Some Concepts
*Estimators: Random variables used to estimate
population parameters.
Example:
p hat is an estimator of p
*Estimates: Specific values of an estimator.
Example:
*Point Estimates: Specify a single value of a
population parameter.
Example: x =120
*Confidence Interval/Interval Estimates:
Calculates 2 numbers based on sample that
form an interval within which the parameter
is expected to lie.
Example: 115=30
Example: a sample with n=50, x =8.1and s=0.24
1) Point estimate of population mean
2) Interval estimate of population mean with a 95%
confidence level
3) Interval estimate of population mean with a 90%
confidence level.
The interval estimate so far is two-sided interval
estimate. Some times we need one-side interval
estimate:
*Lower one-sided confidence interval (LCL)
LCL= (point estimate)- z *(standard error of estimator)
*Upper one-sided confidence interval(UCL)
UCL= (point estimate)+ z *(standard error of estimator)
Where standard error of estimator =
n
Example: a sample with n=50, x =8.1 and s=0.24
4) LCL estimate of population mean
5) UCL estimate of population mean
When sample is large (n is large), we can use
sample standard deviation s as reliable
estimator of population standard deviation
And no matter what distribution the population has,
sampling distribution of sample mean is
normally distributed
But, what if sample is small?
Consequences of small sample size:
1) if population does not possess normal
distribution, sampling distribution of small
sample mean is not normal distribution.
2) s / n is much more variable than n
(It has thicker tails)
What should we do to deal with those problems?
Assumption: 1) population is normally distributed
2) sample is small and is unknown
If we use sample standard deviation s to replace
The statistic t follows one mound-shaped distribution,
which is called t-distribution.
x
t
s/ n
Where s is sample standard deviation
(x i x) 2
s
xi
n 1
Property of t-distribution
1) t-distribution is mound-shaped
2) t-distribution is perfectly symmetric about t=0
3) t-distribution is more variable than z (normal distribution
is also called z-distribution)
4) t-distribution is affected by the sample size n
Because when n changes, s changes, t-distribution
is affected by sample size n. Therefore,
different sample sizes have different t-
distributions, even if sample is picked form
the same population.
We call number n-1 the number of degree of
freedom (d.f.) associated with s2 and thus
the t-statistic.
Example: when n=10, d.f.=10-1=9
So, before we check the t-distribution, we need to
first determine degree of freedom.
How to check the t-distribution table
• tα records the value of t such that an area α lies
to its right. (Appendix II, Table 4, pp611)
• Determine the degree of freedom
• Determine what value α is (usually a is given in
the question)
• Find the tα
Problems:
1) n=10, find t0.05 and t0.025
2) n30, find t0.05 and t0.025
3) Repeat 1 with n=9
Do you remember we said when n>30, we may
think it is a large sample? After doing these
problems, did you get the intuition about why
we set 30 as the dividing line?
Remember the t-distribution based on the assumption that
the sampled population possesses a normal probability
distribution.
This is very restricted assumption.
Fortunately, it can be shown that distribution of the t
statistic possesses nearly the same shape as the theoretical t-
distribution for population that are nonnormal but possesses
a mound-shaped probability distribution. So even
population is not normal but mount-shaped, we can use t-
distribution.
In our class, except for when specifically stated, we will
suppose using t-distribution is proper.
Small-sample (1- )100% confidence interval
estimator of the mean is
s
x t / 2
n
Where, s is sample standard deviation and t has
(n-1) d.f.
s is the estimated standard deviation
n of x
Problem: here is a sample of diamonds weights:
{0.46, 0.61, 0.52, 0.48, 0.57, 0.54}
Find the 95% confidence interval estimate for population
mean.
Estimating the difference between two means
Properties of the sampling distribution of ( x1 x2 ), the
difference two sample means:
When independent random samples of n1 and n2
observations have been selected from population with
means 1 and 2 and variances 12 and 2 2
respectively, the sampling distribution of the difference
will have the following properties:
1) The mean and standard deviation of ( x1 x2 ) will be
( x x ) 1 2
1 2
and
12 2
2
(x x )
1 2
n1 n2
2) If the sampled populations are normally
distributed, then the sampling distribution of
( x1 x2 ) is exactly normally distributed,
regardless of the sample size.
3) If the sample populations are not normally
distributed, then the sampling distribution of
( x1 x2 ) is approximately normally
distributed when n1 and n2 are large, due to
the Central Limit Theorem.
Point Estimation of (1 2 ) Large Sample
1 2 x1 x2
A (1- ) 100% confidence interval for ( 1 2)
12 2
2
( x1 x2 ) z / 2
n1 n2
If population variance are unknown, they can be
approximated by the sample variances.
