Transcript Packing Densities of Permutations

CONFIDENCE INTERVALS
Confidence intervals for means and proportions:
Section 6.1
Section 7.1 ( pages 450 – 453 )
Section 8.1 ( pages 536 - 540 )
Topics:
Point Estimates vs. Interval Estimates
Context / Notation
eight cases
CI’s with z*
assumptions
interpretations
tradeoffs
choosing the right n
CI’s with t* ( thanks to W. S. Gosset )
CI’s for proportions
THE COOKBOOK
THIS ISN’T HARD
Engineers measured the distance between two buildings.
n = number of estimates = 100
x = mean of estimates = 310 m
s = st.dev. of estimates = 0.05 m
They reported their results this way.
The true distance is in the range
309.990 m to 310.010 m
with 95% confidence.
THIS ISN’T HARD
How they got these numbers:
Lower limit
 n   309.990
x  1.96  s
n  310.010
x  1.96 s
Upper limit
More often than not, your confidence intervals will use
exactly this formula. But not always!
Point Estimates vs. Interval Estimates
We want to know .
Point-estimate approach:
“I estimate that  is 310.000.”
The SE is about 0.05.
95% of the time, my error is less than 1.96 SE’s.
Interval-estimate approach:
“I estimate that  is in the range 309.990-310.010.”
95% of the time, my ranges contain the true value.
Interval estimates
Two things you want to know about an interval estimate:
(a) the interval, of course
(b) the CONFIDENCE LEVEL, or
CONTAINMENT PROBABILITY.
What fraction of intervals constructed in this way
contain the true value?
Confidence Levels
We’ll construct interval estimates to achieve
pre-selected confidence levels.
Notation:
C = “Confidence level”
 = “Significance level”
Always:
C = 1 - , and  = 1 – C.
For example: C = 0.95 and
 = 0.05.
Context
There’s a population distribution.
PARAMETERS
 = population mean (we don’t know this)
 = population SD (maybe we know this)
We take a sample with these results:
STATISTICS
n = sample size
(or
x = sample mean
s = sample SD (using n-1)
(or
ˆ
ˆ
)
)
At least 8 cases
n large vs. n small
 known vs.  unknown
population roughly normal vs. non-normal or unknown
(and we’ll consider proportions, too)
Words
You can describe a confidence interval in terms of its
endpoints…
[ 0.61, 0.67 ]
or in terms of its center and radius:
0.64  0.03.
 beware, ambiguous!
MARGIN of ERROR = RADIUS = HALF-WIDTH
Notation for the Interval
Write the interval this way:
[ 309.990 , 310.010 ] or
( 309.990 , 310.010 )
(low end always first)
[x
x – MOE, x
x + MOE ]
MOE = margin of error (text uses “m”)
[x
x – z* SE, xx + z* SE ]
SE = standard error
(or SEE, “standard error of the estimate”)
(or “standard error of the mean”)
z* = “critical value” (1.96 in the example)
Thus:
MOE = z* times SE
Standard error
SE =
or

s
n
n
if you know 
if you have to use s for .
It’s always better to use the population  if you can,
but you usually don’t know it.
Critical value depends on C
Critical value: z* = z*/2
z* is chosen so that /2 of the standard normal values
fall ABOVE z*.
C
z*
0.90
1.645
0.95
1.960
0.99
2.576
In the big table: Find (1 - /2) in the (z) table;
use corresponding z
In Excel:
z*/2 = NORMSINV (1 - /2)
Summary
Critical value from table
  
x   z * / 2  

 n
Or use s if you must
What did we assume?
(1) All this assumes that we really do have a simple
random sample.
-- Observations must be independent
(2) The sampling distribution of x
x must be normal.
-- EITHER n is large (15? 30?)
OR the population is roughly normal
(3) Outliers?
(4) Do we know  ?
Tradeoffs
Things we want:
SMALL INTERVAL
HIGH CONFIDENCE
Unfortunately there’s a tradeoff.
UNLESS we increase n.
Tradeoffs
CHEAP
(low n)
PRECISE
(low MOE)
CORRECT
(high C)
CHOOSE TWO.
How large does n have to be?
We’re testing a cholesterol drug.
The average score in the target population (and in the
very large control group) is 220, with a standard
deviation of 40.
We think our drug will decrease the average score to
210. We don’t think it will change the standard
deviation.
WHEN WE ESTIMATE THE AVERAGE IN OUR TEST
GROUP, WE HAD BETTER HAVE AN MOE OF LESS
THAN 10.
How large does n have to be?
How large does n have to be?
MOE = z* times (  / sqrt(n) )
10 = 1.96 times ( 40 / sqrt(n) )
So… n must be at least 64.
In general:
 z * 
n must be at least  MOE  .
2
C. I. Cookbook (the z* case)
1. Make sure this method is applicable.
Are the cases independent?
Is n large, OR the population normal (so that
the sampling distribution is normal)?
Are outliers an issue?
Do we know  ?
2. Determine n, x
x, s
3. Calculate SE =  / sqrt(n)
(use s for  if necessary)
C. I. Cookbook (the z* case)
4. Decide on the confidence level, C.
Write  = 1 – C.
5. Determine the critical value z*.
C
0.90
0.95
z*
1.645
1.960
0.99
2.576
6. MOE = z* times SE
7. Confidence interval is [ x – MOE, x + MOE ].
Interpretation
 is 310.000.
Interpretation
 is 310.000.
probably  is 310.000.
Interpretation
 is 310.000.
probably  is 310.000.
 is between 309.990 and 310.010.
Interpretation
 is 310.000.
probably  is 310.000.
 is between 309.990 and 310.010.
probably  is between 309.990 and 310.010.
Interpretation
 is 310.000.
probably  is 310.000.
 is between 309.990 and 310.010.
probably  is between 309.990 and 310.010.
We have confidence 95% that  is between 309.990
and 310.010.
Interpretation
 is 310.000.
probably  is 310.000.
 is between 309.990 and 310.010.
probably  is between 309.990 and 310.010.
We have confidence 95% that  is between 309.990
and 310.010.
CONFIDENCE good. PROBABILITY bad.
 superstition
Interpretation
 is 310.000.
probably  is 310.000.
 is between 309.990 and 310.010.
probably  is between 309.990 and 310.010.
We have confidence 95% that  is between 309.990
and 310.010.
CONFIDENCE good. PROBABILITY bad.
 superstition
t distributions
Everything we have done so far assumes that the
sampling distribution is normal.
W. S. Gosset made the same assumption at the Guinness
Brewing Company, but he kept getting bad results.
His population distribution was roughly normal, but his
sample sizes were small. And, like most people, he
had to use s for .
W. S. Gosset
“A. Student”
t distributions
Correct theory:
x 
Z 

