Transcript Confidence Intervals

Chapter 8
Confidence Intervals
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.
Confidence Intervals
8.1
z-Based Confidence Intervals for a
Population Mean: σ Known
8.2
t-Based Confidence Intervals for a
Population Mean: σ Unknown
8.3
Sample Size Determination
8.4
Confidence Intervals for a Population
Proportion
8.5
Confidence Intervals for Parameters of
Finite Populations (Optional)
8.6
A Comparison of Confidence Intervals
and Tolerance Intervals (Optional)
8-2
z-Based Confidence Intervals for a
Mean: σ Known
• The starting point is the sampling
distribution of the sample mean
– Recall that if a population is normally
distributed with mean m and standard
deviation σ, then the sampling distribution
of x is normal with mean mx = m and
standard deviation  x   n
– Use a normal curve as a model of the
sampling distribution of the sample mean
• Exactly, because the population is normal
• Approximately, by the Central Limit Theorem
for large samples
8-3
The Empirical Rule
• 68.26% of all possible sample means
are within one standard deviation of the
population mean
• 95.44% of all possible observed values
of x are within two standard deviations
of the population mean
• 99.73% of all possible observed values
of x are within three standard deviations
of the population mean
8-4
Example 8.1: The Car Mileage Case
• Assume a sample size (n) of 5
• Assume the population of all individual
car mileages is normally distributed
• Assume the population standard
deviation (σ) is 0.8

0.8
x 

 0.358
n
5
• The probability that x with be within ±0.7
of µ is 0.9544
8-5
Example 8.1 The Car Mileage Case
Continued
• Assume the sample mean is 31.3
• That gives us an interval of [31.3 ± 0.7]
= [30.6, 32.0]
• The probability is 0.9544 that the
interval [x ± 2σ] contains the population
mean µ
8-6
Example 8.1: The Car Mileage Case
#3
• That the sample mean is within ±0.7155 of
µ is equivalent to…
• x will be such that the interval [x ± 0.7115]
contains µ
• Then there is a 0.9544 probability that x
will be a value so that interval [x ± 0.7115]
contains µ
– In other words
P(x – 0.7155 ≤ µ ≤ x + 0.7155) = 0.9544
– The interval [x ± 0.7115] is referred to as
the 95.44% confidence interval for µ
8-7
Example 8.1: The Car Mileage Case
#4
Three 95.44% Confidence Intervals for m
• Thee intervals
shown
• Two contain µ
• One does not
8-8
Example 8.1: The Car Mileage Case
#5
• According to the 95.44% confidence interval,
we know before we sample that of all the
possible samples that could be selected …
• There is 95.44% probability the sample mean
is such the interval [x ± 0.7155] will contain
the actual (but unknown) population mean µ
– In other words, of all possible sample means,
95.44% of all the resulting intervals will contain the
population mean µ
– Note that there is a 4.56% probability that the
interval does not contain µ
• The sample mean is either too high or too low
8-9
Generalizing
• In the example, we found the probability that
m is contained in an interval of integer
multiples of x
• More usual to specify the (integer) probability
and find the corresponding number of x
• The probability that the confidence interval
will not contain the population mean m is
denoted by 
– In the mileage example,  = 0.0456
8-10
Generalizing Continued
• The probability that the confidence interval
will contain the population mean m is
denoted by 1 - 
– 1 –  is referred to as the confidence
coefficient
– (1 – )  100% is called the confidence level
• Usual to use two decimal point probabilities
for 1 – 
– Here, focus on 1 –  = 0.95 or 0.99
8-11
General Confidence Interval
• In general, the probability is 1 –  that the
population mean m is contained in the interval
x  z 2 x 

  x  z

2
 

n
– The normal point z/2 gives a right hand tail area
under the standard normal curve equal to /2
– The normal point - z/2 gives a left hand tail area
under the standard normal curve equal to /2
– The area under the standard normal curve
between -z/2 and z/2 is 1 – 
8-12
Sampling Distribution Of All Possible
Sample Means
8-13
z-Based Confidence Intervals for a
Mean with σ Known
• If a population has standard deviation 
(known),
• and if the population is normal or if
sample size is large (n  30), then …
• … a (1-)100% confidence interval
for m is
x  z

  x - z
n 

2

2
n
, x  z
2
 

n
8-14
95% Confidence Level
• For a 95% confidence level, 1 –  = 0.95, so
 = 0.05, and /2 = 0.025
• Need the normal point z0.025
– The area under the standard normal curve
between -z0.025 and z0.025 is 0.95
– Then the area under the standard normal curve
between 0 and z0.025 is 0.475
– From the standard normal table, the area is 0.475
for z = 1.96
– Then z0.025 = 1.96
8-15
95% Confidence Interval
• The 95% confidence interval is

 
x  z0.025 x    x  1.96 
n



 
  x - 1.96
, x  1.96

n
n

8-16
99% Confidence Interval
• For 99% confidence, need the normal
point z0.005
– Reading between table entries in the
standard normal table, the area is 0.495 for
z0.005 = 2.575
• The 99% confidence interval is

 
x  z0.025 x    x  2.575 
n



 
  x - 2.575
, x  2.575

n
n

8-17
The Effect of a on Confidence Interval
Width
z/2 = z0.025 = 1.96
z/2 = z0.005 = 2.575
8-18
Example 8.2: Car Mileage Case
• Given
– x = 31.56
– σ = 0.8
– n = 50
• Will calculate
– 95 Percent confidence interval
– 99 Percent confidence interval
8-19
Example 8.2: 95 Percent Confidence
Interval
x  z0.025

