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One Sample Estimation
Spring 2011
Outline
1
Introduction
2
Outline
3
Mechanics of Computing a Confidence Interval - Reporting
Results
4
Case Studies
Body Temperature (continuous data)
Radiation (discrete data)
Outline
1
Introduction
2
Outline
3
Mechanics of Computing a Confidence Interval - Reporting
Results
4
Case Studies
Body Temperature (continuous data)
Radiation (discrete data)
Confidence Intervals
We will explore confidence intervals in the context of an
unknown
1
population mean µ
2
population proportion p
One Sample Problems
We focus on one sample.
Outline
1
Introduction
2
Outline
3
Mechanics of Computing a Confidence Interval - Reporting
Results
4
Case Studies
Body Temperature (continuous data)
Radiation (discrete data)
Review
We covered most of this material on the board
1
Introduction to confidence intervals. Meaning is a repeated
sampling idea.
2
Confidence interval for µ (t-table)
3
Confidence interval for p (use p̃)
4
Planning a study (in both contexts µ, p); sample size
calculation
Still Left
Reporting results.
Write in the context of the scientific problem.
Outline
1
Introduction
2
Outline
3
Mechanics of Computing a Confidence Interval - Reporting
Results
4
Case Studies
Body Temperature (continuous data)
Radiation (discrete data)
Example (problem 6.10)
Part of a study on the development of the thymus gland:
weights of thymus gland from 5 chick embryos after 14
days of incubation:
29.6
21.5
28.0
34.6
44.9
(mg)
We want to know µ, the mean weight of thymus glands in
the entire population of chick embryos after 14 days of
incubation (in the same incubator).
ȳ = 31.72 is our best estimate for µ. How good is this
estimate? How far is µ from 31.72?
Standard error of the mean
√
We know the standard deviation of Ȳ is σ/ n. But we don’t
know σ. Hopefully, the standard deviation of the data, s, is
close to σ.
s
SEȳ = √
n
is the standard error of the mean.
It is an estimate of the standard deviation of Ȳ . The deviation of
Ȳ from its mean µ, is an error we will make. SEȳ gives us an
idea of how far ȳ is from µ.
What happens to s when the sample size increases?
What happens to SEȳ when the sample size increases?
Standard error of the mean
√
We know the standard deviation of Ȳ is σ/ n. But we don’t
know σ. Hopefully, the standard deviation of the data, s, is
close to σ.
s
SEȳ = √
n
is the standard error of the mean.
It is an estimate of the standard deviation of Ȳ . The deviation of
Ȳ from its mean µ, is an error we will make. SEȳ gives us an
idea of how far ȳ is from µ.
What happens to s when the sample size increases?
s stays about the same, as it gets closer to σ.
What happens to SEȳ when the sample size increases?
Standard error of the mean
√
We know the standard deviation of Ȳ is σ/ n. But we don’t
know σ. Hopefully, the standard deviation of the data, s, is
close to σ.
s
SEȳ = √
n
is the standard error of the mean.
It is an estimate of the standard deviation of Ȳ . The deviation of
Ȳ from its mean µ, is an error we will make. SEȳ gives us an
idea of how far ȳ is from µ.
What happens to s when the sample size increases?
s stays about the same, as it gets closer to σ.
What happens to SEȳ when the sample size increases?
It becomes smaller and smaller, as ȳ gets closer to µ.
Mechanics of a confidence interval
1
2
Choose a confidence level. Typically, 95%. Polls use
90% or 95%.
Find the value t such that Pr {−t ≤ T ≤ t} = confidence
level. It also means
Pr {T ≥ t} = (1 − confidence level)/2
3
Refer to Student distribution on Table 4 (back cover), and
use degree of freedom df= n − 1.
Construct the interval: ȳ ± tSEȳ i.e.
(ȳ − tSEȳ , ȳ + tSEȳ )
4
Conclude:
We are 90% confident that the mean thymus gland weight of all
chick embryos at the age of 14 days of incubation (in the
condition of the experiment) is between 23.41 and 40.03 mg.
Mechanics of a confidence interval: Example
1
2
3
Confidence level. We will do both 90% and 95%.
Find the value t: such that Pr {T ≥ t} = .05 for level 90%
and .025 for level 95%.
Degree of freedom: df=5 − 1 = 4.
t-Table gives: t = 2.13 for 90% confidence and t = 2.77 for
95% confidence. With R:
> qt(.975, df=4)
[1] 2.776445
Interval: We√had ȳ = 31.72, s = 8.73 then
SEȳ = 8.73/ 5 = 3.90.
Add cushion: t ∗ SEȳ = 8.31 (90% confidence) and 10.81
(95% confidence).
The interval is 31.72 ± 8.31 or 31.72 ± 10.81, i.e.
(23.41, 40.03) for 90% confidence
(20.91, 42.53) for 95% confidence
4
Conclude.
Outline
1
Introduction
2
Outline
3
Mechanics of Computing a Confidence Interval - Reporting
Results
4
Case Studies
Body Temperature (continuous data)
Radiation (discrete data)
Case Study
Example
Body temperature varies within individuals over time (it can be
higher when one is ill with a fever, or during or after physical
exertion). However, if we measure the body temperature of a
single healthy person when at rest, these measurements vary
little from day to day, and we can associate with each person an
individual resting body temperture. There is, however, variation
among individuals of resting body temperture. A sample of
n = 130 individuals had an average resting body temperature
of 98.25 degrees Fahrenheit and a standard deviation of 0.68
degrees Fahrenheit.
Case Study: Questions
Example
How can we use the sample data to estimate with
confidence the mean resting body temperture in a
population?
How would we test the null hypothesis that the mean
resting body temperture in the population is, in fact, equal
to the well-known 98.6 degrees Fahrenheit?
How robust are the methods of inference to nonnormality
in the underlying population?
How large of a sample is needed to ensure that a
confidence interval is no larger than some specified
amount?
(some of this is beyond the scope of the course)
Example
Male radiologists may be exposed to much more radiation than
typical people, and this exposure might affect the probability
that children born to them are male. In a study of 30 “highly
irradiated” radiologists, 30 of 87 offspring were male (Hama et
al. 2001). Treating this data as a random sample, find a
confidence interval for the probability that the child of a highly
irradiated male radiologist is male.
Calculation
We find p̂ = 30/87 ≈ .345 and p̃ = 32/91 ≈ .352.
q
The standard error is .352(1−.352)
≈ 0.050.
91
The margin of error is 1.96 × SE ≈ .098.
The confidence interval is .254 < p < 0.450.
Example
We are 95% confident that the proportion of children of highly
irradiated male radiologists that are boys is between .254 and
0.450.
This confidence interval does not contain 0.512, the proportion
of male births in the general population. The inference is that
exposure to high levels of radiation in men may decrease the
probability of having a male child.