Confidence Interval Estimation
Confidence Intervals on
 and p
• An interval estimator is a formula that tells us
how to use sample data to calculate an interval
that estimates a population parameter.
• The confidence coefficient is the probability that
an interval estimator encloses the population
parameter.
• The confidence level is the confidence
coefficient expressed as a percentage.
Normal Distribution for the Mean
Distribution
Revisited
Useful Useful
Probabilities
for Normal
Distributions
68%
95%
99%







• Confidence intervals assume that the sample means
3
are normally distributed.
Confidence Intervals
• A 95% confidence interval represents a range of
values within which you are 95% certain that the true
population mean exists.
– One interpretation is that if 100 different samples
were drawn from the same population and 100
intervals were calculated, approximately 95 of them
would contain the population mean.
4
Confidence Intervals Sometimes Miss
5
Large-Sample Confidence
Interval for 
x  z  x
2
z
• where 2 is the z value with an area 
to its
2
right and    n
• The parameter  is the standard deviation of
the sampled population and n is the sample size.
• When  is unknown (most cases) and n is
large, the value of  can be approximated by
the sample standard deviation, s.
x
Commonly Used Values of
z
2
Confidence Level
100(1   )
90%
95%
99%

.10
.05
.01

2
z
2
.05 1.645
.025 1.96
.005 2.575
Example
• Suppose you sampled 400 college
students to determine the average soft
drink consumption of college students.
The results are a sample mean of 20.1
and a sample standard deviation of 0.5.
What is the 95% CI for the average soft
drink consumption for college students?
Interpretation of a Confidence
Interval for a Population Mean
• We can be 100(1   )% that  lies
between the lower and upper bounds of
the confidence interval.
• The statement reflects our confidence in
the estimation process rather than in the
particular interval that is calculated from
sample data.
Example
• A tire manufacturer is testing a new compound for tread
wear. Tires made with the compound are placed on a
machine that simulates road wear, and the amount of
tread left after the equivalent of 40,000 road miles is
recorded. The mean amount of tread left after a test of
10 tires was 3.6 mm. Construct a confidence interval for
the population mean amount of tread left after 40,000
miles at the 99% level of confidence. Assume the
population distribution is normally distributed with a
standard deviation of 0.2 mm.
Answer
 0.2 
L, U  3.6  2.57

 10 
L  3.44, U  3.76
Example
• An automobile dealer plans to order enough
cars to have a 90-business-day inventory. Over
the past 30 business days the dealer has sold
an average of 20 cars a day. Construct a
Confidence Interval for the population mean
number of cars sold per day at the 95% level of
confidence. Assume that the distribution of cars
sold per day is approximately normal with a
standard deviation of 5 cars.
Answer
 5 
L, U  20  196
. 

 30 
L  18.21,U  2179
.
Small-Sample Estimation of a
Normal Probability
• The sample standard deviation s may
provide a poor approximation of the
population standard deviation when the
sample size is small.
• In these cases the we use the t-statistic
rather than the z-statistic.
t-statistic
• Formula -
x
t 
s
n
• t-statistic is more variable than the zstatistic. The variability depends upon the
sample size, n.
• The t-statistic has (n-1) degrees of
freedom (df).
Small-Sample Confidence
Interval for 
• Formula -
x  t
2
s
n
• The confidence interval using the t-statistic
is wider than the corresponding
confidence interval using the standard zstatistic.
• Curves:
Example
• A sample of monthly sales for 20 Circle K
convenience stores shows mean sales to
be $30,000 with a standard deviation of
$16,000:
– Calculate a 95% confidence interval for
monthly sales using a t-statistic.
– Calculate a 95% confidence interval for
monthly sales using a z-statistic.
Answer FIX THE NOTES
L,U  x  t sx
2
 16,000 
L,U  30,000  2.093

 20 
L  $22,512,U  $37,488
L,U  x  z  x
2
 16,000 
L,U  30,000  1.96

 20 
L  $22,987,U  $37,012
Large-Sample Estimation of a
Binomial Probability
• Often we want to the know the proportion
of a population such as the proportion of
smokers, proportion of smokers that prefer
a specific brand, proportion of viewers that
remember a commercial, etc.
• How do we estimate p, the proportion of
successes in the sample?
Estimation of p
• One logical answer is to calculate p as
x
follows: p  n
– where
– x = number of successes
– n = number of trials
 is an unbiased estimator of p.
– p
• The standard deviation of the sampling
distribution of p is   pq

p
n
Large-Sample Confidence
Interval for p
ˆ  z  pˆ
p
2
• Where:
ˆ 
p
x
n
ˆ  1 p
ˆ
q
• When n is large, we can use p to
approximate the value of p in the formula
for  p .
Example
• A sample of 500 consumers chosen at
random shows that 265 are optimistic
about the state of the economy:
– Calculate a 90% confidence interval to
estimate the proportion of all consumers that
are optimistic about the state of the economy.
Answer
x 265

