Download Lecture15

Review
• Normal Distributions
– Draw a picture.
– Convert to standard normal (if necessary)
– Use the binomial tables to look up the value.
– In the case of a reverse look up we may have to
now solve for x.
Estimator and Point Estimate
An estimator is a “sample statistic” (such as
the sample mean, or sample standard
deviation) used to approximate a population
parameter.
A Point Estimate is a single value or point
used to approximate a population
parameter. A point estimator may be biased
or unbiased.
Central Limit Theorem
Take ANY random variable X and compute m and s
for this variable. If samples of size n are randomly
selected from the population, then:
1) For large n, the distribution of the sample means, x
will be approximately a normal distribution,
2) The mean of the sample means will be the
population mean m and
3) The standard deviation of the sample means will
be
s
n
The Sampling Distribution of X and
the Central Limit Theorem
Assume a population with m = 80, s = 6. If a
sample of 36 is taken from this population,
what is the probability that the sample mean
is larger than 82?
Sketch the curve
of x and identify
area of interest
The Sampling Distribution of X and
the Central Limit Theorem
Convert 82 to z value
First, calculate the standard deviation of the
sampling distribution
s
6
sx 

 1.0
n
36
Then calculate the z value
z
x  mx
sx

82  80
2
1
Use the tables to find probability of interest


P x  82  P( z  2)  .5  .4772  .0228
Practice Problems
• #6.34 page 310
• #6.41 page 311
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
What can you say about the average
balance?
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
What can you say about the average
balance of all VISA accounts?
ANS: That it is approximately $213. (This is
a point estimate.
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts?
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts?
You can create an interval (called a
confidence interval) that you can be 95%
sure the mean is in.
Confidence Interval
A confidence interval (or interval estimate)
is a range of values that estimates the true
value of the population parameter.
This is associated with a degree of
confidence, which is a measure the
probability that a randomly selected
confidence interval encloses the population
parameter.
Confidence Level
The confidence level is equal to 1-  , and is
split between the two tails of the distribution
Usually the confidence level is:
90% (meaning   .10 )
95% (meaning   .05 ) OR
99% (meaning   .01 )
Confidence Intervals
The Confidence Interval is expressed as:
xE
x  z 2s x  x  z
E is called the margin of error.
For samples of size > 30,
 s 
x  z 2 

 n
s
2
n
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 95% confidence
interval.
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 95% confidence
interval.
213 21.95
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 90% confidence
interval.
 112 
213  1.645

 100 
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 90% confidence
interval.
 112 
213  1.645
  213  18.42
 100 
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 90% confidence
interval.
 112 
213  2.575

 100 
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 90% confidence
interval.
 112 
213  2.575
  213  28.84
 100 
Age of STFX Students
Find the 90% confidence interval for the
mean.
x  21.4,
s  2.6,
n  50   0.10
Age of STFX Students
Find the 90% confidence interval for the
mean.
x  21.4,
x  z s
2
21.4  z0.05
s  2.6,
n  50   0.10
 m  x  z s
n
2.6
20.8  m  22.0
2
50
n
 m  21.4  z0.05 2.6
50
Sample Size
The sample size needed to estimate m so as
to be (1-)*100 % confident that the sample
mean does not differ from m more than E is:
 z 2s
n  
 E
…round up



2
Overview
• Confidence Intervals
Practice Problems
• #7.11 page 329
• #7.19 page 331
• #7.21 page 331
Homework
• Review Chapter 7.1, 7.2
• Read Chapters 7.3-7.5
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