Transcript Estimating the Population Mean

Estimating the Population Mean
Assumptions
1. The sample is a simple random sample
2. The value of the population standard deviation
(σ) is known
3. Either the population is normally distributed or
n > 30
The sample mean x is the best point estimate for
the population mean.
Confidence Interval estimate of the
Population Mean μ (with σ known)
E  z 2

n
xE  xE
xE
 x  E, x  E 
Example
Suppose an insurance company studies repair
costs after rear collisions, and finds the
mean repair cost to be $2300 based on a
sample of 40 accidents. Suppose the
standard deviation is $1025.
Find the 95% Confidence Interval
x  2300
z 2  1.96
Example
E  z 2

1025
 1.96
 317.65
n
40
x  E  $2300  $317.65  $1982.35
x  E  $2300  $317.65  $2617.65
So our confidence interval is:
$1982.4 < μ < $2617.7
Example
$1982.4 < μ < $2617.7
We are 95% confident that the population mean
repair cost is contained in this interval.
Problem
Usually σ is not known
Solution:
Use s (the sample standard deviation) to approximate σ.
Since this is not as accurate, we can no longer use our
good friend the normal distribution. Now we’re going
to need the Student’s t-distribution
Estimating the Population Mean
(σ unknown)
Assumptions:
1. The sample is a simple random sample
2. Either the sample is from a normally
distributed population or n > 30
The sample mean x is the best point estimate
for the population mean.
CI for the Population Mean
(σ unknown)
E  t 2
s
n
Where tα/2 has n-1 degrees
of freedom
Looking up critical t-values from the table
Since our confidence level is centered in the distribution, α is
the area in the two tails (far right and far left sides). So in the
table:
Find the column listing for α in the Area in Two Tails
Find the row (degrees of freedom) that is the closest to n-1
Example
Suppose an insurance company studies repair
costs after rear collisions, and finds from a
sample of 28 accidents the mean repair cost
to be $2300 and the standard deviation to be
$1025.
Find the 95% Confidence Interval
x  2300
Example
27 degrees of freedom, 0.05 area in two tails:
tα/2 = 2.052
E  t 2
s
1025
 2.052
 $397.49
n
28
So the 95% confidence interval is:
$2300  $397.5
Estimating Sample Size
 z 2   
n

 E 
2
For sigma not known:
range
• Estimate using the range rule of thumb  
4
• Do pilot study
• Estimate value using previous study
Example
Suppose you were testing whether the mean
GPA of a group of students is greater than
2.0. What should your sample size be?
Depends on what we want the margin of error
to be. Suppose we wanted to be 95%
confident that the sample mean is within 0.1
of the population mean.
Example
Suppose we wanted to be 95% confident that
the sample mean is within 0.1 of the
population mean.
z 2  1.96
4.0  0.0
1
σ unknown, estimate:  
4
1.96 1
n

384

0
.
1


2
We would need a
sample size of 384
Homework
6.3: 9, (11), 13, 17, 19, (21)
6.4: 1, 5, 9, 13, (19)