 0.53
n 500
qˆ  1  pˆ  1  0.53  0.47
pˆ 
L,U  pˆ  z  pˆ
2

L,U  0.53  1.645 0.53 * 0.47
500 

L  0.493,U  0.567
Determining Sample Size:
Population Mean Estimation
4( z ) 2 
n 
2
2
W
2
• Where:
– W is the desired width of the confidence interval.
– The population standard deviation must be
estimated.
– n is rounded up to ensure that the sample size will
be sufficient to achieve the desired confidence
interval.
Example
• Footballs inflated to a mean pressure of
13.5 pounds:
– Due to machine calibration, individual
footballs vary in pressure from 13.3 to 13.7
pounds.
– What sample size is necessary for a 99%
confidence interval that is only 0.05 wide?
4( z ) 2  2
n
2
W2
( 4)( 2.575) 2 (0.1) 2

 106.09
2
(0.05)
Determining Sample Size:
Binomial Probability
4( z ) 2 ( pq )
n 
2
W
2
• Where:
– W is the desired width of the confidence interval.
– The population standard deviation is estimated by
p*q.
– n is rounded up to ensure that the sample size will
be sufficient to achieve the desired confidence
interval.
Example
• The probability that a consumer will feel
optimistic about the economy is 0.53:
– What sample size is necessary for a 90%
confidence interval that is only 0.05 wide?
4( z ) 2 ( pq)
n
2
W2
4(1.645) 2 (0.53)( 0.47)

 1078.5
2
(0.05)
Confidence Interval Review
• Large-Sample Confidence Interval:
x  z  x
2
• Small-Sample Confidence Interval:
s
n
x  t
2
• Large-Sample Binomial Confidence Interval:
ˆ  z
p
2
ˆ qˆ
p
n
Commonly Used Values of
z
2
Confidence Level
100(1   )
90%
95%
99%

.10
.05
.01

2
.05
.025
.005
z
2
1.645
1.96
2.575
Example
• Example:
– We want to estimate the mean number of
unoccupied seats per flight for a major airline.
Specifically, we want to construct a 90%
confidence interval for the population mean.
– 225 flights sampled.
– Sample mean is 11.6 empty seats.
– Sample standard deviation is 4.1 seats.
Solution
• Large sample size so use z-statistic rather
than t-statistic.
x  z  x
2
• Formula:
– where z-value is at 10/2=5% level. Z=1.645.
– the population standard deviation is estimated by
the sample standard deviation.
4.1
x  z  x  11.6  1.645(
)  11.6  0.45  (11.15,12.05)
225
2
– We are 90% confident that the mean number of
unoccupied seats per flight (population mean) lies
between 11.15 and 12.05 seats.
Example
• Earnings per share example:
– We want to estimate the earnings per share of a type
of stock so we ask five portfolio analysts what their
projections are for the upcoming year. Specifically,
we want to construct a 95% confidence interval for
the mean projected earnings estimate for all analysts.
– 5 analysts sampled.
– Sample mean is $2.63 per share.
– Sample standard deviation is $0.72 per share.
Solution
• Small sample size so use t-statistic rather than zstatistic.
s
x  t
• Formula:
n
2
– where t-value is at 5/2=2.5% level and there are 5-1=4
degrees of freedom. t=2.776.
– the population standard deviation is estimated by the
sample standard deviation.
s
0.72
x  t
 2.63  2.776(
)  2.63  0.89  (1.74,3.52)
n
5
2
– We are 95% confident that the mean of all analysts’
earnings per share projections (population mean) for
this type of stock is between $1.74 and $3.52.
Example
• Smokers brand preference example:
– Philip Morris wants to determine the proportion of
smokers who prefer Marlboro. Specifically, they want
to construct a 95% confidence interval for the
proportion of smokers in the smoking population that
prefer Marlboro.
– 1000 smokers interviewed.
– X, the number of smokers (out of 1000 sampled) that
prefer Marlboro is a binomial random variable.
– 313 out of the 1000 smokers prefer Marlboro:
313
pˆ 
 0.313
1000
qˆ  1  pˆ  1  0.313  0.687
Solution
• Binomial distribution so use the following
formula:
ˆ qˆ
p
ˆ  z
p
2
n
– where z-value is at 5/2=2.5% level. Z=1.96.
– the population standard deviation is estimated by the
sample standard deviation in the formula above.
pˆ qˆ
0.313 * 0.687
pˆ  z
 0.313  1.96
 0.313  0.029  (0.284,0.342)
2
n
1000
– We are 95% confident that the proportion of smokers
(population proportion) that prefer Marlboro is
between 28.4% and 34.2%.
Determining Sample Size
• Population Mean Estimation:
4( z ) 2 
n 
2
2
W
2
• Binomial Probability:
4( z ) 2 ( pq )
n 
2
W
2
Population Mean Example
• We want to determine the sample size necessary
such that a 95% confidence interval for the mean
overdue amount for all delinquent accounts is
within $5 of the population mean:
– A previous sample of delinquent accounts shows a
standard deviation of $90.
– Note that information concerning mean is NOT
necessary.
4( z )2 2
2
2
n
2
W2
(4)(1.96) (90)

 1244.68
2
(10)
– Use 1245 as the sample size.
Binomial Probability Example
• The probability that a consumer will choose
brand X is 0.20. What sample size is necessary
for a 95% confidence interval that is only 0.08
wide?
4( z )2 ( pq)
n
2
W2
4(1.96)2 (0.2)( 0.8)

 384.16
2
(0.08)
– Use 385 as the sample size.