Download Topic 8 Estimating Single Parameter

zA
point estimateis a single number, used
to estimate an unknown population
parameter
z a confidence intervalprovides additional
information about variability
z How
much uncertainty is associated with
a point estimate of a population
parameter?
z An interval estimateprovides more
information about a population
characteristic than does a point estimate
z Such interval estimates are called
confidence intervals
z An
interval gives a rangeof values:
Takes into consideration variation in sample
statistics from sample to sample
z Based on observation from 1 sample
z Gives information about closeness to
unknown population parameters
z Stated in terms of level of confidenceNever
100% sure
z
z Confidence
z
Level
Confidence in which the interval will contain
the unknown population parameter
zA
percentage (less than 100%)
z Suppose confidence level = 95%
z Also written (1 -α) = .95
z
A relative frequency interpretation:In the
long run, 95% of all the confidence intervals
that can be constructed will contain the
unknown true parameter
zA
specific interval either will contain or
will not contain the true parameter
z
No probability involved in a specific interval
z Assumptions
Population
Standard deviation σ is known
z Population is normally distributed
z If population is not normal, use large sample
z
z
z
z
A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
Determine a 95% confidence interval for the
true mean resistance of the population.
A sample of 11 circuits from a large normal
population has a mean resistance of 2.20
ohms. We know from past testing that the
population standard deviation is 0.35 ohms.
z
z
z
We are 95% confident that the true mean
resistance is between 1.9932 and 2.4068
ohms
Although the true mean may or may not be in
this interval, 95% of intervals formed in this
mannerwill contain the true mean
An incorrectinterpretation is that there is 95%
probability that this interval contains the true
population mean. (This interval either does or
does not contain the true mean, there is no
probability for a single interval)
z If
the population standard deviation σis
unknown, we can substitute the sample
standard deviation, s
z This introduces extra uncertainty, since s
is variable from sample to sample
z So we use the t distributioninstead of the
normal distribution
z Assumptions
Population standard deviation is unknown
z Population is normally distributed
z If population is not normal, use large sample
z
z Use
Student’s t Distribution
z
z
z
Since t approaches z as the sample size
increases, an approximation is sometimes
used when n is very large
The text t-table provides t values up to 500
degrees of freedom
Computer software will provide the correct tvalue for any degrees of freedom
z The
required sample size can be found
to reach a desired margin of error (e)
and level of confidence (1 - α)
z If
unknown, σcan be estimated when
using the required sample size formula
Use a value for σthat is expected to be at
least as large as the true σ
z Select a pilot sampleand estimate σwith the
sample standard deviation, s
z Use the range R to estimate the standard
deviation using σ= R/6 (or R/4 for a more
conservative estimate, producing a larger
sample size)
z
z An
interval estimate for the population
proportion ( π) can be calculated by
adding an allowance for uncertainty to
the sample proportion ( p )
z A random sample of 100 people shows
that 25 are left-handed.
z Form a 95% confidence interval for the
true proportion of left-handers
z Recall
that the distribution of the sample
proportion is approximately normal if the
sample size is large, with standard
deviation
z We
will estimate this with sample data:
z Upper
and lower confidence limits for the
population proportion are calculated with
the formula
z How
large a sample would be necessary
to estimate the true proportion defective
in a large population within 3%,with 95%
confidence? (Assume a pilot sample
yields p = .12)
z The
sample is a simple random sample.
z The population must have normally
distributed values (even if the sample is
large).
(n – 1) s2
2
Where
The values of chi-square can be zero or
positive, but they cannot be negative.
z The chi-square distribution is different for
each number of degrees of freedom, which
is df = n – 1 in this section. As the number
increases, the chi-square distribution
approaches a normal distribution.
z
σ2
n = sample size
s 2 = sample variance
σ 2 = population variance
z The
chi-square distribution is not symmetric,
unlike the normal and Student t distributions.
As the number of degrees of freedom
increases, the distribution becomes more
symmetric.
z
Find the critical
values of χ2 that
determine critical
regions containing an
area of 0.025 in each
tail. Assume that the
relevant sample size
is 10 so that the
number of degrees of
freedom is 10 – 1, or
9.
(n – 1)s 2
χ
Right-tail CV
2
<σ <
2
R
(n – 1)s 2
χ
2
L
Left-tail CV
Confidence Interval for the Population Standard Deviation σ
(n – 1)s 2
χ
2
R
< σ <
(n – 1)s 2
χ
2
L