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Introduction to Probability and Statistics
Chapter 7Ammar M. Sarhan,
[email protected] of Mathematics and Statistics,
Dalhousie UniversityFall Semester 2008
Chapter 7
Statistical IntervalsBased on a
Single Sample
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Confidence Intervals
An alternative to reporting a single value for the parameter being estimated is to calculate and report an entire interval of plausible values – a confidence interval (CI). A confidence level is a measure of the degree of reliability of the interval.
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7.1 Basic Properties of Confidence Intervals
If after observing X1 = x1,…, Xn = xn, we compute the observed sample mean , then a 95% confidence interval for μ can beexpressed as
⎟⎠
⎞⎜⎝
⎛ ⋅+⋅−n
xn
x σσ 96.1,96.1
x
The population has a N(μ, σ2) and σ is known.
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Other Levels of Confidence
A (1- α)100% confidence interval for the mean μ of a normal populationwhen the value of is known α is given by
⎟⎠
⎞⎜⎝
⎛ ⋅+⋅−n
zxn
zx σσαα 2/2/ ,
Other Levels of Confidence
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Sample Size
The general formula for the sample size n necessary to ensure an interval width w is
2
2/2 ⎟⎠⎞
⎜⎝⎛ ⋅=
wzn σα
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Deriving a Confidence Interval
Let X1,…, Xn denote the sample on which the CI for the parameter θ is to be based. Suppose a random variable satisfying the following properties can be found:
1. The variable depends functionally on both X1,…,Xn and θ.2. The probability distribution of the variable does not depend on θ
or any other unknown parameters.
For any α, 0 < α < 1, constants a and b can be found to satisfy
( ) αθ −=<< 1);,,( 1 bXXhaP nL
θ̂
Let h(X1,…, Xn; θ) denote this random variable. In general, the form of h is usually suggested by examining the distribution of an appropriate estimator .
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( ) αθ −=<< 1),,(),,( 11 nn XXuXXlP LL
Now suppose that the inequalities can be manipulated to isolate θ
lower confidencelimit
upper confidencelimit
For a 100(1- α)% CI.
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Examples: 7.5 p. 260. will be given in the class.
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7.2 Large-Sample Confidence Intervals for a Population Mean and ProportionLarge-Sample Confidence Interval
Let X1,…, Xn denote the sample from a population having a mean μ and standard deviation σ. If n is sufficiently large, then
This implies that
is a large-sample confidence interval for μ with level 100(1- α)%.
)1,0(~/
Nn
XZσ
μ−=
⎟⎠⎞
⎜⎝⎛ ⋅+⋅−
nzX
nzX σσ
αα 2/2/ ,
This formula is valid regardless of the shape of the population distribution.
For practice: n > 40.
( )nNX 2,~ σμ
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Examples: 7.6 p. 264.Suppose a random sample with size 48 from a population with unknown mean μ and unknown variance σ2 with the following information:
( )2.56,2.534823.596.17.54,
4823.596.17.54 =⎟
⎠
⎞⎜⎝
⎛ +−
23.5and7.54 == sx
950,144and,2626 2 == ∑∑ ii xx
Find the 95% confidence interval of μ?Solution: From the information given in the problem, we have:
Then, the 95% confidence interval of μ is
That is, with a confidence level of approximation 95%, 2.562.53 << μ
⎟⎠⎞
⎜⎝⎛ ⋅+⋅−
nszX
nszX 2/2/ , αα
That is, Notice, if σ is unknown, replace it with the sample standard deviation s.
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Confidence Interval for a Population Proportion
Let p denote the proportion of “successes” in a population, where success identifies an individual or object that has a specified property. A random sample of n individuals is to be selected, and X is the number of successes in the sample. A confidence interval for a population proportion p with level 100(1- α)% is:
( ) ( )⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
+
+++
+
+−+
nzn
znqpz
nzp
nzn
znqpz
nzp
/14
ˆˆ2
ˆ,
/14
ˆˆ2
ˆ2
2/
2
22/
2/
22/
22/
2
22/
2/
22/
α
αα
α
α
αα
α
where, .ˆ1ˆ,ˆ pqnXp −==
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Notice, if the sample size is quite large, the approximate CI limits become
⎟⎟⎠
⎞⎜⎜⎝
⎛+−
nqpzp
nqpzp
ˆˆˆ,ˆˆˆ 2/2/ αα
Since is negligible compared to .p̂)2/(22/ nzα
Sample Size
The general formula for the sample size n necessary to ensure an interval width w is
2
22/ ˆˆ4
wqpzn α≈
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Examples: 7.8 p. 267.Suppose that in 48 trials in a particular laboratory, 16 resulted in ignition of a particular type of substrate by a lighted cigarette.
