Download Nature of Estimation

Nature of
Estimation
• Point Estimate – a
single number
designed to estimate a
quantitative
parameter of a
population, usually
the value of the
corresponding
statistic.
Values from a SAMPLE.
• Examples:
a. A sample mean
is the point
estimate for the
population
mean.
b. A sample variance
is the pt. estimate
for the
population
variance.
Now do the following….
• Consider repeating the following
process:
Roll a die 5 times and find the mean x .
• Continue to do this for a VERY LARGE
number of samples.
• Notice the behavior of all sample
means that are generated as the
process continues to 10,000 trials.
Notice the mean of all of the 10,000 samples
is 3.49 (very close to 3.5) and the shape of
the sampling distribution is approximately
normal.
Example 2….
• Find the population variance of
rolling a die 5 times.
  1.707825128  2.9
2
2
Now do the following….
• Consider repeating the following
process:
Roll a die 5 times and find the
2
sample variance s .
• Continue to do this for a VERY LARGE
number of samples.
• Notice the behavior of all sample
variances that are generated as the
process continues to 10,000 trials.
Notice the variance of all of the 10,000
samples is 2.88 (very close to 2.9) and the
shape of the sampling distribution is
skewed to the right.
Example 3…
• Find the population proportion of
rolling a die and landing on an odd
number.
• Because the values of 1, 2, 3, 4, 5, 6
are all equally likely, the proportion
of odd numbers in the population is
0.5.
Now do the following….
• Consider repeating the following
process:
Roll a die 5 times and find the
proportion of odd numbers.
• Continue to do this for a VERY LARGE
number of samples.
• Notice the behavior of all sample
proportions that are generated as
the process continues to 10,000
trials.
Notice the proportion of all of the 10,000 samples is
0.50 (exactly the same as the population
proportion) and the shape of the sampling
distribution is approx normal.
In General, the mean of the population
= mean of the sample and the shape of
the sample is approx normal.
In General, the population variance = the
sample variance and the shape of the
sampling distribution is skewed.
In General, the population proportion = the
sample proportion and the shape of the
sampling distribution is approx normal.
Unbiased Estimators…
• The preceding 3 examples show that the sample
means, variances, and proportions tend to target
the corresponding population parameters.
• We call these 3 statistics unbiased estimators.
• In other words, their sampling distributions have
a mean that is equal to the mean of the
corresponding population parameters.
• If we want to use a sample statistic
(such as a sample proportion) to estimate
a population parameter (such as the
population proportion), it is important that
the sample statistic used as the estimator
targets the population parameter.
• Using a biased estimator might
underestimate or overestimate the
population parameter.
• A good sample should be less
variable and unbiased.
• Variability – measured by the
standard error of the mean, which
requires a larger sample. (Bigger
sample size, smaller variation)
UNBIASED ESTIMATORS: Can
be called point estimates.
• These target the population
parameters.
Mean x
2
Variance s
proportion pˆ
BIASED ESTIMATORS:
• These do NOT target the population parameter.
Median
Range
STANDARD DEVIATION , s
• NOTE: The bias is relatively small when using s,
therefore, it is sometimes used to estimate the
standard deviation of the population.
Price is Right – Range Game
• Credit for the idea of the Price is Right clip goes to
Summer Abney, our fellow reader.
• Watch the clip of the Price Is Right Range Game on you
tube.
– Students watch the directions for the game. Pause
video.
– What should the strategy be?
– Watch conclusion of clip.
– https://www.youtube.com/watch?v=00a4pPyIwes
Important Points…….
• Sampling mean is unbiased since it
equals the population mean.
• Each individual sample may not
equal the population mean, but it
should be close.
• When we say 95% of the data is
within 2 st. deviations we mean 95%
of data values are   2 .
Speaking specifically about
the mean…..
• When speaking of samples we mean
that 95% of the intervals captured
will contain the population mean.
Definitions…….
• Interval Estimate – An interval bounded
by two values & used to estimate the
value of a population parameter.
Statisticians prefer to use an interval
estimate rather than a point estimate.
• Level of Confidence –(1   ); the
proportion of all interval estimates that
involve the parameter being estimated.
• Confidence Interval – An interval
estimate with a specified level of
confidence.
What does the answer
mean?
• If the answer of a confidence interval
is (918.23, 930.23), and the
confidence level was 95%, then it
means that I captured the population
mean of the samples 95% of the
time.
EXAMPLE
• The next figure shows typical
confidence intervals resulting from
20 different samples. With 95%
confidence, we expect that 19 out
of 20 samples should result in
confidence intervals that contain the
true value of p=0.75.
1 Confidence Interval did NOT have the predicted
proportion. Therefore, 19 out or 20 did.
Remember…..
• Sample measures are called
statistics.
• Population measures are called
parameters.
Using 6, 8, 12, 16, 18, 21,
22, 25, 28, 29…….
• Find a point
estimate for
a. The mean
x  18.5
b. Variance
s  64.06
c. St. Deviation
s  8.0
2
Use the correct symbols!!!!!
Confidence
Interval for the
Mean – Pop. St.
Dev. KNOWN
Z-Interval
Z Confidence Interval……
• Use When:
a.
Sampling Distribution is normal
OR
b.
Population standard deviation is known OR
the sample size is greater than or equal to
30. ( n  30 )
Confidence Interval for
Estimation of the Mean……
Pt. Estimate  Confidence Coefficient ( St. Error of the Mean)
x  z (
2

n
)
This produces the lower and upper confidence levels.
Definition……
• Confidence Level: the probability
that the interval estimate will contain
the parameter.
• The most common levels of
confidence are: 90%, 95%,
99%
Alpha (  )……
• Alpha (  ) = the total area in both
tails.
• Alpha/2 (  2 ) = the total area in one
tail.
• Example:
Confidence Level = 90%
 = 1 - .90 = .10
 = .10/2 = 0.05
2
Example……
• The president of a university
wants to estimate the average
age of students. It is known that
the standard deviation is 2 years.
A sample of 50 is selected and
the mean age is found to be 23.2
years. Find the 95% confidence
interval.
Answer……
• This is what the distribution looks like with
a 95% confidence level.
• Remember:
  1  .95  .05
  .05  .025
2
z
2
  
x  z 

2 n 
2
2 

 invnorm(1  .025)  1.96 23.2  1.96 

n  50,   2, x  23.2
 50 
Answer : 22.65,23.75
What does this answer
mean?......
• “We can say with 95% confidence
that the average age of students at
the university is between 22.65
years and 23.75 years.”
• Note: A 95% confidence interval
would be a wider interval than a
90% confidence interval. Why do
you think this is?
Example……
• A certain medication increases the
pulse rate. The variance is 25
beats/minute. A sample of 30 users
has an average rate of 104
beats/minute. Find the 99%
confidence interval.
Answer……
• Values:
n  30
 2  25 So,   25  5
x  104
  1  0.99  0.01
  0.01  0.005
2
2
z  invnorm(1  0.005)  2.58
2
  
x  z 

2
n
 5 
104  2.58

 30 
Answer :101.64,106.36
Example……
• A sample of 50 days showed a
store served an average of 182
customers. The standard
deviation was 8. Find the 90%
confidence interval.
Answer……
• Values:
n  50
 8
x  182
  1  0.90  0.10
  0.10  0.05
2
z
2
2
 invnorm(1  0.05)  1.64
  
x  z 

2
n
 8 
182  1.64

 50 
Answer :180.14,183.86