0.8
 31.56  1.96
n
50
 31.56  0.222
 31.34, 31.78
8-20
Example 8.2: 99 Percent Confidence
Interval
x  z0.005

0.8
 31.56  2.575
n
50
 31.56  0.294
 31.27, 31.85
8-21
Notes on the Example
• The 99% confidence interval is slightly wider
than the 95% confidence interval
– The higher the confidence level, the wider the
interval
• Reasoning from the intervals:
– The target mean should be at least 31 mpg
– Both confidence intervals exceed this target
– According to the 95% confidence interval, we can
be 95% confident that the mileage is between
31.33 and 31.78 mpg
• We can be 95% confident that, on average, the mean
exceeds the target by at least 0.33 and at most 0.78 mpg
8-22
t-Based Confidence Intervals for a
Mean:  Unknown
• If  is unknown (which is usually the
case), we can construct a confidence
interval for m based on the sampling
distribution of
t
x -m
s
n
• If the population is normal, then for any
sample size n, this sampling distribution
is called the t distribution
8-23
The t Distribution
• The curve of the t distribution is similar to
that of the standard normal curve
– Symmetrical and bell-shaped
– The t distribution is more spread out than
the standard normal distribution
– The spread of the t is given by the number
of degrees of freedom
• Denoted by df
• For a sample of size n, there are one fewer
degrees of freedom, that is,
df = n – 1
8-24
Degrees of Freedom and the
t-Distribution
As the number of degrees of freedom increases, the spread
of the t distribution decreases and the t curve approaches
the standard normal curve
8-25
The t Distribution and Degrees of
Freedom
• As the sample size n increases, the
degrees of freedom also increases
• As the degrees of freedom increase, the
spread of the t curve decreases
• As the degrees of freedom increases
indefinitely, the t curve approaches the
standard normal curve
– If n ≥ 30, so df = n – 1 ≥ 29, the t curve is
very similar to the standard normal curve
8-26
t and Right Hand Tail Areas
• Use a t point denoted by t
– t is the point on the horizontal axis under
the t curve that gives a right hand tail equal
to a
– So the value of t in a particular situation
depends on the right hand tail area a and
the number of degrees of freedom
• df = n – 1
•  = 1 – a , where 1 – a is the specified
confidence coefficient
8-27
t and Right Hand Tail Areas
8-28
Using the t Distribution Table
• Rows correspond to the different values of df
• Columns correspond to different values of a
• See Table 8.3, Tables A.4 and A.20 in
Appendix A and the table on the inside cover
– Table 8.3 and A.4 gives t points for df 1 to 30, then
for df = 40, 60, 120 and ∞
• On the row for ∞, the t points are the z points
– Table A.20 gives t points for df from 1 to 100
• For df greater than 100, t points can be approximated by
the corresponding z points on the bottom row for df = ∞
– Always look at the accompanying figure for
guidance on how to use the table
8-29
Using the t Distribution
• Example: Find t for a sample of size
n=15 and right hand tail area of 0.025
– For n = 15, df = 14
–  = 0.025
• Note that a = 0.025 corresponds to a
confidence level of 0.95
– In Table 8.3, along row labeled 14 and
under column labeled 0.025, read a table
entry of 2.145
– So t = 2.145
8-30
Using the t Distribution
Continued
8-31
t-Based Confidence Intervals for a
Mean:  Unknown
• If the sampled population is normally
distributed with mean m, then a (1)100%
confidence interval for m is
x  t
s
2
n
• t/2 is the t point giving a right-hand tail area
of /2 under the t curve having n1 degrees of
freedom
8-32
Example 8.4 Debt-to-Equity Ratios
• Estimate the mean debt-to-equity ratio of the
loan portfolio of a bank
• Select a random sample of 15 commercial
loan accounts
– x = 1.34
– s = 0.192
– n = 15
• Want a 95% confidence interval for the ratio
• Assume all ratios are normally distributed but
σ unknown
8-33
Example 8.4 Debt-to-Equity Ratios
Continued
• Have to use the t distribution
• At 95% confidence, 1 –  = 0.95 so  = 0.05
and /2 = 0.025
• For n = 15, df = 15 – 1 = 14
• Use the t table to find t/2 for df = 14, t/2 =
t0.025 = 2.145
• The 95% confidence interval:
x  t0.025
s
0.192
 1.343  2.145
n
15
 1.3433  0.1064  1.2369,1.4497
8-34
Sample Size Determination (z)
If  is known, then a sample of size
 z 2 

n  

B


2
so that x is within B units of m, with
100(1-)% confidence
8-35
Sample Size Determination (t)
If σ is unknown and is estimated from s,
then a sample of size
 t 2 s 

n  
 B 
2
so that x is within B units of m, with
100(1-)% confidence. The number of
degrees of freedom for the t/2 point is
the size of the preliminary sample
minus 1
8-36
Example 8.6: The Car Mileage Case
• What sample size is needed to make
the margin of error for a 95 percent
confidence interval equal to 0.3?
• Assume we do not know σ
• Take prior example as preliminary
sample
– s = 0.7583
– z/2 = z0.025 = 1.96
– t/2 = t0.025 = 2.776 based on n-1 = 4 d.f.
8-37
Example 8.6: The Car Mileage Case
Continued
 t / 2 s   2.7760.7583 
n
 
  49.24  50
0.3

 E  
2
2
8-38
Example 8.7: The Car Mileage Case
• Want to see that the sample of 50 mileages
produced a 95 percent confidence interval
with a margin or error of ±0.3
s  
0.798 

 x  t0.025
  31.56  2.010

n 
50 

 31.56  0.227  31.33,31.79
• The margin of error is 0.227, which is smaller
than the 0.3 desired
8-39