667.0,333.031
4816ˆ ===== q
nXp
Solution: n = 48 Let p denote the long-run proportion of all such trails that would result in ignition.
Then, the approximately 95% CI of p is
Find the 95% confidence interval of the long-run proportion of all such trails that would result in ignition?
( ) 48/96.11)48(4)96.1(
48)667.0)(333.0(96.1
)48(2)96.1(333.0
2
2
22
+
+±+
)474.0,217.0(08.1
139.0333.0=
±=
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Notice, the traditional 95%CI is
)466.0,200.0(48
)667.0)(333.0(96.1333.0 =±
Examples: 7.9 p. 267.Find the sample size necessary to ensure a width of 0.10 for the 95% confidence interval of the long-run proportion of all such trails that would result in ignition?
2
22/ ˆˆ4
wqpzn α≈
=4*(1.96) 2 * 0.333*0.667 / .01 = 341.305
n =341
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7.3 Intervals Based on a Normal Population Distribution
The population of interest is normal, so that X1,…, Xn constitutes a random sample from a normal distribution with both μ and σ2 unknown.
has a probability distribution called a t distribution with n-1 degrees of freedom (df).
t Distribution
Let X1,…, Xn ~ N(μ, σ2), then the rv
nSXT
/μ−
=
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Properties of t Distributions
Let tv denote the density function curve for v df.
1. Each tv curve is bell-shaped and centered at 0.
2. Each tv curve is spread out more than the standard normal (z) curve.
3. As v increases, the spread of the corresponding tv curve decreases.
4. As v →∞, the sequence of tv curves approaches the standard normalcurve (the z curve is called a t curve with df =∞)
0
t25 curve
t5 curve
z curve
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t Critical Value
Let tα,v = the number on the measurement axis for which the area under the t curve with v df to the right of tα,v is α. tα,v is called t critical value.
t critical value
Table A.5, p. 671, gives the t critical value for given α, v.
t0.025,15 =2.131, t0.05,22 =1.717 , t0.01,22 =2.508.
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⎟⎠
⎞⎜⎝
⎛ ⋅+⋅− −− nStX
nStX nn 1,2/1,2/ , αα
Now, let X1,…, Xn be a random sample from a normal distribution with both μ and σ2 unknown, then the (1- α)100% CI of μ is
or
nstX n ⋅± −1,2/α
t Confidence Interval
XHere, and S are the sample mean and sample variance.
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Examples: 7.12 p. 274.Consider the following sample of fat content (in percentage) on 10 randomly selected hot dogs:
25.2 21.3 22.8 17.0 29.8 21.0 25.5 16.0 20.9 19.5
Find the 95% confidence interval of the population mean fat content, assuming the population is normal.?Solution:
025.02/,134.4and90.21,10 ==== αsxnWe have:
Then, the 95% confidence interval of μ is
⎟⎠⎞
⎜⎝⎛ ⋅+⋅− −− 10
134.490.21,10134.490.21 110,025.0110,025.0 tt
⎟⎠⎞
⎜⎝⎛ ⋅+⋅−=
10134.4262.290.21,
10134.4262.290.21
( )86.24,94.18=
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Summary
X1,…, Xn ~ N(μ, σ2) σ is known n
zX σα ⋅± 2/
X1,…, Xn ~ N(μ, σ2) σ is unknown n
StX n ⋅± −1,2/α
X1,…, Xn ~ any population with mean μ and variance σ2
σ is known (n ≥ 30) nzX σα ⋅± 2/
σ is unknown (n ≥ 30) nSzX ⋅± 2/α
(1- α)100% CI of μThe random sample