Transcript Estimation 2

Estimation – More
Applications
Estimation 2: 1
So far …
• defined appropriate confidence interval
estimates for a single population mean, .
• Confidence interval estimators are valuable
because they provide:
• Indicate the width of the central (1-alpha)% of
the sampling distribution of the estimator
• Provide an idea of how much the estimator
might differ if another study was done.
Estimation 2: 2
Next step:
• extend the principles of confidence interval
estimation to develop CI estimates for other
parameters.
The important things to keep track of FOR EACH
PARAMETER:
• What is the appropriate probability distribution
to describe the spread of the point estimator of
the parameter?
• What underlying assumptions about the data are
necessary?
• How is the confidence interval calculated?
Estimation 2: 3
The confidence interval estimates of interest are:
1. Confidence Interval calculation for the difference
between two means, 1 – 2 , for comparing two
independent groups.
2. Confidence Interval calculation for the mean
difference, d , for paired data.
3. Population variance, s2, when the underlying
distribution is Normal.
We will introduce the 2 (Chi-square) distribution.
4. The ratio of two variances for comparing
variances of 2 independent groups:
– introducing the F-distribution.
s 12
s 22
Estimation 2: 4
Confidence Interval Estimation for:
5. Population proportion, , using the Normal
approximation for a Binomial proportion.
6. The difference between two proportions, 1 – 2
two independent groups.
Estimation 2: 5
1. Confidence Interval calculation for the
difference between two means, 1 – 2,
for Two Independent groups
We are often interested in comparing two groups:
1. What is the difference in mean blood pressure
between males and females?
2. What is the difference in body mass index (BMI)
between breast cancer cases versus noncancer patients?
3. How different is the length of stay (LOS) for
CABG patients at hospital A compared to
hospital B?
Estimation 2: 6
We are interested in the similarity of the two groups.
Statistically, we focus on
•
the difference between the means of the two
groups.
•
Similar groups will have small differences,
or no difference, between means.
Thus, we focus on estimating the difference in means:
1-2
An obvious point estimator is the difference between
sample means,
x1 – x2
Estimation 2: 7
To compute a confidence interval for this difference,
we need to know the standard error of the difference.
Suppose we take independent random samples from
two different groups:
x1 ~ N ( 1 , s 12 )
x2 ~ N ( 2 , s 22 )
We know the sampling distribution of the mean for
each group:
x1 ~ N ( 1 ,
s
2
1
n1
)
x2 ~ N (  2 ,
s 22
n2
)
Estimation 2: 8
What is the distribution of the difference between the
sample means, (x1 – x2) ?
• The sum (or difference) of normal RVs will be
normal. What will be the mean and variance?
• This is a linear combination of two independent
random variables. As a result,
E  X1 - X 2   E  X1  - E  X 2 
var  X 1 - X 2   var  X 1   var  X 2 
Estimation 2: 9
In general, for any constants a and b, :
2
2





s
s
2
2
1
2
ax1  bx2 ~ N  a 1  b 2 , a 
b 

 n1 
 n2  

That is, the distribution of the sum of ax1 and bx2,
• The mean is the sum of a1 and b2
• The variance is the sum of (a2)(var of
sampling distribution of x1) and
(b2)(var of sampling distribution of x2)
Estimation 2: 10
Letting a = 1, and b = -1, we have:

 s 12 s 22  
( x1 - x2 ) ~ N  1 -  2 , 


 n1 n2  

Thus, the standard error of the difference between
means:
s x -x 
1
2
s 12
n1

s 22
n2
Estimation 2: 11
Once we have
•
a point estimate
•
its standard error,
→ we know how to compute a confidence interval
estimate.
Confidence
Interval
Estimate
s2
 Point
Estimate
known 
s2 estimated
x1 – x2
from samples 
 Confidence
Coefficient
Percentile
From N(0,1)
Percentile
from tdf
Std
Error
s 12
n1

