Lecture 3: Confidence Intervals in Statistical Inference, Study notes of Statistics

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Statistics 431:

Statistical Inference

Lecture 3: Confidence intervals

Introduction

• A point estimate (eg, sample mean^ estimating population mean ) could be very precise, or not at all. Can’t tell from just the number.
• Instead of reporting single estimate of^ , can report a range of plausible values based on data: a confidence interval for.
• Each CI has an associated^ confidence level , like 90%, 95%, ...
  • the higher the confidence level, the more likely the CI is to contain
• A wide interval implies we don’t have a good handle on^ ; a narrow interval implies is known precisely.
• To find the CI for a given confidence level, we need assumptions plus a probability calculation.

X ¯

μ μ μ μ μ μ

• Then:
• Interlude: derivation at the board.
• Therefore, a 100^ % confidence interval for the mean^ of a normal population ( known) is given by
• eg, a 95% CI is

P

X^ ¯ − z α 2 ·^

σ √ n

< μ < X ¯ + z α 2 ·

σ √ n

= γ

γ μ σ 2

X^ ¯ − 1. 96 · √σ n

, X ¯ + 1. 96 ·

σ √ n

X^ ¯ − z α 2 ·^

σ √ n

, X ¯ + z α 2 ·

σ √ n

• Before^ the data are observed:
  • the CI is a random interval (in this case, centered at )
  • there is probability that the observed CI will cover
  • note: center is random but width is not
• After^ the data are observed:
  • the CI is a fixed interval , determined by
  • this fixed interval either covers or it doesn’t (no probability statement applies)

X ¯

γ μ

x (^) 1 ,... , x (^) n μ

Confidence vs. width

• Higher confidence (good) = wider interval (bad)
• The only way to get higher confidence and a narrower interval is to increase the sample size.
• For confidence 100^ % and width^ we need

(Again, we don’t know : we’ll come back to this.)

• Example: Fisher’s iris data had^ ,^ , 95% CI

CI width. To achieve on a new sample,

n γ w

n (w) =

2 z α 2 ·

σ w

σ

w = 0. 5 n = ( 2 · 1. 96 · 3. 5 / 0. 5 ) 2 ≈ 753

n = 50 σ^ =^3.^5

5. 0 ± 1. 96 · 3. 5 /

w = 2 · 0. 97 = 1. 94

• If the answer^ were small, we’d treat it as very approximate, because we don’t know the true.

n σ

Derivation of CI: example

• ;^.
• Can show^ has chi-square distribution with degrees of freedom,. Since this is a known distribution (in particular, it doesn’t depend on ), is a pivot.

chspdftb.gif 380!280 pixels 09/09/2005 05:22 PM

a b

X (^) 1 ,... , X (^) n ∼ Exp(λ) p ( x ) = λ e −λ x^ , x > 0 , λ > 0 h ( X (^) 1: n , λ) = 2 n λ X ¯ 2 n χ^22 n λ h ( X (^) 1: n , λ)

• So

implies

• We pivoted^ to get the 100(1-^ )% CI

for.

P

a 2 n X ¯

< λ <

b 2 n X ¯

= 1 − α

P ( a < 2 n λ X ¯ < b ) = 1 − α

h ( X (^) 1: n , λ) α ( a 2 n X ¯

b 2 n X ¯

λ

• What about^? Large sample also justifies substituting sample variance for. (When n is not large, this doesn’t work : next lecture.)
• Result: approximate 100(1-^ )% CI for^ is

and this holds regardless of shape of popn distribution.

• What is a “large”^? Unfortunately, no universal answer. Certainly is “small,” not “large.” will sometimes (but not always) suffice.
• The shape of the popn distribution^ does^ affect how large^ must be for the approximation to work. - eg, we already know that the approximation is exact if popn distribution is normal and is known

σ 2 s^2 σ 2

α μ

X ¯ ± z 2 ·^

S

n

n n < 50 n > 50

n

σ 2

• When we knew^ , we could achieve CI width^ using

samples.

• With^ unknown, we are out of luck: can’t substitute^ until we observe the sample, but we need to pick sample size before sample is observed.
• Alternatives:
  • make a guess about based on other information
  • draw a sample of arbitrary size, compute , compute , draw a second sample of size
  • other such sequential methods (not part of this course)

Large-sample CI widths

σ 2 w

n (w) =

2 z α 2 ·

σ w

σ 2 s^^2

σ 2 s^2 n (w)

n (w)