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Transcript of McGrawHill/IrwinCopyright © 2009 by The McGrawHill Companies, Inc. All Rights Reserved....
McGrawHill/Irwin
Copyright © 2009 by The McGrawHill Companies, Inc. All Rights Reserved.
Confidence Intervals
Chapter 8
82
Confidence Intervals
8.1 zBased Confidence Intervals for a Population Mean: σ Known
8.2 tBased 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)
83
zBased 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 and standard deviation σ, then the sampling distribution of is normal with mean = and standard deviation
– 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
nx
84
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
85
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
• The probability that with be within ±0.7 of µ is 0.9544
358.05
8.0
nx
86
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 [ ± 2σ] contains the population mean µ
87
Example 8.1: The Car Mileage Case #3• That the sample mean is within ±0.7155 of
µ is equivalent to…• will be such that the interval [ ± 0.7115]
contains µ
• Then there is a 0.9544 probability that will be a value so that interval [ ± 0.7115] contains µ– In other words
P( – 0.7155 ≤ µ ≤ + 0.7155) = 0.9544
– The interval [ ± 0.7115] is referred to as the 95.44% confidence interval for µ
88
Example 8.1: The Car Mileage Case #4
Three 95.44% Confidence Intervals for
• Thee intervals shown
• Two contain µ
• One does not
89
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 [ ± 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
810
Generalizing
• In the example, we found the probability that is contained in an interval of integer multiples of
• More usual to specify the (integer) probability and find the corresponding number of
• The probability that the confidence interval will not contain the population mean is denoted by– In the mileage example, = 0.0456
811
Generalizing Continued
• The probability that the confidence interval will contain the population mean is denoted by– 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
812
General Confidence Interval
• In general, the probability is 1 – that the population mean is contained in the interval
– 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 –
nzxzx x 22
813
Sampling Distribution Of All Possible Sample Means
814
zBased Confidence Intervals for aMean 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 is
nzx,
nzx
nzx 222
815
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
816
95% Confidence Interval
• The 95% confidence interval is
n.x,
n.x
n.xzx x.
961961
9610250
817
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
n.x,
n.x
n.xzx x.
57525752
57520250
818
The Effect of a on Confidence Interval Width
z/2 = z0.025 = 1.96 z/2 = z0.005 = 2.575
819
Example 8.2: Car Mileage Case
• Given
– = 31.56
– σ = 0.8
– n = 50
• Will calculate
– 95 Percent confidence interval
– 99 Percent confidence interval
820
Example 8.2: 95 Percent Confidence Interval
78.31,34.31
222.056.3150
8.096.156.31025.0
n
zx
821
Example 8.2: 99 Percent Confidence Interval
85.31,27.31
294.056.3150
8.0575.256.31005.0
n
zx
822
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
823
tBased Confidence Intervals for aMean: Unknown
• If is unknown (which is usually the case), we can construct a confidence interval for based on the sampling distribution of
• If the population is normal, then for any sample size n, this sampling distribution is called the t distribution
ns
xt
824
The t Distribution
• The curve of the t distribution is similar to that of the standard normal curve
– Symmetrical and bellshaped
– 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
825
Degrees of Freedom and thetDistribution
As the number of degrees of freedom increases, the spread of the t distribution decreases and the t curve approaches the standard normal curve
826
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
827
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
828
t and Right Hand Tail Areas
829
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
830
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
831
Using the t Distribution Continued
832
tBased Confidence Intervals for aMean: Unknown
• If the sampled population is normally distributed with mean , then a (1 )100% confidence interval for is
• t/2 is the t point giving a righthand tail area of /2 under the t curve having n 1 degrees of freedom
n
stx 2
833
Example 8.4 DebttoEquity Ratios
• Estimate the mean debttoequity ratio of the loan portfolio of a bank
• Select a random sample of 15 commercial loan accounts– = 1.34
– s = 0.192
– n = 15
• Want a 95% confidence interval for the ratio
• Assume all ratios are normally distributed but σ unknown
834
Example 8.4 DebttoEquity RatiosContinued
• 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:
4497.1,2369.11064.03433.115
192.0145.2343.1025.0
n
stx
835
Sample Size Determination (z)
If is known, then a sample of size
so that is within B units of , with 100(1)% confidence
22
B
zn
836
Sample Size Determination (t)
If σ is unknown and is estimated from s, then a sample of size
so that is within B units of , with 100(1)% confidence. The number of degrees of freedom for the t/2 point is the size of the preliminary sample minus 1
2
2
B
stn
837
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 n1 = 4 d.f.
838
Example 8.6: The Car Mileage Case Continued
5024.49
3.0
7583.0776.222
2/
E
stn
839
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
• The margin of error is 0.227, which is smaller than the 0.3 desired
79.31,33.31227.056.31
50
798.0010.256.31025.0
n
stx