s 22
n2
Est. of
std error
Estimation 2: 12
Example: Data are available on the weight gain of
weanling rats fed either of two diets. The weight
gain in grams was recorded for each rat, and the
mean for each group computed:
Diet Group #1
Diet Group #2
n1 = 12 rats
n2 = 7 rats
x1 = 120 gms
x2 = 101 gms
What is the difference in weight gain between rats
fed on the 2 diets, and a 99% CI for the difference?
Estimation 2: 13
We will assume
a. The rats were selected via independent simple random samples
from two populations.
b. the variance of weight gain of weanling rats is known, and is the
same for both diet groups:
s12 = s22 = 400
Construct a 99% confidence interval estimate of the
difference in mean weight gain, 1 – 2
1. Point Estimate: x1 – x2 = 120 – 101 = 19 gms
2. Std error of point estimate:
s x -x 
1
2
s 12
s 22
400 400



 9.51
n1 n2
12
7
Estimation 2: 14
3. With known variance, use a percentile of N(0,1):
For (1 – a = .99, z.995 = 2.576
4. The 99% CI for 1 – 2 is:
( x1 - x2 )  ( z1-a / 2 )
s 12
n1

s 22
n2
 19  (2.576)(9.51)
= (–5.5, 43.5) gms
Estimation 2: 15
How do we interpret this interval?
(–5.5, 43.5) gms
With different samples, we will have different
estimates of the true difference in gains. The
endpoints of the confidence interval indicate how
wide the difference in estimates is expected to be
99% of the time.
Alternatively, if we repeatedly selected samples and
computed a CI, then for 99% of the intervals
computed would include the true difference in gains.
Estimation 2: 16
How do we compute a confidence interval when
• we don’t “know” the population variance(s),
• but must estimate them from our samples?
If s12 and s22 are UNknown:
Is it reasonable to assume that the variances of
the two groups are the same?
That is, is it OK to assume unknown s12  s22 ?
Questions to consider:
• Do data arise from the same measurement
process?
• Have we have sampled from the same population?
• Does difference in groups lead us to expect
different variability as well as different mean
levels?
Estimation 2: 17
If OK to assume variances equal: s12 = s22 = s2
•
We have 2 estimates of same parameter, s2
•
One from each sample: s12 and s22
We can create a pooled estimate: sp2
2
2
(
n
1)
s

(
n
1)
s
2
1
2
2
sp  1
(n1 - 1)  (n2 - 1)
•
This is a weighted average of the 2 estimates
of the variance
•
Weighting by (ni–1) for the ith sample
Estimation 2: 18
The standard error of the difference in means, x1 – x2
is then:
se( x1 - x2 ) 
s 2p
n1

s 2p
n2
That is,
Sp2 is used as an estimator of the variance of x1
and of x2 rather than the two sample estimates
Estimation 2: 19
Use the t-distribution to compute percentiles:
•
we are estimating the variance from the
samples
•
One degree of freedom is lost for each
sample mean we estimated, resulting in
df = (n1 – 1) + (n2 – 1) = n1 + n2 – 2
Thus, our confidence interval estimator when the
variance is Unknown, but assumed equal for the two
groups:
( x1 - x2 )  (tn1  n2 -2;1-a / 2 )
s
2
p
n1

s
2
p
n2
Estimation 2: 20
Example: Weanling Rats Revisited
Diet Group #1
Diet Group #2
n1 = 12 rats
n2 = 7 rats
x1 = 120 gms
x2 = 101 gms
s12 = 457.25 g2
s22 = 425.33 g2
Is there a difference in mean weight gain among rats
fed on the 2 diets?
• Use sample estimates of the variance
• Assume that the variances are equal since
• the rats in each group come from the same
breed
• were fed the same number of calories on their
different diets
• Used same scale in weighing.
Estimation 2: 21
Assuming s12  s22, equal but unknown, construct a
99% CI for the difference in means, 1 – 2 .
1. Point Estimate: x1 – x2 = 120 – 101 = 19 gms
2. Std error of point estimate:
Step 1: sp2
2
2
(
n
1)
s

(
n
1)
s
1
2
2
s 2p  1
(n1 - 1)  (n2 - 1)
(12 - 1)(457.25)  (7 - 1)(425.33)

