Download Confidence Interval for µ - Department of Statistics and Probability

Caution!!!
• Unless n is extremely large, the large sample
procedure performs poorly when p is near 0
or near 1.
• For instance ; p= 0.001 and n=100 so that
np=0.1 <15
• Unless we have an extremely large n we can
not satisfy the sample size assumption.
• However, making minor adjustment on the
sample proportion can handle that problem.
Adjusted (1 – )100%
Confidence Interval for a
Population Proportion, p
p  z 2
x2
p
n4
p 1  p 
n4
the adjusted sample proportion of
observations with the characteristic of
interest, x is the number of successes in
the sample, and n is the sample size.
Example
• According to the Bureau of Labor Statistics
(2012), the probability of injury while working
at a jewelry store is less than 0.01. Suppose
that in a random sample of 200 jewelry store
workers, 3 were injured on the job. Estimate
the true proportion of jewelry store workers
who are injured on the job using 95%
confidence interval.
Example-Solution
• We are 95% confident that the true proportion of jewelry
store workers who are injured while on the job falls between
0.004 and 0.046.
• If we had applied the standard large-sample confidence
interval formula, we would have obtained the following
interval which has negative lower bound.
Sample Size Determination for 100(1 – ) %
Confidence Interval for p
In order to estimate p with a fixed margin of error and with
100(1 – )% confidence, the required sample size is found
by solving the following equation for n:
When true proportion is unknown, we use sample
proportion from a prior sample instead.
• The value of p(1-p) has its maximum when
p=0.5 or values close to 0.5.
• So, we can obtain conservatively large values
of n by approximating p by 0.5 or values close
to 0.5.
• In any case we should round the value of n
obtained upward to ensure that the sample
size will be sufficient to achive specified
reliability.
Example
• A cellular telephone manufacturer that entered the
postregulation market too quickly has an initial problem
with excessive customer complaints and consequent
returns of the cell phones for repair or replacement. The
manufacturer wants to determine the magnitude of the
problem in order to estimate its warranty liability.
• How many cellular telephones should the company
randomly sample from its warehouse and check in order to
estimate the fraction defective, p, to within 0.01 with 90%
confidence?
• Assume proportion of number of defective cellular phones
is 0.10.
Example-solution
6.2
Confidence Interval for a Population
Mean:
Normal (z) Statistic
Estimation Process
Population
Mean
x = 50
Mean, , is
unknown


Sample 




Random Sample





I am 95% confident
that  is between 40 &
60.
Confidence Interval
According to the Central Limit Theorem, the sampling
distribution of the sample mean is approximately normal
for large samples. Let us calculate the interval estimator:
1.96
x  1.96 x  x 
n
That is, we form an interval from 1.96 standard
deviations below the sample mean to 1.96 standard
deviations above the mean. Prior to drawing the sample,
what are the chances that this interval will enclose µ,
the population mean?
Confidence Interval
If sample measurements yield a value of x that falls
between the two lines on either side of µ, then the
interval x  1.96 x will contain µ.
The area under the
normal curve between
these two boundaries is
exactly .95. Thus, the
probability that a
randomly selected
interval will contain µ is
equal to .95.
95% Confidence Level
If our confidence level is 95%, then in the long run, 95% of
our confidence intervals will contain µ and 5% will not.
For a confidence coefficient of 95%, the area in the two tails
is .05. To choose a different confidence level we increase or
decrease the area (call it ) assigned to the tails. If we place
/2 in each tail
and z/2 is the z-value, the
confidence interval with
coefficient (1 – ) is
 
x  z 2  x .
Conditions Required for a Valid LargeSample
Confidence Interval for µ
1. A random sample is selected from the target
population.
2. The sample size n is large (i.e., n ≥ 30). Due to the
Central Limit Theorem, this condition guarantees
that the sampling distribution of x is approximately
normal. Also, for large n, s will be a good estimator
of .
Large-Sample (1 – )% Confidence Interval for µ
  
x  z 2  x  x  z 2 
 n 
where z/2 is the z-value with an area /2 to its right and
in the standard normal distribution. The parameter  is
the standard deviation of the sampled population, and n
is the sample size.
Note: When  is unknown and n is large (n ≥ 30), the
confidence interval is approximately equal to
 s 
x  z 2 
 n 
where s is the sample standard deviation.
 
Thinking Challenge
You’re a Q/C inspector for Gallo.
The  for 2-liter bottles is .05
liters. A random sample of 100
bottles showed x = 1.99 liters.
What is the 90% confidence
interval estimate of the true
mean amount in 2-liter bottles?
2 liter
2 liter
© 1984-1994 T/Maker Co.
Confidence Interval
Solution*
x  z /2 
1.99  1.645

n
.05
   x  z /2 

n
   1.99  1.645
100
.05
100
1.982    1.998
We can be 90% confident that the mean amount in 2-liter bottles
between 1.982 and 1.998. Our confidence is derived from the fact that
90% of the intervals formed in repeated applications of this procedure
would contain 
Exercise
• A random sample of 70 observations from a normally
distributed population possesses a sample mean equal
to 26.2 and a sample standard deviation equal to 4.1
• A) Find an approximate 95% confidence interval for .
• B) What do you mean when you say that a confidence
level is 95%?
• C) Find an approximate 99% confidence interval for .
• D) What happens to the width of a confidence interval
as the value of the confidence coefficient is increased
while the sample size is held fixed?
Thinking Challenge
• We have a random sample of customer order totals
with an average of $78.25 and a population standard
deviation of $22.5.
• A) Calculate a 90% confidence interval for the mean
given a sample size of 40 orders.
• B) Calculate a 90% confidence interval for the mean
given a sample size of 75 orders.
• C) Explain the difference in the 90% confidence
intervals calculated in A and B.
• Calculate the minimum sample size needed to identify
a 90% confidence interval for the mean assuming a $5
margin of error.