8. Ch Condence Intervals
Ch 8. Condence Intervals Modules 33. Normal Mean CIs (variance known)
34. Normal Mean CIs (variance unknown)
35. CIs for Other Parameters
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8.33 Normal Mean CIs (var known)
33. Normal Mean Condence Intervals (variance known) Intro to CIs
CIs for Normal Mean (variance known)
Sample-Size Calculation
CIs for Dierence of Two Normal Means (vars known)
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8.33 Normal Mean CIs (var known)
Introduction to Condence Intervals Instead of estimating a parameter by a point estimator alone, give a (random) interval that contains the unknown parameter with a certain probability. Example: X is a point estimator for the parameter . A 95% condence interval for might look like X z.025 2/n, X + z.025 2/n ,
where z.025 is the Nor(0, 1)s 0.975 quantile. This means that is in the interval with probability 0.95.
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8.33 Normal Mean CIs (var known)
Denition: A 100(1)% condence interval for an unknown parameter is given by two random variables L and U satisfying Pr(L U ) = 1 . L is the lower condence limit (and is a RV). U is the upper condence limit (and is a RV). 1 is the condence coecient, specied in advance. There is a 1 chance that actually lies between L and U .
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8.33 Normal Mean CIs (var known)
Example: Were 95% sure that President Bushs popularity is 56% 3%. Since L U , we call [L, U ] a two-sided CI for . If L is such that Pr(L ) = 1 , then [L, ) is a 100(1 )% one-sided lower CI for . Similarly, if U is such that Pr( U ) = 1 , then (, U ] is a 100(1 )% one-sided upper CI for .
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8.33 Normal Mean CIs (var known)
Example: Here are some results from 10 independent samples, each consisting of 100 dierent observations. From each sample, we use the 100 obsns to re-calculate L and U . Is the unknown in [L, U ]?
Sample # L U CI covers ? 1 1.86 2.23 2 Yes 2 1.90 2.31 2 Yes 3 3.21 3.86 2 No 4 1.75 2.10 2 Yes 5 1.72 2.03 2 Yes . . . . . . . . 10 1.62 1.98 2 No
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8.33 Normal Mean CIs (var known)
As the number of samples gets large, the proportion of CIs that cover the unknown approaches 1 . Sometimes CIs miss too high, i.e., > U . Sometimes CIs miss too low, i.e., < L. But 1 of the time, theyre OK.
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8.33 Normal Mean CIs (var known)
CIs for Normal Mean (variance known) Set-up: Sample from a normal distribution with unknown mean and known variance 2. Goal: Obtain a CI for . Remark: This is an unrealistic case, since if we didnt know in real life, then we probably wouldnt know 2 either. But its a good place to start the discussion.
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8.33 Normal Mean CIs (var known)
Details. . . Suppose X1, . . . , Xn Nor(, 2), where 2 is known. Use X = X
n i=1 Xi/n iid
as our point estimator. Recall Z X 2/n Nor(0, 1).
Nor(, 2/n)
The quantity Z is called a pivot. point for us.
Its a starting
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8.33 Normal Mean CIs (var known)
The denition of Z implies that 1 = Pr z/2 Z z/2 = Pr z/2 X 2/n z/2
= Pr z/2 2/n X z/2 2/n = Pr X z/2 2/n X + z/2 2/n
L U
= Pr L U .
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8.33 Normal Mean CIs (var known)
Remarks: Notice how we used the pivot to isolate all by itself to the middle of the inequalities. After you observe X1, . . . , Xn, you calculate L and U . Nothing is unknown, since L and U dont involve . Sometimes well write the CI as X H, where the half-width is H z/2 2/n.
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8.33 Normal Mean CIs (var known)
Example: Suppose we take n = 25 i.i.d. obsns from a Nor(, distribution 2) where we somehow know that = 30. Further suppose that X turns out to be 278. Lets nd a 1 = 0.95 CI for . X z/2 2/n = 278 z.025(30/5) = 278 11.76 (z.025 = 1.96).
So a 95% CI for is 266.24 289.76.
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8.33 Normal Mean CIs (var known)
Sample-Size Calculation If we had taken more obsns, the CI would have gotten shorter, since H = z/2 2/n. In fact, how many obsns should be taken to make the half-length (or error) ? z/2 2/n i n (z/2/ )2.
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i 2/n ( /z/2)2
8.33 Normal Mean CIs (var known)
Example: Suppose, in the previous example, that we want the half-length to be 10, i.e., X 10. What should n be? n (z/2/ )2 =
2
(30)(1.96)/10
= 34.57.
Just to make n an integer, round up to n = 35.
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8.33 Normal Mean CIs (var known)
Remark: We can similarly obtain one-sided CIs for (if were just interested in one bound). . . 100(1 )% upper CI for : X + z 2/n. 100(1 )% lower CI for : X z 2/n. Note that we use the 1 quantile (not 1 /2).
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8.33 Normal Mean CIs (var known)
CIs for Di. of Two Normal Means (vars known) Idea: Compare the means of two competing alternatives or processes by getting a CI for their dierence. Example: Do Georgia Tech students have higher avg. IQs than Univ. of Georgia students? Answer: Yes. Well again assume that the variances of the two competitors are somehow known. Well do the more16
realistic unknown variance case in the next module.
8.33 Normal Mean CIs (var known)
Suppose we have samples of sizes nx and ny from the two competing populations. X1, X2, . . . , Xnx Y1, Y2, . . . , Yny
iid
2 Nor(x, x ) 2 Nor(y , y )
(population 1) (population 2),
iid
2 where the means x and y are unknown, while x and 2 y are somehow known.
Also assume that the Xis are indep of the Yis. Lets nd a CI for the dierence in means, x y .
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8.33 Normal Mean CIs (var known)
Dene the sample means from popns 1 and 2, 1 ny 1 nx Xi and Y Yi. X nx i=1 ny i=1 Obviously,
2 X Nor(x, x /nx) and 2 Y Nor(y , y /ny ),
so that
2 2 y x X Y Nor x y , + ...