 445.98 gm 2
12  7 - 2
Estimation 2: 22
Step 2: SE of point estimate:
se( x1 - x2 ) 
s 2p
s 2p
445.98 445.98



 10.04
n1 n2
12
7
3. Confidence Coefficient for 99% CI:
df = n1+n2 – 2 = 12 + 7 – 2 = 17
1-a = .99  a/2 = .005  1 – a/2=.995
t17;.995= 2.898
Estimation 2: 23
3. Confidence interval estimate of (1 – 2):
( x1 - x2 )  (t17;.995 ) se( x1 - x2 )  19  (2.898)(10.04)
 (-10.1, 48.1)
Again, we can conclude that if we repeated the
study many times, and looked at how widely the
sample mean differences were spread out (99%
of the time), then the width would be equal to
the confidence interval width.
Notice that the width is wider than when we know
the variance.
Estimation 2: 24
What if it is not reasonable to assume that the
variances of the two groups are the same?
When it seems likely that s12  s22
For example
•
we have used a different measuring process
•
we have other reasons to believe both the mean
level and variability are different between the two
populations
Then
•
Use separate estimates of the variance from each
sample, s12 and s22
•
Compute Satterthwaite’s df – and appropriate tvalue
Estimation 2: 25
Satterthwaite’s Formula for Degrees of freedom:
2
s
s 
  
n1 n2 

f 
2
2
2
2
 s1   s2 
   
 n1    n2 
n1 - 1 n2 - 1
2
1
2
2
horrible … avoid computing by hand!
Note
• it is a function both of the sample sizes and the
variance estimates.
• When in fact the variances and sample sizes are
similar – the df will be similar to the pooled
variance df.
Estimation 2: 26
Putting it all together yields
the CI estimator when UNknown s12  s22 :
2
1
2
2
s
s
( x1 - x2 )  (t f ;1-a / 2 )

n1 n2
Note: use separate estimates of standard error of
sample means for each sample
Estimation 2: 27
Example: Weanling Rats Once Again!
Assume that the population variances are not equal –
s12  s22 .
Diet Group #1
Diet Group #2
n1 = 12 rats
n2 = 7 rats
x1 = 120 gms
x2 = 101 gms
s12 = 457.25 g2
s22 = 425.33 g2
Is there a difference in mean weight gain among rats
fed on the 2 diets?
Compute at 99% CI for the difference in the group
means, assuming s12  s22 .
Estimation 2: 28
1. Point Estimate: x1 – x2 = 120 – 101 = 19 gms
2. Std error of point estimate:
s12 s22
457.25 425.33
se( x1 - x2 ) 



 9.94
n1 n2
12
7
3. Confidence Coefficient for 99% CI:
df = f = … = 13.08  use 13
1-a = .99  a/2 = .005  1 – a/2=.995
t13;.995= 3.012
Estimation 2: 29
4. Confidence interval estimate of (1 – 2):
( x1 - x2 )  (t13;.995 ) se( x1 - x2 )  19  (3.012)(9.943)
= (-11.0, 49.0)
The interpretation of this confidence interval is
the same. Notice that it is somewhat wider than
the previous two intervals- indicating a wider
variation in the sample mean difference when
variances are not equal between groups.
Estimation 2: 30
Since the unit is common to the two measures, we
expect
• the two responses to the unit to be similar in
some respects
• We expect the 1st and 2nd responses within a
unit to be related.
Studies use this design to reduce the effects of
subject-to-subject variability
• This variability can be reduced by subtracting
the common part out.
• We do this by taking the difference between the
2 measures, on the same subject or unit.
Estimation 2: 31
Analysis of Paired Data Focuses on:
difference = Response 2 – Response 1
for each subject, or paired unit.
Work with the differences –
• as if never saw the individual paired responses
• and see only the differences as our data set
• The data set comprised of differences has been
reduced to a one sample set of data.
• We already know how to work with this.
Estimation 2: 32
1st Response
1
…
n
2nd Response
Difference
2nd – 1st
x1 = 10
y1 = 12
d1 = 12–10 = 2
xi
yi
di = xi – yi
xn = 14
y1 = 11
dn = 11–14 = -3
Note:
• The order in which you take differences is
arbitrary, but it must be consistent. If you choose
yi – xi , then compute that way for all pairs.
• Direction is important. Keep track of positive and
negative differences.
Estimation 2: 33
Confidence Interval Calculations
for the mean difference, d
Preliminaries:
1. Compute sample of differences, d1, …, dn , where
n = # of paired measures.
2. Obtain sample mean and sample variance of the
differences
1 n
d   di
n i 1
n
1
2
sd2 
(
d
d
)

