grade sample:
4, 12, 7, 2, 10, 7, -3, 0, 0,
-3, 2, 7, 10
3.0
2.5
the value(s) with highest frequency
2.0
1.5
Median: X(i) increasing sequence
1.0
0.5
0.0
 X ( n 1) 2
~ 
X   X n 2  X n 21

2

2
n odd
n even
-3
0
2
4
7
10
12
x  55 / 13  4.23
m7
Median : ~
x 4
Mean :
Mode :
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Sample
Range & variance
Range:
R = X(n) - X(1)
Example cont.:
Range: r = 15
1 15
2
(
X

4
.
23
)
 25.03
Variance: s 

i
13  1 i  1
2
Sample varinace:
1 n
2
2
S 
(
X

X
)
Standard div: s  25.03  5.00

i
n 1 i1
Sample standard deviation:
S  S2
Notice!!
3
Lower case letters: Based on observations
Upper case letters: Based on random variables
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Sample mean
Normal distribution
Theorem:
Let X1, X2, ... Xn be independent normal distributed random
variables with same mean m and same finite variance s2,
that is
X ~ N ( m , s 2 ), i  1, 2,, n iid.
i
4
It follows that
 s2 
X ~ N  m , 
 n 
and then
X m
~ N 0,1
s n
lecture 6
Sample mean
Distribution
The Central Limit Theorem (CLT):
Let X1, X2, ... Xn be independent identically distributed
random variables with same mean m and same finite
variance s2. Then the distribution of
X m
Z
s
n
will tend towards the standard normal distribution as n → ∞.
How large should n be before the approximation is good?
• Most distributions:
• Normal distribution :
5
n > 30
for all n
lecture 6
Sample mean
Example
Problem: Production of light bulbs
A company produces bulbs with a life time X,
which is approximately normal distributed with
mean:
m = 800 hours
standard diviation:
s = 40 hours
(a) Find the probability that a sample
consisting of 16 bulbs has a mean life time
less than 775 hours?
(b) If you observe a sample mean life time of 775 hours,
would you believe that the population mean is in fact 800
hours?
6
lecture 6
Two sample means
Comparison
Theorem:
Assume two independent samples are taken from two
populations with means m1 and m2, respectively, and finite
variances s12 and s22, respectively.
Then for the difference between the two sample means
X 1  X 2 , we have

s 12 s 22 

X 1  X 2 ~ N  m1  m2 ,

n1 n2 

and hence X 1  X 2   m1  m 2 
~ N 0, 1
2
2
s1 s 2
n1
7

n2
lecture 6
Sample variance
Distribution
Theorem:
Lad X1, X2, ... Xn be independent normal distributed
random variables with mean m and variance s2.
Then
(n  1) S 2
s2

1
s2
n
2
2
(
X

X
)
~

(n  1)
 i
i 1
When calculating S2, it’s usually more convenient to use
2

n
n

 
1
2
2
n  X i    X i  
S 

 
n (n  1)  i  1
i 1

 

8
lecture 6
Sample variance
Example
Problem: Car batteries
A producer of car batteries claims that the life time of
their batteries are normal distributed with
mean:
standard deviation:
m = 3 year
s = 1 year
Sample of 5 batteries: 1.9 2.4 3.0 3.5 4.2
(a) Calculate sample standard deviation.
(b) Do you believe that the standard deviation is 1 year?
9
lecture 6
Sample mean
Distribution (unknown varinace)
Typically the variance s2 is unknown.
If we replace the unknown variance by s2 we obtain:
Theorem:
Lad X1,X2, ... Xn be independent normal distributed
random variables with mean m and variance s2 (unknown).
Then
10
X m
~ t n  1
S
n
lecture 6
Sample mean
Example (unknown variance)
Problem cont.: Car batteries
Producer claims that the life time of
their batteries are normal distributed with
mean:
standard deviation:
m = 3 years
unknown
Sample of 5 batteries: 1.9 2.4 3.0 3.5 4.2
Do you believe that mean life time is 3 years?
11
lecture 6
Two sample variances
Comparison
Theorem:
If two independent samples are taken from two normal
populations with variances s12 and s22, respectively, then
2
S1
s 12
~ F (n 1  1, n 2  1)
2
S2
s 22
Notice!!
12
1
f 1   (n 1 , n 2 ) 
f  (n 2 , n 1 )
Eg.
f 0.95 (6, 10) 
1
f 0.05 (10, 6)
lecture 6