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```Chapter 8
Confidence Intervals on
 and p
Confidence Interval Estimation
Normal Distribution for the Mean
• 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.
Confidence Intervals
Distribution
Revisited
Useful Useful
Probabilities
for Normal
Distributions
68%
95%
99%







• Confidence intervals assume that the sample means
3
are normally distributed.
• 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
1
Chapter 8
Confidence Intervals Sometimes Miss
Large-Sample Confidence
Interval for 
x  z 
x
2
z
• where 2 is the z value with an area 2 to its
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
5
Commonly Used Values of
z

2
C o n fid e n c e L e v e l
1 0 0 (1   )

90%
95%
99%
.1 0
.0 5
.0 1

2
.0 5
.0 2 5
.0 0 5
z
2
1 .6 4 5
1 .9 6
2 .5 7 5
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?
2
Chapter 8
Interpretation of a Confidence
Interval for a Population Mean
• We can be 1 0 0 ( 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.
 0.2 
L,U  3.6  2.57

 10 
L  3.44,U  3.76
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
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.
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.
3
Chapter 8
 5 
L,U  20  196
. 

 30 
L  18.21,U  2179
.
t-statistic
• Formula -
t 
x  
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 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.
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:
4
Chapter 8
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.
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?
L,U  x  t s x
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
Estimation of p
• One logical answer is to calculate p as
follows: p  nx
– where
– x = number of successes
– n = number of trials
– p is an unbiased estimator of p.
• The standard deviation of the sampling
distribution of p is   p q
p
n
5
Chapter 8
Large-Sample Confidence
Interval for p
pˆ  z  
2
• Where:
pˆ 
x
n
pˆ
qˆ  1  pˆ
• 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.
Determining Sample Size:
Population Mean Estimation
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
4 ( z
n 

)
2

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.
6
Chapter 8
Example
Determining Sample Size:
Binomial Probability
• 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
n 
2

2
W
2
( 4 )( 2 . 575 ) 2 ( 0 . 1 ) 2
 106 . 09
( 0 . 05 ) 2
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?
2
W
2
)

2
( pq )
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.
Confidence Interval Review
• Large-Sample Confidence Interval:
x  z  x
2
• Small-Sample Confidence Interval:
4 ( z  ) 2 ( pq )
n 
4 ( z
n 

4 ( 1 . 645 ) 2 ( 0 . 53 )( 0 . 47 )
 1078 . 5
( 0 . 05 ) 2
s
n
x  t
2
• Large-Sample Binomial Confidence Interval:
pˆ  z 
2
pˆ qˆ
n
7
Chapter 8
Commonly Used Values of
z
Example
• Example:

2
C o n fid e n c e L e v e l
1 0 0 (1   )

90%
95%
99%
.1 0
.0 5
.0 1

2
.0 5
.0 2 5
.0 0 5
z
2
1 .6 4 5
1 .9 6
2 .5 7 5
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.
– 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.
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.
8
Chapter 8
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:
pˆ 
Solution
• Binomial distribution so use the following
formula:
pˆ qˆ
pˆ  z 
2
2
pˆ qˆ
0.313 * 0.687
 0.313  1.96
 0.313  0.029  (0.284,0.342)
n
1000
– We are 95% confident that the proportion of smokers
(population proportion) that prefer Marlboro is
between 28.4% and 34.2%.
qˆ  1  pˆ  1  0.313 0.687
Determining Sample Size
• Population Mean Estimation:
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  z
313
 0.313
1000
4 ( z
n 
)


2
2
2
2
W
• Binomial Probability:
4 ( z
n 
)

2
( pq )
2
W
2
9
Chapter 8
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
(10)2
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
W
2

4 (1 . 96 ) 2 ( 0 . 2 )( 0 .8 )
 384 . 16
( 0 . 08 ) 2
– Use 385 as the sample size.
– Use 1245 as the sample size.
10
```