Download Confidence Interval

o Make Copies of:
o Areas under the Normal Curve
o Appendix B.1, page 784
o (Student’s t distribution)
o Appendix B.2, page 785
o Binomial Probability Distribution
o Appendix B.9, pages 794,798
n x
p( x)  P( X  x)    p (1  p) n  x
 x
Confidence Interval
 Definition:
A range of values constructed from sample data so that the
population parameter is likely to occur within that range at a
specified probability.
The specified probability is called the level of confidence.
 Ex.
We are 90% sure that the mean yearly income of
construction workers in the New York area is between $61,000
and $69,000.
Confidence Interval
If normal and σ known
The CI is given by:
[X  z

n
,X z

n
]
X , σ and n are known. We can obtain z from Appendix B.1 by
looking for a value of z that satisfies:
(Area from 0 to z) = K/2 = (level of confidence)/2
Confidence Interval
If (normal, σ unknown and n>30) or If (not normal and n>30)
The CI is given by:
s
s
[X  z
,X z
]
n
n
X , σ and n are known. We can obtain z from Appendix B.1 by
looking for a value of z that satisfies:
(Area from 0 to z) = K/2 = (level of confidence)/2
Confidence Interval
If (normal, σ unknown and n≤30)
The CI is given by:
s
s
[X t
, X t
]
n
n
X , σ and n are known. We can obtain t from Appendix B.2 by
looking for a value of t that satisfies the level of confidence
(K%) and the degrees of freedom (df=n-1).
Confidence Interval, c
df = n-1
.
9
10
90%
95%
98%
99%
When to Use the t Distribution
Is the
population
normal?
No
Yes
Is the
population SD
known?
Is n 30 or
more?
No
Use a
nonparametric
test
Yes
Use the z
distribution
No
Use the t
distribution
Yes
Use the z
distribution
When to Use the t Distribution
Is the
population
normal?
No
Yes
Is the
population SD
known?
Is n 30 or
more?
No
Use a
nonparametric
test
Yes
Use the z
distribution
No
Use t if n less than
or equal to 30,
Use z if n is more
than 30
Yes
Use the z
distribution
Example(ex14, page309): Develop a 90% CI interval for the population
mean. Develop a 98% CI for the population mean. Interpret the result.
29,38,38,33,38,21,45,34,40,37,37,42,30,29,35
o
o
o
o
o
n=15 < 30
Assume Normal distribution otherwise we can not use the t
distribution.
Sample Mean = 35.1,
Sample SD = 6.0
For K=90% and df=n-1 =14, t = 1.761
s
X t
n
6
 35.1  1.761
 35.1  2.7
15
 CI  [32.4,37.8]
For K=98% and df=n-1 =14, t = 2.624
s
6
X t
 35.1  2.624
 35.1  4.1
n
15
 CI  [31,39.2]
As expected, the length of the CI for a 90% confidence level
is smaller than the CI for a confidence level of 98%.
Confidence Interval
 When
looking for a value for z in the expression,
X z

n
 95%
of the sample means selected from a population will
be within 1.96 SD’s of the population mean µ. (The z
value for a confidence level of 95% is 1.96).
 99%
of the sample means will lie within 2.58 SD’s of the
population mean. (The z value for a confidence level of
99% is 2.58).
Confidence Interval

How did we get the 1.96 and the 2.58 for the 95% and 99% confidence
intervals?

For the 95% CI:


Probability area is 0.95/2=0.475.

In Appendix B.1, the z value for .475 is 1.96.
Use same reasoning and calculations for the 99% CI.

Probability area is 0.99/2=0.495.

In Appendix B.1, the z value for .495is 2.58.
Finite-Population Correction Factor
If the population size N is not very large, then we use a
population correction factor when computing the CI.
If (n/N > 0.05) then use :
s
X t
n
N n
N 1

N n
N 1
X z
n
,
OR
s
X z
n
N n
N 1
Example: There are 250 families in a certain area. A poll of 40
families reveals the mean spending per week is $450 with a
standard deviation of $75. Construct a 90% CI for the mean
spending.
n/N = 40/250 = 0.16 > 0.05.
s
X z
n
N n
75
 450  1.65
N 1
40
 450  17.97
250  40
250  1
Finite-Population Correction Factor

A population that has a fixed upper bound limit. A small population.
 Ex. Students registered for this class.
 When population is finite, need to make adjustments in the way we
compute the standard error of the sample means, and the standard error
of the sample proportion.
 Reduce the Standard Error
 The adjustment is called the finite-population correction factor.
Sample Size
Fraction of
population
(size 1000)
Correction
Factor
10
.01
.9955
50
.05
.9752
100
.1
.9492
500
.5
.7075
Sample Size for Estimating Population Mean
 zs 
n 
E
2
n is the sample size;
z is the standard normal value corresponding to the desired
level of confidence;
s is an estimate of the population SD;
E is the maximum allowable error (1/2 length of the CI).
If the result is not a whole number, round up.
Example: A student wants to determine the mean amount of
earnings per month of city council members. The error in
estimating the mean is to be less than $100 with a 95% level of
confidence. The student found a report by the Department of
Labor that estimated the SD to be $1,000. What is the required
sample size?
K=95% => z=1.96
Error in the estimating mean is to be less than 100 =>E=100
SD=1000
2
2
 zs   (1.96)(1000) 
n  
  384.16
100
E 

