Transcript Lecture 12 Confidence intervals Par 1x

Review of normal distribution
• If x is normally distributed with mean 
and standard deviation 𝜎, then with
probability 0.95, x will lie in the interval
[𝜇 − 1.96𝜎, 𝜇 + 1.96𝜎]
Exercise
Assume x is normal distributed with
mean𝜇 and standard deviation 𝜎.
Find an interval in which x is
guaranteed to lie with probability
0.90.
Solution
Assume x is normal distributed with mean and
standard deviation 𝜎. Find an interval in which x is
guaranteed to lie with probability 0.90.
We want 10% of the area under the curve to lie
outside this interval, leaving 5% to lie above the
interval. We therefore determine z so that the
area under the standard normal curve lying
between 0 and z equals 0.45 (a picture helps).
The result is z = 1.645.
Conclusion: With probability 0.90, the value of x
will lie in the interval [𝜇 − 1.645𝜎, 𝜇 + 1.645𝜎]
Review of sampling distributions
Review
Review of the Central Limit Theorem
Take ANY random variable X and compute  and s
for this variable. If samples of size n are randomly
selected from the population, then:
1) For large n, the distribution of the sample means,
will be approximately a normal distribution,
2) The mean of the sample means will be the
population mean  and
3) The standard deviation of the sample means will
be s
n
Restatement
The Sampling Distribution of X and
the Central Limit Theorem
Assume a population with  = 80, s = 6. If a
sample of 36 is taken from this population,
what is the probability that the sample mean
is larger than 82?
Sketch the curve
of x and identify
area of interest
The Sampling Distribution of X and
the Central Limit Theorem
Convert 82 to z value
First, calculate the standard deviation of the
sampling distribution
s
6
sx 

 1.0
n
36
Then calculate the z value
z
x  x
sx

82  80
2
1
Use the tables to find probability of interest


P x  82  P( z  2)  .5  .4772  .0228
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
What can you say about the mean balance
of ALL Visa accounts.
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
What can you say about the mean balance
of all VISA accounts?
ANS: That it is approximately $213. (This is
called a point estimate).
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213.
What can you say about the average balance
of all VISA accounts?
ANS: That it is approximately $213. (This is
called a point estimate).
We would LIKE an interval estimate, i.e., an
interval in which we will have confidence that
the true value of the population mean lies.
Example continued
Example continued
Interpretation
• If samples of size 100 were taken
respectively, then with probability 0.95
(i.e., 19 times out of 20), the true value of
the population mean would be included in
the interval constructed using the
procedure outlined above.
Possible was in which this would
be stated in a news report
A recent survey of 100 Visa accounts
showed that the average Visa account is
$213. This finding is valid with a margin
of error of $21.95 with 95% confidence.
This means that if the survey were
repeated many times, 19 times out of 20
the true mean would lie in the interval
obtained.
General Confidence Intervals
A confidence interval (or interval estimate)
is a range of values that estimates the true
value of the population parameter.
This is associated with a degree of
confidence, which is a measure the
probability that a randomly selected
confidence interval encloses the population
parameter.
Confidence Level
The confidence level is equal to 1-  , and is
split between the two tails of the distribution
Usually the confidence level is:
90% (meaning   .10 )
95% (meaning   .05 ) OR
99% (meaning   .01 )
Confidence Intervals
The Confidence Interval is often expressed
as:
xE
x  z 2s x  x  z
E is called the margin of error.
For samples of size > 30,
 s 
x  z 2 

 n
s
2
n
Typical values for 𝑧𝛼/2
 95% confidence interval
 𝛼=
𝛼
0.05,
2
= 0.025, 𝑧.025 = 1.96
 𝜇 = 𝑥 ± 1.96
𝜎
𝑛
 90% confidence interval
 𝛼=
𝛼
0.10,
2
= 0.05, 𝑧.05 = 1.645
 𝜇 = 𝑥 ± 1.645
𝜎
𝑛
NOTE
If you increase the confidence level, the
size of the confidence interval increases.
This makes sense. If the a sample of 100
students is taken and the mean age
computed, I would probably be very
confident the mean age is between 18 and
24, but less confident the mean age is
between 20 and 22.
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 95% confidence
interval.
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 95% confidence
interval.
213 21.95
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 90% confidence
interval.
 112 
213  1.645

 100 
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 90% confidence
interval.
 112 
213  1.645
  213  18.42
 100 
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 99% confidence
interval.
 112 
213  2.575

 100 
Example
Suppose you took a sample of 100 accounts
from Visa and found the mean balance was
$213. If in addition you know the standard
deviation for all accounts is $112, what can
you say about the average balance of all
VISA accounts? Build a 99% confidence
interval.
 112 
213  2.575
  213  28.84
 100 
Age of STFX Students
Find the 90% confidence interval for the
mean.
x  21.4,
s  2.6,
n  50   0.10
Age of STFX Students
Find the 90% confidence interval for the
mean.
x  21.4,
x  z s
2
21.4  z0.05
s  2.6,
n  50   0.10
   x  z s
n
2.6
20.8    22.0
2
50
n
   21.4  z0.05 2.6
50
Sample Size
The sample size needed to estimate  so as
to be (1-)*100 % confident that the sample
mean does not differ from  more than E is:
 z 2s
n  
 E
…round up



2