Chapter 9 Slides
Chapter 9 Slides
Chapter 9 Slides
Chapter 9
McGraw-Hill/Irwin Copyright © 2010 by The McGraw-Hill Companies, Inc. All rights reserved.
GOALS
9-2
Sampling and Estimates
Why Use Sampling?
1. To contact the entire population is too time
consuming.
9-4
Confidence Interval Estimate
9-5
Interval Estimates - Interpretation
For a 95% confidence interval about 95% of the similarly constructed intervals will contain
the parameter being estimated. Also 95% of the sample means for a specified sample
size will lie within 1.96 standard deviations of the hypothesized population mean.
9-6
INTERPRETATION OF:
95% Confidence Interval for Population Mean for
a sample size of 300
INTERPRETATION:
– If you construct 100 confidence intervals for a sample size of 300, 95 of
those confidence intervals will contain the population mean.
– You could take many samples of size 300 (based upon what we studied
previously) and if we construct confidence intervals for all those samples,
95% of those confidence intervals will contain the population mean.
SAMPLING ERROR:
– But, a particular confidence interval either contains the population
mean or it does not.
9-9
9-10
Confidence Level and Z-value
9-11
Point Estimates and Confidence Intervals for a
Mean – σ Known
x sample mean
9-13
Width of Confidence Interval
9-14
Example
The American Management Association wishes to have information on the mean income of
middle managers in the retail industry. A random sample of 256 managers reveals a
sample mean of $45,420. The standard deviation of this population is $2,050. The
association would like answers to the following questions:
2. What is a reasonable range of values for the population mean? (Use 95%
confidence level)
9-15
Population Standard Deviation
σ Unknown – The t-Distribution
In most sampling situations the population standard deviation (σ) is not known.
Below are some examples where it is unlikely the population standard
deviations would be known.
1. The Dean of the Business College wants to estimate the mean number
of hours full-time students work at paying jobs each week. He selects a
sample of 30 students, contacts each student and asks them how many
hours they worked last week.
3. The manager of a store wants to estimate the mean amount spent per
shopping visit by customers. A sample of 20 customers is taken.
9-16
t-Distribution
CHARACTERISTICS OF THE t-Distribution
4. The t distribution is more spread out and flatter at the center than the
standard normal distribution As the sample size increases, however, the t
distribution approaches the standard normal distribution
9-17
Z-distribution vs. t-distribution
9-18
Confidence Interval Estimates for the Mean
9-19
When to use Z or t-distribution
9-20
Confidence Interval for the Mean – Example
using the t-distribution
EXAMPLE
A tire manufacturer wishes to investigate the tread life of its
tires. A sample of 10 tires driven 50,000 miles revealed a
sample mean of 0.32 inch of tread remaining with a
standard deviation of 0.09 inch.
Construct a 95 percent confidence interval for the population
mean.
Would it be reasonable for the manufacturer to conclude that
after 50,000 miles the population mean amount of tread
remaining is 0.30 inches?
9-21
9-22
9-23
A Confidence Interval for a Proportion (π)
9-24
Proportion
Proportion:
– Fraction, ratio or percent indicating the part of the
sample having a particular trait of interest
9-25
Confidence Interval for Population Proportion
Using the Normal Distribution to Approximate the Binomial Distribution
To develop a confidence interval for a proportion, we need to meet the following assumptions.
1. The binomial conditions, discussed in Chapter 6, have been met. Briefly, these conditions are:
a. The sample data is the result of counts.
b. There are only two possible outcomes.
c. The probability of a success remains the same from one trial to the next.
d. The trials are independent. This means the outcome on one trial does not affect the outcome on
another.
2. The values n π and n(1-π) should both be greater than or equal to 5. This condition allows us to
invoke the central limit theorem and employ the standard normal distribution, that is, z, to
complete a confidence interval.
9-26
Confidence Interval for a Population
Proportion- Example
First, compute the sample proportion :
x 1,600
EXAMPLE
p 0.80
n 2000
The union representing the Bottle
Blowers of America (BBA) is
considering a proposal to merge Compute the 95% C.I.
with the Teamsters Union.
According to BBA union bylaws, at p( 1 p )
least three-fourths of the union C.I. p z / 2
membership must approve any n
merger. A random sample of 2,000 .80( 1 .80 )
current BBA members reveals 0.80 1.96 .80 .018
1,600 plan to vote for the merger 2,000
proposal. What is the estimate of
the population proportion? ( 0.782 , 0.818 )
`
Develop a 95 percent confidence
interval for the population Conclude : The merger proposal will likely pass
proportion. Basing your decision because the interval estimate includes values greater
on this sample information, can
you conclude that the necessary than 75 percent of the union membership .
proportion of BBA members favor
the merger? Why?
9-27
Selecting an Appropriate Sample Size
9-28
Example: Finding Sample Size for Estimating A
Population Mean
9-29
Sample Size for Estimating a
Population Proportion
2
EXAMPLE 1
Z The American Kennel Club wanted to
n p (1 p)
E estimate the proportion of children that
have a dog as a pet. If the club wanted
the estimate to be within 3% of the
population proportion, how many children
where: would they need to contact? Assume a
n is the size of the sample 95% level of confidence and that the club
estimated that 30% of the children have a
z is the standard normal value dog as a pet.
corresponding to the desired level 2
of confidence 1.96
n (. 30)(. 70) 897
.03
E is the maximum allowable error
NOTE:
use p = 0.5 if no initial information on
the probability of success is available
9-30
Sample Size for Estimating a Population
Proportion
EXAMPLE 2
A study needs to estimate the
proportion of cities that have private
refuse collectors. The investigator 2
1.65
n (.5)(1 .5) 68.0625
wants the margin of error to be . 10
within .10 of the population n 69 cities
proportion, the desired level of
confidence is 90 percent, and no
estimate is available for the
population proportion. What is the
required sample size?
N n p (1 p ) N n
x p
n N 1 n N 1
However, if n/N < .05, the finite-population correction factor may be ignored.
Why? See what happens to the value of the correction factor in the table below
when the fraction n/N becomes smaller
Applied, when:
– N is known
– (n/N)>0.05 i.e. the sample is more than 5% of the
population
N n p (1 p ) N n
x p
n N 1 n N 1
9-33
CI for Mean with FPC - Example
s N n
EXAMPLE X t
There are 250 families in Scandia, n N 1
Pennsylvania. A random sample of 40 of
these families revealed the mean annual
church contribution was $450 and the
standard deviation of this was $75.
Construct a 95% confidence interval for $75 250 40
the population mean? $450 t.10 / 2, 401
40 250 1
Could the population mean be $445 or
$425? $75 250 40
$450 2.023
40 250 1
Given in Problem:
N – 250
n – 40 $450 $22.03
s - $75
($427.96, $472.03)
Since n/N = 40/250 = 0.16, the finite
population correction factor It is likely th at the population mean is more than $427.96
must be used.
but less than $472.03
The population standard To put it another way, could the population mean be $445?
deviation is not known, Yes, but it is not likely tha t it is $425 because the value
hence we use the t-
distribution $445 is within th e confidence interval and $425 is not within
the confidence interval.
9-34