# Chapter 7 Sampling Distribution I Basic Definitions II Sampling Distribution of Sample Mean III Sampling Distribution of Sample Proportion IV Odds and Ends I Basic Definitions • Pop by MalsD5xC

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```									        Chapter 7 Sampling Distribution
I. Basic Definitions
II. Sampling Distribution of Sample Mean
III. Sampling Distribution of Sample Proportion
IV. Odds and Ends
I. Basic Definitions
• Population and sample (N presidents and n presidents)
• Parameter and statistic         p.258&p.264
Based on all (N) possible values of X: ,  and p
Based on n values of X: X , S and p
• Statistic as a point estimator for parameter
• Sample statistic becomes a new random variable
• Sampling distribution: probability distribution of a
statistic
II. Sampling Distribution of Sample Mean X
1. Summary measures         p.270
E( X )   x       x
x
x               - If N is infinite or N >> n p.271
n         - n  x
If X is normally distributed, X is normally distributed.
2. Central Limit Theorem          p.272
For any distribution of population, if sample size is large,
the sampling distribution of sample mean is
approximately a normal distribution.

3. Applications
X  P( X )                  P( X )  X
Example. P.278 #18
A population has a mean of 200 and a standard
deviation of 50. A simple random sample of size 100
will be taken and the sample mean will be used to
estimate the population mean.
a. What is the expected value of X ?
b. What is the standard deviation of X ?
c. Show the sampling distribution of X ?
d. What does the sampling distribution of X show?
a. E ( X )   x = 200   b.  x         5
n   100
c. X  N(200, 5)
d. Probability distribution of X
3. Applications

(1) X  P( X )
Procedure: X  Z  P(Z)  P( X ) = P(Z)
x  x                    x
z            x  x   x 
x                        n
Example.
A population has a mean of 100 and a standard
deviation of 16. What is the probability that a sample
mean will be within 2 of the population mean for
each of the following sample sizes?
a. n = 50
b. n = 200
c. What is the advantage of larger sample size?
Solution.
Given x = 100, x = 16
a. n = 50, find P (98  X  102 )
x  x     98  100                      98 100 102
z1                          .88
x       16 / 50
z1   0   z2
x  x  102  100
z2                          .88
x       16 / 50
From Z-Table: When z = -.88, the left tail is .1894.
P(98  X  102 )  P(.88  Z  .88 )  1  (2)(. 1894 )  .6212
b. n = 200, find P (98  X  102 )
x  x     98  100
z1                            1.77
x      16 / 200
z2 = 1.77. Use Z-Table: P(98  X  102 )  1  (2)(. 0384 )  .9232
c. Larger sample size results in a smaller standard
error - a higher probability that the sample mean
will be within 2 of x
(2) P( X )  X
Procedure: P ( X ) P(Z)  Z  X  x  Zx

Example. The diameter of ping-pong balls is
approximately normally distributed with a mean of
1.30 inches and a standard deviation of 0.04 inches.
a. What is the probability that a randomly selected
ping-pong ball will have a diameter between 1.28
and 1.30 inches?
b. If a random sample of 16 balls are selected, 60% of
the sample means would be between what two
values that are symmetrically around population
mean.
Random variable X: diameter of a ping-pong ball.
a. P(1.28  X  1.3) = .1915
z = (1.28-1.3)/.04 = .5              1.28 1.3   .5-.3085=.1915

z=-0.5 0

b. P( x1  X  x2 )  .6
From Z table: When the left tail is .2005,    x1 1.3 x2
z1 = -.84  z2 = .84                        .2
.04
x1  1.3  (.84 )(      )  1.2916                 z1   0 z2
16
.04
x2  1.3  (. 84 )(     )  1.3084
16
Homework:
Sample Mean
p.278 #18, #19
p.278 #25
III. Sampling Distribution of Sample Proportion p
1. Summary measures
E ( p)  p                    p.280

p (1  p )
p                  - If N is infinite or N >> n   p.280
n        - n  p

If n and p satisfy the rule of five, p is approximately
normally distributed.            P.282

2. Applications
p  P( p )                 P( p )  p
2. Applications
(1) p  P( p )
Procedure:    p  Z  P(Z)  P( p ) = P(Z)
p  p                    p(1  p)
z             p  p   p 
p                          n

Example.
Assume that 15% of the items produced in an assembly
line operation are defective, but that the firm’s
production manager is not aware of this situation.
Assume further that 50 parts are tested by the quality
assurance department to determine the quality of the
assembly operation. Let p be the sample proportion
found defective by the quality assurance test.
a. Show the sampling distribution for p.
b. What is the probability that the sample proportion
will be within .03 of the population proportion
that is defective?
c. If the test show p = .10 or more, the assembly line
operation will be shut down to check for the cause
of the defects. What is the probability that the
sample of 50 parts will lead to the conclusion that
the assembly line should be shut down?

a. p ~ N (  p ,  p )
 p  p  .15
p (1  p )   (. 15 )(1  .15 )
p                                    .0505
n               50
b. P(.12  p  .18)
.12  .15                       .12 .15 .18
z1             .59        .2776
.0505
.18  .15                       z1   0 z2
z2             .59
.0505
P(.12  p  .18)  1  (2)(.2776)  .4448

c. P( p  .10)
.10  .15
z             .99
.0505
P( p  .10)  1  .1611  .8389           .10      .15
.1611

z      0
Homework:
Sample Proportion
p.283 #31
p.284 #35
p.285 #40

Compare wordings in Chapter 6 homework
problems, homework problems for sample mean
and homework problems for sample proportion:
X? or X ? or p?
IV. Odds and Ends
1. Three properties of a good estimator p.286&p.287
ˆ
(1) Unbiasedness E ( )  
Mean of a sample statistic is equal to the population
parameter. ( E ( X )   , E ( p )  p )
(2) Efficiency
The estimator with smaller standard deviation.
( X for  , p for p )
(3) Consistency If n  ˆ 
(The values of sample statistic tend to become closer
to the population parameter. For example,
x                  p (1  p )
x                p 
n                      n      )
2. Sampling Methods (p.260) and fpc
(finite population correction factor) p.271&p.280
(1) Infinite population, or finite population sampling
with replacement, or n/N  0.05:
x                   p (1  p )
x               p 
n                       n
(2) Finite population sampling without replacement and
n/N > 0.05:
N n
fpc           (fpc  1. If N >> n, fpc  1)
N 1
x              p (1  p )
 x  fpc          p  fpc
n                  n

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