Download Estimating population mean

Determining Sample Size
E=
zα / 2
Slide 1
pˆ qˆ
n
(solve for n by algebra)
n=
( zα / 2)2 pˆ qˆ
E2
Sample Size for Estimating
Proportion p
ˆ
When an estimate of p is known:
n=
( zα / 2 )2 pˆ qˆ
Formula 6-2
E2
When no estimate of p is known:
n=
( zα / 2)2 0.25
E2
Formula 6-3
Slide 2
Slide 3
Example: Suppose a sociologist wants to determine
the current percentage of U.S. households using e-mail.
How many households must be surveyed in order to be
95% confident that the sample percentage is in error by no
more than four percentage points?
a) Use this result from an earlier study: In 1997, 16.9% of U.S.
households used e-mail (based on data from The World
Almanac and Book of Facts).
b) Assume that we have no prior information suggesting a
possible value of p.
ˆ
1- α=95%=.95 , α=0.05
z α/2=z.025=1.96.
E=4%=0.04
(a) phat=16.9%=.169
(b) Use phat=0.5
Slide 4
a) Use this result from an earlier study: In 1997, 16.9% of U.S.
households used e-mail (based on data from The World
Almanac and Book of Facts).
ˆˆ
n = [za/2 ]2 p q
E2
= [1.96]2 (0.169)(0.831)
0.042
= 337.194
= 338 households
To be 95% confident that our
sample percentage is within
four percentage points of the
true percentage for all
households, we should
randomly select and survey
338 households.
Slide 5
b) Assume that we have no prior information suggesting a
possible value of p.
ˆ
n = [za/2 ]2 • 0.25
E2
= (1.96)2 (0.25)
0.042
= 600.25
= 601 households
With no prior information,
we need a larger sample to
achieve the same results
with 95% confidence and an
error of no more than 4%.
Finding the Point Estimate
and E from a
Confidence Interval
ˆ
(upper confidence limit) + (lower confidence limit)
Point estimate of p:
ˆ
p=
2
Margin of Error:
E = (upper confidence limit) — (lower confidence limit)
2
Slide 6
Section 6-3
Estimating a Population Mean:
σ Known
Slide 7
Assumptions:
1. The sample is a simple random
sample.
2. The value of the population standard
deviation σ is known.
3. Either or both of these conditions is
satisfied: The population is
normally distributed or n > 30.
Point estimate
Slide 8
! Population mean: µ (unknown)
! Point Estimate
is a single value (or point) used to approximate
a population parameter.
Example:
The sample mean x is the best point estimate of the
population mean µ.
Slide 9
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the point
estimate of the population mean µ of all body temperatures.
Because the sample mean x is the
best point estimate of the population
mean µ, we conclude that the best
point estimate of the population
mean µ of all body temperatures is
98.20o F.
Interval Estimate
Slide 10
Level of Confidence
! The confidence level =1 - α,
!where α is the complement of the confidence level.
! For a 0.95(95%) confidence level, α = 0.05. For
a 0.99(99%) confidence level, α = 0.01.
Slide 11
Margin of Error
is the maximum likely difference observed
between sample mean x and population
mean µ, and is denoted by E.
E = zα/2 •
σ
n
Standard Error of the sample mean
Confidence Interval
Slide 12
(or Interval Estimate) for
Population Mean µ when σ is known
x –E <µ< x +E
x +E
(x – E, x + E)
Slide 13
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and the
sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval for µ.
n = 106
x = 98.20o
s = 0.62o
α = 0.05
α /2 = 0.025
z α/ 2 = 1.96
E = z α/ 2 • σ
n
= 1.96 • 0.62
106
= 0.12
x –E <µ< x +E
98.08 < µ < 98.32
o
o
98.20o – 0.12
98.08o
<µ<
< µ <
98.20o + 0.12
98.32o
Sample Size for Estimating
Mean µ
n=
(zα/2) • σ
E
2
Slide 14
Round-Off Rule for
Sample Size n
Slide 15
When finding the sample size n, if the use of
Formula 6-5 does not result in a whole number,
always increase the value of n to the next larger
whole number.