Round up => sample size n = 385
What if the student wanted to increase the level of confidence to
99%?
The corresponding z value for k=99% is 2.58.
2
2
 zs   (2.58)($ 1,000) 
2
n  

(
25
.
8
)
 665.64

$100
E 

The recommended sample size is now 666. Notice the change in the required
sample size for the different levels of confidence.
There is an increase of 281 observations. This could greatly increase the cost
and the time of the study. Therefore, the level of confidence should be
considered carefully.
Sample Size for Estimating Population Mean
 zs 
n 
E
2
o Sample size increase if we need a smaller CI (more precision).
o Sample size increases if we need a higher Confidence Level (as
K% increase then z increases and as a result, n increases).
o If the SD increases (higher variability), then the sample size
increases.
Proportion
o
So far the populations characteristics we considered have
numerical values.
o
If the population characteristic we are looking for can only take
on two values (true or false), (yes or no), (1 or 0),… then we
use a proportion to represent the population characteristic.
o
Proportion: The fraction, ratio, or percent indicating the part of
the sample or the population having a particular trait of interest
Proportion
Example: A recent survey indicated that 92 out of 100 surveyed
favored the continued use of daylight savings time in the summer.
The sample proportion is 92/100, or .92, or 92%.
o Define p as the sample proportion, and π as the population
proportion.
o p is the point estimate of π.
o To estimate π, we can also construct a CI.
Assumptions for Proportion CI Construction
1.
2.
The binomial conditions have been met:
a.
Sample data is a result of counts.
b.
There are only 2 possible outcomes (Success and Failure).
c.
The probability of a success remains the same from one trial to
the next.
d.
The trials are independent.
The values nπ and n(1-π) should be both ≥5. (π is the population
proportion) so that we can use the CLT (z-distribution)
Sample Proportion
X
p
n
If π is the population
proportion, then p is a point
estimator for π.
X is the number of (successes) obtained from the sample.
Confidence Interval for a Population
Proportion
p  z p
Standard Error of the Sample
Proportion
p 
p(1  p)
n
Confidence Interval for a Population
Proportion
p(1  p)
pz
n
Example: The union representing ABC company is considering a merger with
Teamsters Union. According to ABC union bylaws, at least three-fourth
of the union membership must approve any merger. A random sample of
2,000 current ABC members reveal 1,600 plan to vote for the merger
proposal. What is the estimate of the population proportion? Develop a
95% confidence interval for the population proportion. Basing your
decision on this sample information, can you conclude that the
necessary proportion of ABC members favor the merger? Why?
Sample size is N=2000,
Number that approve the merger is X=1600.
The sample proportion p=X/N=1600/2000 = 0.8.
We determine the 95% CI. The z value is 1.96.
p(1  p )
.80(1  .80)
pz
 .80  1.96
 .80  .018
n
2,000
Example (cont’d)
The endpoints are .782 and .818. The lower limit is greater than .75. So, we
conclude that the merger proposal will likely pass because the interval
estimate includes values greater than 75% of the union membership.
Sample Size for the
Population Proportion
Three items need to be specified:
1. The desired level of confidence.
2. The margin of error in the population proportion.
3. An estimate of the population proportion.
z
n  p (1  p )  
E
2
If an estimate of π is not available, use p=0.5 to approximately
estimate the sample size.
Example: A group of students want to estimate the proportion of
cities with subsidized transportation systems. They want the
estimate to be within .10 of the population proportion. The
desired level of confidence is 90%. No estimate for the
population proportion is available. What is the required sample
size?



E= .10
The level of confidence is 90%. The corresponding z value is 1.65.
No estimate for p is available, so we use .50.
2
2
z
 1.65 
n  p (1  p )    (.5)(1  .5) 
  68.0625
E
 .10 
Round up, so a random sample of 69 cities is needed.
Finite-Population Correction Factor
If the population size N is not very large, then we use a
population correction factor when computing the CI.
If (n/N > 0.05) then use :
p(1  p) N  n
pz
n
N 1
Example (ex22,page315): There are 300 welders at MSC. A
sample of 30 welders revealed that 18 graduated from a
registered welding course. Construct a 95% CI for the
proportion of all welders who graduated from a registered
welding course.
p = X/n =18/30 = 0.6.
n/N = 30/300=0.1>0.05
K=95% => z=1.96.
pz
p(1  p) N  n
n
N 1
0.6(1  0.4) 300  30
 0.6  1.96
30
300  1
 CI  0.6  0.167  [0.433, 0.767]