Example: n1=30 n2=30, sample mean are 1.32 and
1.04 respectively, sample variance are 0.9734
and 0.7291 respectively.
1) Point estimate of difference of mean
2) 90% confidence interval estimate of difference
of mean
In small sample
Assumption 1: both samples are picked from
population with normal distribution.
Assumption 2: both population possess equal
variances.
Then t statistic follows t-distribution
( x1 x2 ) ( 1 2 )
t
1 1
s
n1 n2
Where s is sample standard deviation
Degrees of freedom = n1+n2-2
Both assumptions are very restrictive
Assumption 1: As before, even if the population has a
non-normal distribution, but have a mound-shaped
distribution, a t-statistic found will be very close
to the theoretical t-statistic.
Assumption 2: Usually we don’t know population
distribution. In our class, we just can suppose they
are same when we have such questions
As to the question which s to use, we pool the
information from two samples and get the so-called
“pooled estimator of population variance”
S is called pooled estimator of population variance
n1 n2
( x1i x1 ) 2 ( x2i x2 ) 2
s2 i 1 i 1
n1 n2 2
or
(n1 1)s12 (n2 1) s2
2
s2
(n1 1) (n2 1)
Estimating the Difference Between Two Mean:
Small Sample
If the two samples are small (<30)
Point estimate of the difference:
1 2 x1 x2
A (1- )100% small-sample confidence interval
for ( 1 2 )
1 1
( x1 x2 ) t / 2 s
n1 n2
Where s is obtained from the pooled estimate, given before.
Example: here are two samples
{32, 37, 35, 28, 41, 44, 35, 31, 34}
{35, 31, 29, 25, 34, 40, 27, 32, 31}
What is 95% confidence interval estimate of
difference between two sample means?
Properties of sampling distribution of the
sample proportion
1. If a random sample of n observations is selected from
a binomial population with parameter p, the sampling
^
distribution of the sample proportion p x n
will have a mean ^ p
p
pq
and a standard deviation: ^
p n
where q=1-p
2. when the sample size n is large, the sampling
distribution of sample proportion will be
approximately normal. Remember the rule is np and
nq both greater than or equals to 5
Estimating a Binomial proportion
Point estimator:
The sample proportion is an unbiased estimator of
population proportion, so
^
*Point estimator of p: px n
*Interval estimator depend on what sampling
distribution of sample proportion is
So, If
^
1) we use sample proportion p as reliable
estimate of population proportion p, and
^ ^
2) n* p and n*(1- p ) both greater than 5, then
A (1- )100% confidence interval for p is
ˆˆ
pq
p z / 2
ˆ
n
Example: n=100, x=59
1) What is the point estimate of p?
2) What is 95% confidence interval of the
estimate for p?
Estimating the Difference Between Two
Binomial Proportions
• Point estimate
( p1 p2 ) ( p1 p2 )
ˆ ˆ
• Confidence interval for the difference
ˆ ˆ ˆ ˆ
p1q1 p2 q2
( p1 p2 ) z / 2
ˆ ˆ (
n1 n2
Example
• A recent survey of girls between 9 and 14 years old
showed that 44% of the white girls surveyed and 28% of
the African-American girls reported that they had tried to
lose weight. Assume 100 girls of each ethnicity were
surveyed.
– What is the estimated difference in population
proportions.
– Construct a 99% confidence interval of the difference
in proportions.
Choosing sample size
Questions to ask before determining optimal n:
1) What level of confidence do you want to have
(i.e., the value of 100(1- )?
2) What is the maximum difference (D) you want
to permit between the estimate of the population
parameter and the true population parameter.
Procedure
• Once you have chosen D and alpha solve
the following equation for sample size n:
z / 2 (standard error of the estimator) D
So this can be used for any of the estimators that we have
talked about thus far, you just need the proper standard
error
estimation of the mean
σ
σx
n
estimation of differenceof means
12 2
2
x x
1 2
n1 n2
estimation of a proportion
ˆˆ
pq
p
ˆ
n
estimation of differenceof proportions
ˆ ˆ
p1q1 ˆ ˆ
p2 q2
(p p
ˆ ˆ )
1 2
n1 n2
Additional info
• When you are using two different samples
calculate just one n and use it for both
• Choosing the proper standard deviation and
proportion can be hard, use any information
from past experiments you may have
• When using proportions choosing a p of .5
will give you the largest possible n value
you may want to use it to be safe
Additional info continued
• If you are concerned with the n to choose
when using means and you have some idea
of what the range (R ) is you can estimate
the standard deviation as
4 R
Example: we know the population standard
deviation is 0.24 and we want the error of
the estimation to be less than 0.06 with
95% confidence.
At least how many observations do we need in
the sample?