i
n - 1 i 1
3. Treat like any other 1-sample case for estimating a
mean, , (here a mean difference.)
Estimation 2: 34
Example: Reaction times in seconds to 2 different
stimuli are given below for 8 individuals. Estimate
the average difference in reaction time, with a 95%
CI. Does there appear to be a difference in reaction
time to the 2 stimuli?
Difference (X2 – X1)
Subject
X1
X2
1
1
4
3
2
3
2
-1
3
2
3
1
4
1
3
2
5
2
1
-1
6
1
2
1
7
3
3
0
8
2
3
1
Estimation 2: 35
We have paired data
•
each subject was measured for each stimuli
•
we focus on the within-subject difference.
Since I have subtracted in the direction X2 – X1 :
•
a positive difference means longer reaction time
for stimulus 2
•
a negative difference means a longer reaction
time for stimulus 1.
We can compute the mean and standard deviation of
the differences:
d = .75
and
Sd = 1.39
Estimation 2: 36
•
For a 95% confidence interval,
•
using my sample estimate of standard error,
•
use the t-distribution.
The confidence interval is:
d ± tn-1; .1-a/2(sd/n) = .75 ± t 7; .975(sd/8)
= .75 ± 2.36 (1.39/8)
95% CI is
(-0.41, 1.91)
The results indicate that repeating the study may
produce an estimate quite different from that
observed, and even possibly a negative estimate.
Estimation 2: 37
Notes:
• It is a common error to fail to recognize paired
data, and therefore fail to compute the
appropriate confidence interval.
• The mean difference d is equal to the difference
in means, 2 – 1 if we ignore pairs – your point
estimate will be correct.
• However, the variance of the mean difference
does NOT equal the variance of the difference in
means – so the confidence interval will not be
correctly estimated if you neglect to use a paired
data approach.
Sd2/n = (S12/n) + (S22/n)-2Cov/n
Estimation 2: 38
Confidence Interval Estimation
of the Variance, s2
Standard Deviation, s
and Ratio of Variances of 2 groups
Estimation 2: 39
3. Confidence Interval for the variance, s2:
Introducing the 2 Distribution
What if our interest lies in estimation of the variance,
s2 ?
Some common examples are:
• Standardization of equipment
– repeated measurement of a standard should
have small variability
• Evaluation of technicians
– are the results from person i “too variable”
• Comparison of measurement techniques
– is a new method more variable than a
standard method?
Estimation 2: 40
We have an obvious point estimator of s2  s2,
which we have shown earlier is an unbiased
estimator (when using Simple random with
replacement sampling).
How do we get a confidence interval?
We will define a new standardized variable, based
upon the way in which s2 is computed:
 
2
(n - 1) s
s
2
2
~  n2-1
That is, [(n-1)s2 / s2] follows a chi-square
distribution with n-1 degrees of freedom
Estimation 2: 41
A quick and dirty derivation:
We defined the sample variance as:
n
1
2
s2 
(
x
x
)

i
n - 1 i 1
Multiplying each side by (n-1):
n
(n - 1) s   ( xi - x )
2
2
i 1
Note this is the
numerator from
the 2 variable.
This side is the sum of
squared deviations from
the mean.
Estimation 2: 42
Recall, for X ~ N(, s2)
We can standardize as:
X -
s
~ N (0,1)
If we square this, we have a squared standard
normal variable:
 X - 
2
~

