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Slide 1
Section 7-3, 7.4
Estimating a Population Mean
Assumptions:
1. The sample is a simple random sample.
2. The population is normally distributed
or n > 30.
Point estimate
™ Population mean: µ (unknown)
™ Point Estimate:
™ The sample mean: x
™ Exact standard error:
σ
n
™ Estimated standard error (se):
s
n
Slide 2
(i) 100(1-α)% Confidence Interval forSlide 3
Population Mean µ if σ is known
point estimate ± margin of error
x±E
(x – E, x + E)
E = zα / 2 •
σ
n
Example: Constructing confidence interval
Slide 4
A study found the body temperatures of 106 healthy adults. The
sample mean was 98.2 degrees and the standard deviation was 0.62
degrees. Find
(a) the point estimate of the population mean μ of all body
temperatures. 98.2 degree
(b) the margin of error E
(c) the 95% confidence interval for µ.
Slide 5
Solution: (a) 98.20
(b) E =
n = 106
x = 98.20o
σ = 0.62o
z α/ 2 • σ
n
= 1.96 • 0.62
106
= 0.12
(c) Recall
α = 0.05
α /2 = 0.025
z α/ 2 = 1.96
x –E <μ< x +E
98.20o – 0.12
98.08o
<μ<
< μ <
98.20o + 0.12
98.32o
Sample Size for Estimating
Mean μ
n=
(zα/2) • σ
Slide 6
2
E
Sample size formula with
95% confidence level and
margin of error E
is approximated by if σ is
unknown
2
4s
n= 2
E
Round-Off Rule for
Sample Size n
Slide 7
When finding the sample size n, if it does not
result in a whole number, always increase the
value of n to the next larger whole number.
Finding the Sample Size n Slide 8
when σ is unknown
1. Use the range rule of thumb 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.
Example:
Slide 9
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 0.025 = 1.96
E
σ
= 2
= 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.
(ii) σ is not known
Use Student t
distribution
Slide 10
Important Properties of the
Student t Distribution
Slide 11
1. The Student t distribution is different for different sample sizes
(see Figure 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.
Table B
Slide 12
Student t -score
Slide 13
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
™ Degrees of Freedom (df)=n-1
Margin of Error E
for Estimating μ
Slide 14
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.
•100(1-α)%Confidence Interval for µ
Slide 15
s
x ± t a/2 ( ); df = n - 1
n
• tα/2 found in Table B
•
zA
Based on an Unknown σ and a Small Simple Random
Sample from a Normally Distributed Population
95% confidence interval for the population mean µ is:
s
x ± t ( ); df = n - 1
n
.025
Procedure for Constructing a
Confidence Interval for µ
when σ is not known
Slide 16
1. Verify that the required assumptions are met.
2. Using n — 1 degrees of freedom, refer to Table B:
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.
The Standard Normal Distribution Slide
is 17
the t-Distribution with df = ∞
Example:
Slide 18
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 0.025 = 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 Flow
the Normal
and
Chart
t Distribution
Slide 19
Summary: Sections 7.1-7.4
• Point estimates:
•
Slide 20
p̂ − − − −− > p
x − − − −− > μ
Margin of Error: E=(critical value)(standard error)
• Confidence
Interval:
(point estimate) ± E
• Sample size n: a solution of E=(critical value)(standard error)