Topic 5: Confidence Intervals (Chapter 9)
1. Introduction
• The two general areas of statistical inference are: 1) estimation of parameter(s), ch. 9
2) hypothesis testing of parameter(s), ch. 10
• Let X be some random variable with unknown meanµ. Suppose we take a sample of size . One could find the sample mean,n x, and use it to estimateµ. Such a single number used to estimate a parameter is called a point estimate. • Question: Are there other reasonable point estimates (other thanx) for µ? • Claim: The “best” point estimate of µ isx.
• The point estimate does not give us a sense of how close x might be toµ, i.e. it doesn’t tell us about the inherent uncertainty in the process. Therefore, interval estimates are used, which provide a range of values that contain the parameter, sayµ, with some specified degree of confidence. This range is called a confidence interval.
2. Two-Sided Confidence Interval
• Consider now some random variable X with mean µ and standard deviationσ. Suppose we want to estimate µ (the estimation of σ follows later). Using the central limit theorem
~ ( , X N n) σ µ Therefore, (X Z n ) µ σ− = (1) is a standard normal. • Recall that by definition
2 2 ( ) P z− α < <Z zα = −1 α (2) e.g. because z.025 =1.96 ( 1.96 1.96) 1 .05 .95 P − < <Z = − =
• Substituting (1) into (2), and using some algebra, one can show that (2) implies 2 2 ( P X z X z n n α α ) 1 σ µ σ α − < < + = − (3) e.g. letting α =.05 ( 1.96 1.96 ) .95 P X X n n σ µ σ − < < + =
• The interpretation of (3) is tricky. Notice that we calculate a random interval (X 1.96 ,X 1.96 )
n n
σ σ
− + .
If we do this repeatedly, we expect that 95% of the intervals contain the unknown parameterµ. It does not mean that µ assumes a value within the interval with probability 0.95. An illustration of this is given in Figure 9.1. • In general, a 100 (1−α)% confidence interval for µ is
2 2 (X z ,X z n n α α ) σ σ − + . (4)
• The confidence interval could be written as X E± where 2 E z n α σ = . (5)
E is also called the bound on error (BOE).
• Suppose one wants to reduce E. This could obviously be done by increasing α or increasing . In fact, may be determined from E. Notice from (5), n n
2 n z E α σ = . Therefore, 2 2 [ n z E α ] σ = . (6)
• Consider an example. Let X be the cholesterol level of a U.S. male who smokes. Suppose µ is unknown, but that we know σ =46 (we actually assume this value because we presumably know σ =46 for population of all U.S. adult males). Suppose we take a sample ofn=12, and find
217
x= .
Then the best point estimate of µ is 217. Also, a 95% confidence interval for µ is (using X =217)
1.96(46) 217 12 ± 217 26 (191, 243). = ± = A 99% confidence interval for µ is
2.58(46) 217 12 (183, 251) ± = Why is it wider?
• Suppose we are designing the study, and we want to know µ within 10 units. (Note that E=10 and the interval’s width is 20). Substitutingσ, E, and α =.01 into (6), one has
2 2.58(46)
( ) 140.8
10
n= = .
Therefore our recommendation is to obtain a sample for n=141 males. 3. One-Sided Confidence Intervals
• Sometimes (though not often), one desires just an upper or a lower bound onµ. Using algebraic manipulations as before, one has:
An 100(1-α )% upper (one-sided) confidence bound on µ is X z
n α
σ
+ , (7)
and a corresponding lower confidence bound is X z
n α
σ
− . (8)
e.g. for U.S. males, to find a 95% upper bound, recall that . Therefore the upper bound from the sample is
.05 1.65 z = 1.65(46) 217 12 217 21.9 238.9 ± = + = 4. Students’ t distribution
• Consider a random variable X which is known to be normally distributed, but where neither µ nor σ are known. To make inferences aboutµ, we could use the Z transformation, provided σ is known. However, in case of unknownσ, which is the common case, we replace σ by to find s
(x ) t s n µ − = .
This statistic has the t distribution with n-1 “degrees of freedom” (df). • Several facts of interest are
1) The t-distribution is also symmetric about 0, but it has “thicker tails”, i.e. it’s a bit flatter. This is because there is added variability introduced by using the random variable rather than a constant s σ in the denominator (see Figure 9.2).
2) The t-distribution approaches Z as becomes large. This is because as increases, we have more information and approaches
n
n s σ. Why?
• Recall we had an extensive Z table, but it’s impractical to give a table for each statistic. Therefore, tables typically give only upper percent points. For example, from Table A-4, one has
1 n t − 10,.025 2.228 t = . By symmetry, 10,.975 2.228 t = − ,
and thus one can find upper and lower percentiles.
• One can construct confidence intervals based on the t-distribution in the same way as before. The expression comparable to (3) is
2 2 ( s s ) P X t X t n n α µ α 1 α − < < + = − .
In general, a 100(1-α )% confidence interval for µ, with approximately normal
x, is 2 2 (X t s ,X t n n α α − + s ). (9)
• For example, let X denote an infant’s plasma aluminum level, which is assumed to be approximately normal. Suppose we take a random sample of
infants, and find 10
n= x =37.2 and s=7.13. To find a 95% confidence interval, we find
9,.025 2.262
t = .
Hence, the interval estimate is
2.262(7.13) 37.2 10 37.2 5.1 (32.1, 42.3) ± = ± =
• Notice that in this problem, the interval width is a random variable. Hence, if two experimenters took different samples, their interval widths would differ, unlike the case of knownσ . In general, the interval based on t would be wider than that based on Z. Why? Though the widths differ, we expect 100(1-α )% of the intervals to contain µ. See Figure 9.3.
5. Case Study
• This has a nice case study on the efficacy of using a particular drug for treating ADD children.