Finding the Sample Size n
when σ is unknown
Slide 16
1. Use the range rule of thumb (see Section 2-5) to
estimate the standard deviation as follows: σ ≈
range/4.
2. Conduct a pilot study by starting the sampling
process. Based on the first collection of at least
31 randomly selected sample values, calculate the
sample standard deviation s and use it in place of
σ.
3. Estimate the value of σ by using the results of
some other study that was done earlier.
Slide 17
Example:
Assume that we want to estimate the mean
IQ score for the population of statistics professors. How
many statistics professors must be randomly selected for
IQ tests if we want 95% confidence that the sample mean is
within 2 IQ points of the population mean? Assume that σ
= 15, as is found in the general population.
α = 0.05
α /2 = 0.025
z α/ 2 = 1.96
E = 2
σ = 15
n =
1.96 • 15 2= 216.09 = 217
2
With a simple random sample of only
217 statistics professors, we will be
95% confident that the sample mean
will be within 2 points of the true
population mean µ.
6-4: σ Not Known
Assumptions
Slide 18
1) The sample is a simple random sample.
2) Either the sample is from a normally
distributed population, or n > 30.
Use Student t distribution
Student t Distribution
Slide 19
If the distribution of a population is
essentially normal, then the distribution of
t =
x-µ
s
n
! is essentially a Student t Distribution for all
samples of size n, and is used to find critical
values denoted by tα/2.
Slide 20
Degrees of Freedom (df )
corresponds to the number of sample values
that can vary after certain restrictions have
been imposed on all data values
df = n – 1
in this section.
Margin of Error E
for Estimate of µ
Slide 21
Based on an Unknown σ and a Small Simple Random
Sample from a Normally Distributed Population
E = tα/
s
2
n
where tα / 2 has n – 1 degrees of freedom.
Student t Distributions for
n = 3 and n = 12
Slide 22
Important Properties of the
Student t Distribution
Slide 23
1. The Student t distribution is different for different sample sizes
(see Figure 6-5 for the cases n = 3 and n = 12).
2. The Student t distribution has the same general symmetric bell
shape as the normal distribution but it reflects the greater
variability (with wider distributions) that is expected with small
samples.
3. The Student t distribution has a mean of t = 0 (just as the
standard normal distribution has a mean of z = 0).
4. The standard deviation of the Student t distribution varies with
the sample size and is greater than 1 (unlike the standard normal
distribution, which has a σ = 1).
5. As the sample size n gets larger, the Student t distribution gets
closer to the normal distribution.
Confidence Interval for
the Estimate of E
Slide 24
Based on an Unknown σ and a Small Simple Random
Sample from a Normally Distributed Population
x–E <µ<x +E
where
E = tα/2 s
n
tα/2 found in Table A-3
Procedure for Constructing a
Confidence Interval for µ
when σ is not known
Slide 25
1. Verify that the required assumptions are met.
2. Using n — 1 degrees of freedom, refer to Table A3 and find the critical value tα/2 that corresponds
to the desired degree of confidence.
3. Evaluate the margin of error E = tα/2 • s / n .
4. Find the values of x - E and x + E. Substitute those
values in the general format for the confidence
interval:
x –E <µ< x +E
5. Round the resulting confidence interval limits.
Slide 26
Example:
A study found the body temperatures of 106
healthy adults. The sample mean was 98.2 degrees and
the sample standard deviation was 0.62 degrees. Find the
margin of error E and the 95% confidence interval for µ.
n = 106
x = 98.20o
s = 0.62o
α = 0.05
α /2 = 0.025
t α/ 2 = 1.984
E = t α/ 2 • s = 1.984 • 0.62 = 0.1195
n
106
x–E <µ< x +E
98.20o – 0.1195 < µ < 98.20o + 0.1195
98.08o <
µ < 98.32o
The interval is the same here as in Section 6-2, but in some
other cases, the difference would be much greater.
Using the Normal and
t Distribution
Figure 6-6
Slide 27