Download Confidence Interval

Confidence Interval for a Mean
Confidence
Intervals 
_
Sample Size
 30
•
•
•
 30
Sigma
  
x  z 

2
n


known
sknown
( unknown)
Topics:
• Essentials
• Inferential Statistics
• Terminology
• Margin of Error
• Z/2
• Decision Grid
• Examples
_
X %C.I .  ( x E    x E )
Large Sample
Small Sample – Student’s t Distribution
Proportion
x  t
2
, df
  
x  z 

2
n


 s 


n


 s 
x  z 

2 n 
Confidence Interval for a Proportion


X %C.I .  ( p  E  p p  E )


p  z 
2



 
pq 

n 

Essentials: Confidence Intervals
(How sure we are.)

Inferential statistics, precision and the margin of error.

Obtaining a confidence interval.

Z/2

Guinness, Gosset & the Student’s t Distribution

Confidence Intervals for large and small samples, and proportions.
Inferential Statistics:

INFERENTIAL STATISTICS: Uses sample data to
make estimates, decisions, predictions, or other
generalizations about the population.


The aim of inferential statistics is to make an inference
about a population, based on a sample (as opposed to a
census), AND to provide a measure of precision for the
method used to make the inference.
An inferential statement uses data from a sample and
applies it to a population.
Some Terminology


Estimation – is the process of estimating
the value of a parameter from information
obtained from a sample.
Estimators – sample measures (statistics)
that are used to estimate population
measures (parameters).
Terminology (cont’d.)


Point Estimate – is a specific numerical
value estimate of a parameter.
Interval Estimate – of a parameter is an
interval or range of values used to estimate
the parameter. It may or may not contain
the actual value of the parameter being
estimated.
Terminology (cont’d.)


Confidence Level – of an interval estimate
of a parameter is the probability that the
interval will contain the parameter.
Confidence Interval – is a specific interval
estimate of a parameter determined by using
data obtained from a sample and by using a
specific confidence level.
Margin of Error, E

  
z 

2 n 
The term
is called the maximum error
of estimate or margin of error. It is the
maximum likely difference between the point
estimate of a parameter and the actual value of
the parameter. It is represented by a capital E.
  
E  z 

2 n 
Z/2 : Areas in the Tails
Obtaining :Convert the
Confidence Level to a
decimal, e.g. 95% = .95.
Then:
  1  C.L. / 100
  1 95 / 100
  1 .95
  .05

  .025
2
2

95%
.025
.025

  
z 

2 n 
-z (here -1.96)
  
z 

2 n 
z (here 1.96)
2
Decision Grid
Confidence Interval for a Mean
_
_
X %C.I .  ( x E    x E )
Sample Size
Sigma
 known
sknown
( unknown)
 30
  
x  z 

2
 n
x  t
 s 


, df 
2
n


 30
  
x  z 

2
n


 s 
x  z 

2 n 
Confidence Interval for a Proportion


X %C.I .  ( p  E  p p  E )


p  z 
2



 
pq 

n 

t or z????
Is

Known?
yes
no
Is n greater than
or equal to 30?
yes
Use z-interval formula values
no matter what the sample
size is.*
Use z-interval formula values
and replace  in the formula
with s (sample std. dev.).
no
Use t-values and
s in the formula.**
*Variable must be normally distributed when n<30.
**Variable must be approximately normally distributed.
Situation #1: Large Samples or
Normally Distributed Small Samples




A population mean is unknown to us, and we wish to
estimate it.
Sample size is > 30, and the population standard
deviation is known or unknown.
OR sample size is < 30, the population standard
deviation is known, and the population is normally
distributed.
The sample is a simple random sample.
Confidence Interval for
(Situation #1)



1 
  
x  z 

2 n 
 s 
x  z 

2 n 
Consider

The mean paid attendance for a sample of 30
Major League All Star games was $46,970.87,
with a standard deviation of $14,358.21.
Find a 95% confidence interval for the mean
paid attendance at all Major League All Star
games.
95% Confidence Interval for the
Mean Paid Attendance at the Major
League All Star Games
 $14,358.21 
$46,970.87  1.96

30


 $46,970.87  $5,138.02
($41,832.85    $52,108.89)
Minimum Sample Size Needed

For an interval estimate of the population
mean  is given by
 z   
n 2 
 E 


2
Where E is the margin of error (maximum error
of estimate)
Situation #2: Small Samples




A population mean is unknown to us, and we wish to
estimate it.
Sample size is < 30, and the population standard
deviation is unknown.
The variable is normally or approximately normally
distributed.
The sample is a simple random sample.
Student t Distribution







Is bell-shaped.
Is symmetric about the mean.
The mean, median, and mode are equal to 0 and
are located at the center of the distribution.
Curve never touches the x-axis.
Variance is greater than 1.
As sample size increases, the t distribution
approaches the standard normal distribution.
Has n-1 degrees of freedom.
Student t Distributions for
n = 3 and n = 12
Student t
Standard
normal
distribution
distribution
with n = 12
Student t
distribution
with n = 3
0
Confidence Interval for
(Situation #2)

A 1   confidence interval for
given by
 s 
x  t  ,n1 

2
 n


is
Consider

The mean salary of a sample of n=12
commercial airline pilots is $97,334, with a
standard deviation of $17,747. Find a 90%
confidence interval for the mean salary of all
commercial airline pilots.
90% Confidence Interval for the
Mean Salary of Commercial Airline Pilots
 $17,747 
$97,334  1.796

12 

 $97,334  $9,201.12
($88,132.88    $106,535.12)
Situation #3: Confidence
Interval for a Proportion

A confidence interval for a population proportion
p, is given by
pˆ  z 

2
pˆ qˆ
n
Where p̂ is the sample proportion .
qˆ  1  pˆ
n = sample size
np and nq must both be greater than
or equal to 5.
Consider


In a recent survey of 150 households, 54 had
central air conditioning. Find the 90% confidence
interval for the true proportion of households that
have central air conditioning.
Here
pˆ  54
 .36
150
qˆ  1  pˆ  1  .36  .64
n  150

(NOTE both np and nq > 5)
pˆ  z 
2
pˆ qˆ
n
 (.36)(. 64) 

.36 1.645

150


90% C.I. = .36 ± .065
or
90% C.I. = (.295, .425)
or
90% C.I. = (.295 < p < .425)
We can be 90% confident that the true proportion, p, of
all homes having central air conditioning is between 29.6%
and 42.5%
Minimum Sample Size Needed

For an interval estimate of a population proportion
is given by
2
 z 
n  pˆ qˆ  2 
 E 


Where E is the maximum error of estimate (margin
of error)
End of slides