n
But:
x 
T 
s
n
has a standard normal distribution.
has a different distribution, which
depends on n. Gosset worked it
out, and called it the t
distribution.
t distributions
Specifically,
x 
T 
s
n
has a “t distribution with
n-1 degrees of freedom.”
Number of “degrees of freedom” = n-1
t distributions
When n is large (30 or more?) the t distributions are
very close to normal. Otherwise…
t distributions
This means that when we’re using s for , we need
different critical values.
t*/2, n-1 = value such that /2 of values lie above t*,
if the values are distributed according to
the t distribution with n-1 degrees of
freedom.
Use Table D, inside back cover of text.
In Excel: t*/2, n-1 = TINV (, n-1 )
(Yes, Excel wants both tails in the argument.)
CI’s with t distributions
Use z* with , t* with s:
MOE = z* times (/sqrt(n))
OR
t* times (s/sqrt(n))
If n is large enough (say, 30 or more) it doesn’t matter
which you use, but it’s still good to observe the above
distinction.
NOTE: Gosset’s theory requires that the population be
roughly normal. But even if this isn’t true, if you’re
using s for , it’s better to use the t* critical values.
C. I. Cookbook (z* and t*)
1. Make sure this method is applicable.
Are the cases independent?
Is n large, OR the population normal (so that
the sampling distribution is normal)?
Are outliers an issue?
2. Decide whether you know .
3. Determine n, x
x, s.
C. I. Cookbook (z* and t*)
4. Decide on the confidence level, C.
Write  = 1 – C.
5. Determine the critical value, z* OR t*
6. MOE = z*/2 times (  / sqrt(n) )
OR
MOE = t* /2, n-1 times ( s / sqrt(n) )
7. Confidence interval is [ x – MOE, x + MOE ].
What about proportions?
Suppose that the population contains all 0’s and 1’s
(“hits and misses”, “succeses and failures”)
PARAMETER:
p = Success rate in population (=fraction of 1’s)
STATISTICS
n = Sample size
k = Number of “hits” in sample
pˆ = k / n = Success rate is sample
What about proportions?
pˆ
pˆ
STANDARD ERROR:
SE =
p (1  p )
n
But you don’t know p. So you have two practical choices:
(If you’re a pollster)
SE =
(0.5)
(because 0.5 is an upper bound)
n
(For most scientific applications)
SE =
pˆ(1  pˆ)
n
pˆ
pˆ
Various ways to write SE
SE  pˆ(1  pˆ)

1
n
pˆ(1  pˆ) 
n
1
n
pˆ(1  pˆ) 
These are all the same thing.
pˆ(1  pˆ)
n
Another choice for SE
The Wilson method
(our text calls it the “plus four” method)
Write:
SE =
where
p (1  p )
p
n
k 2 .
n 4
Why would anyone do this? It only matters when p is close to 0 or
1; in this case the usual SE is too small. The Wilson method
makes it larger.
This is NOT standard.
CI’s for proportions
MOE = z* times SE
(always use z* for proportions; it depends on C)
Confidence interval:
[ pˆ - MOE, pˆ + MOE ] .
Proportions are really just means
If the population consists of 0’s and 1’s, then
p, the fraction of 1’s,
is the same as the population mean, .
The population standard deviation (usually called )
is equal to
p (1  p ) .
So, SE’s for proportions are just like SE’s for means.
The “Quick Method”
95% CI for p:
 pˆ  1 , pˆ  1 

n
n 
(We’re using z*  2.00 and using 0.5 for
p (1  p ) )
Cookbook: CI’s for proportions
1. Decide whether method applies
Observations independent?
Each with same (unknown) probability?
At least 5 hits and 5 misses?
n large enough? (15? 30?)
2. SE =
pˆ(1  pˆ)
n
(or one of the alternatives)
3. MOE = z* times SE
4. CI = [ pˆ – MOE, pˆ + MOE ]