ESTIMATION
• Estimation: process of using sample
values to estimate population values
• Point Estimates: parameter is estimated
as single point
– Examples: x, s, p
• Careful statisticians dislike point estimates
Interval Estimates
• Example: there is a 90% probability that
somewhere between 58 and 68% of
Americans oppose same-sex marriage
• Draws explicit attention to the fact of
variability in the sample results; avoids
putting too much weight on one number
Think about the interval 42 – 1.64 * 1 to 42 + 1.64 * 1: This
interval contains 90% of the sample means that could ever
be drawn from this population
37
42
47
• The interval 40.36 to 43.64 contains 90%
of the sample means possible from this
population
• No sample mean in this interval differs
from μ by more than 1.64 grams
• Hence, there is a 90% probability that any
arbitrary x differs from μ by no more
than 1.64
• Thus, there is a 90% probability that μ is in
the interval x  1.64
• Look at this interval
• x  1.64 * 1
• 1.64 is a z value, chosen to correspond to
90%, the confidence level
• 1 is the standard error of the mean
• So the width of the interval is set by the
confidence level (which determines
number of standard errors in interval) and
the standard error, the measure of
variation in sample means
A C% Confidence Interval for the
Population Mean When σ Is Known
x  zC   x
Examples:
• A population of Christmas trees has
unknown mean with σ = 4. For a sample
of 25 trees, the sample mean = 16.6 ft.
Calculate a 95% confidence interval for
the population mean.
• Same data: calculate a 90% confidence
interval
• Same data: suppose that we increase
sample size to 81
Width of Confidence Interval
Depends On
• Confidence level: as C increases, width
decreases
• Sample size: as n increases, width
decreases
• Variability in population: as σ increases,
standard error increases and width of
interval increases
• The quantity zC * σX is called the maximum
error in the estimate
• The quantity 2 * zC * σX is called the
precision in the estimate
– this quantity is the width of the confidence
interval
FINDING THE RIGHT SAMPLE
SIZE
• Sometimes we wish to hold the error in the
estimate within some limit
• Define e = zC * σX or substituting
e

n
Solve this expression for n, yielding
 zC   
n

 e 
2
Example: With σ = 4 and 95% confidence level,
we require that the maximum error in the estimate
be no more than 0.5 ft. What sample size is
necessary?
Examples:
• Expectations of inflation are known to be
normally distributed with standard
deviation = 1.2%. A survey of sixty
households found a sample average
expectation of 4% inflation for the coming
year. Calculate a 98% confidence interval
for the population’s expectation of inflation
in the coming year.
• If we require a maximum error in the
estimate of 0.1%, how large a sample
must we take?
• Cigarette filters have a “process” standard
deviation of 0.3 mm with normal
distribution. The current mean is
unknown, but a sample of 25 filters have a
mean of 20 mm.
– Calculate a 90% confidence interval for the
population mean
– Find the sample size necessary to hold the
error in the estimate to 0.04 mm
Student’s t distribution
• Suppose σ is NOT known; then we are not
entitled to use a z value in calculating
confidence intervals
• If, however
– The population is known to be normally
distributed OR
– The sample size is large enough to invoke the
Central Limit Theorem, then we use
• A value drawn from the t distribution
Hey, Prof, what’s a t distribution?
• Characteristics
– Symmetric about its mean of zero
– Values tend to cluster in the center, producing
a bell shaped curve
• Differences from z:
– Fatter tails and less mass in the center
– There is a family of t distributions, based on
“degrees of freedom”
• Degrees of freedom: the sample size minus
number of parameters to be estimated before
estimating a variance
( x  x )
s 
n 1
2
2
Before estimating the variance, we must first
calculate x-bar, an estimate of the population
mean: we lose one degree of freedom, leaving us
with n – 1 degrees of freedom
Confidence Intervals with the t
distribution:
x  t  sx
s
sx 
n
Where t is chosen for the desired confidence level
and has n – 1 degrees of freedom
Examples:
• Seven male students are allowed to
imbibe their favorite beverage until they
are visibly inebriated. The amounts
consumed in ounces are: 3.7, 2.9, 3.2, 4.1,
4.6, 2.3, 2.5. Calculate a 95% confidence
interval for the amount of the drink it would
take to get the average member of the
population drunk.
• Calculate x-bar and s
• Then calculate the sample standard error
• Find t for 6 degrees of freedom and  =
0.025
• Finally, calculate the confidence interval
• In a sample of 41 students who work, the
sample mean is 16.561 hours and s =
5.7128 hours. The distribution appears to
be somewhat skewed upwards. Find a
90% confidence interval for the average
hours worked by all ASU students who
work.
USE OF THE t DISTRIBUTION
• Footnote: Who was “Student”? A pseudonym for
William Gosset
• The t is often thought of as a small-sample
technique
• But, STRICTLY SPEAKING, the t should be
used whenever the population standard
deviation σ is NOT KNOWN
• Some practitioners use z whenever the sample
is large
– Central Limit Theorem
– There isn’t much difference between t and z
Population standard
deviation known?
Yes
No
Population normal?
Yes
Population normal?
No
Yes
No
Sample Size
Sample Size
z value
n >= 30
n < 30
z or t (see
t value
note)
ERROR
n >= 30
n < 30
z or t (see
note)
ERROR
Notes:
• For large samples with σ unknown, different
practitioners may proceed differently. Some
argue for using a z, appealing to CLT. Others
use a t since it gives a less precise estimate.
For this course: use a t whenever the
population standard deviation is not known.
• Small samples from non-normal populations are
beyond the scope of this course
Confidence intervals for the
population proportion 
• Sample proportion p = x/n
• E(p) =  and
p 
  (1   )
n
In general  is not know, so must be estimated
with p and we use
sp 
p  (1  p )
n
• Then the confidence interval is
• p  zC  sp
• Note that proportion problems always use
a z value
– Normal approximates binomial
• EXAMPLE: Of 112 students in a sample,
70 have paying jobs. Calculate a 95%
confidence interval for the proportion in the
population with paying jobs.
• p = 70/112 = 0.625
0.625  0.375
sp 
 0.045745315
112
• 0.625  1.96 * 0.045 etc.
• 0.625  0.089660819 or 0.625  0.09
• We are 95% confident that
0.54    0.71
• EXAMPLE:
• In a sample of 320 professional
economists, 251 agreed that “offshoring”
jobs is good for the American economy.
Calculate a 90% confidence interval for
the proportion in the population of
professional economists who hold this
view.
Finding the Right Sample Size
• The error in the estimate is given by
zC  σp or, substituting
e  zC 
  (1   )
n
Solving for n yields:
n
2
zC
   (1   )
2
e
• In general  is not known
• Two solutions:
– Assume  = 0.5
• Result is the largest sample that would ever be
needed
– Conduct a pilot study and use the resulting p
as an estimate of 
• May give a somewhat smaller sample size if p is
much different from 0.5
• Saves sampling cost
Example:
• Above we had a 95% confidence interval with n =
112 of 0.625  0.09 or a 9% error. Suppose we
require a maximum error of 3%.
• Approach 1: let  = 0.5
1.96  0.5  0.5
n

1067
.
11

1068
2
.03
2
• Approach 2: assume  = 0.625
1.96  0.625  0.375
n

1000
.
41

1001
2
.03
2
The difference is more dramatic if p is much
different from 0.5. In a random sample of 300
students in NC, 30 have experienced “study”
abroad. A 95% confidence interval for the
population proportion is 10%  3.4%. Suppose we
require a maximum error of 2%. Approach 1 gives
_______ and approach 2 gives _________.