1


 s 
2
That is, a squared standard normal variable follows
a chi- square distribution, with 1 degree of freedom
– this is the definition of a chi-square, df=1
Estimation 2: 43
If we sum n such random variables, we define a chisquare distribution with n degrees of freedom:
n

( xi -  )2
i 1
s2
~  n2
However, if we first estimate  from the data: x,
we reduce the degrees of freedom:
n

i 1
( xi - x )2
s2

(n - 1) s 2
s2
~  n2-1
Estimation 2: 44
Features of the Chi Square Distribution
•
Chi-squared variables are sums of squared
Normally distributed variables.
•
Chi-squared random variables are always positive.
(Why? –square is always positive)
•
The distribution is NOT symmetric. A typical
shape is:
0
Estimation 2: 45
Features of the Chi Square Distribution
• Each degree of freedom defines a different
distribution.
• The shape is less skewed as n increases.
df = 1
df = 2
df = 4
df = 6
df =10
df = 100
Estimation 2: 46
How to Use the Chi Square Table – Table 6, Rosner
The format is the same as for the Student t-tables:
d
1
2
…
5
.005 .01
2
2
.025
2
…
.995
2
7.88
10.60
…
16.75
Each row gives information for a separate chi square
distribution, defined by the degrees of freedom.
The column heading tells you which percentile will be
given to you in the body of the table.
The body of the table is comprised of the values of
the percentile
Estimation 2: 47
This area
= .995
2 distribution
with 5 df
16.750
Pr[ 25  16.750]=.995
Note: Because the 2 distribution is not symmetric
• will often need to look up both upper and lower
percentiles of the distribution
Estimation 2: 48
Confidence Interval for s2
For
X2 
(n - 1) s 2
s2
~  n2-1
To obtain a (1-a) confidence interval, we want to find
percentiles of the 2 distribution so that:
Pr[ 
2
n -1,a / 2
X 
2
This area
Is a/2
2
n -1,1-a / 2
]  1-a
This area
is a/2
(1 – a)
2
a/2
2

1- a/2
Estimation 2: 49
Substitute for X2 in the middle of the inequality:
Pr[ n2-1,a / 2  X 2   n2-1,1-a / 2 ]  1 - a
Pr[ 
2
n -1,a / 2

(n - 1) s
s2
2
  n2-1,1-a / 2 ]  1 - a
A little algebra yields the confidence interval formula:
2
 (n - 1) s 2
(n - 1) s 
2
Pr  2
s  2
  1-a
 n -1,a / 2 
  n -1,1-a / 2
Estimation 2: 50
Confidence Interval for s2
Lower limit of
the (1– a) CI:
(n - 1) s 2
Upper limit of
the (1– a) CI:
(n - 1) s 2
 n2-1,1-a / 2

2
n -1,a / 2
Estimation 2: 51
Exercise
A precision instrument is guaranteed to read
accurately to within  2 units.
A sample of 4 readings on the same object yield
353, 351, 351, and 355.
Find a 95% confidence interval estimate for the
population variance, s2 and also for the population
standard deviation, s.
Estimation 2: 52
Solution
1. Point Estimate: We must first estimate the mean,
x = 352.5, and then the variance, s2 = 3.67
2. Since n=4, the correct chi-square distribution has
df = n-1 = 3.
3. For a (1-a) = .95 CI
a = .05  a/2 = .025
and
1- a/2 = .975
We want 23;.025 and 23;.975
This area = .025
This area = .025
.95
23;.025
23;.975
Estimation 2: 53
4. Using Table 6 in Rosner (page 758) :
(i) Using column labeled .025 read down to
df= 3 row
 23;.025 = .216
(ii)Using column labeled .975 read down to
df= 3 row
 23,.925 = 9.35
(or use Minitab or other program…)
Estimation 2: 54
Using Minitab:
Calc 
Prob Dist  Chi sq
Inverse Cumulative
Probability
Degrees of
freedom df = n-1
Input desired
percentiles
e.g., .025, .975
Inverse Cumulative Distribution Function
Chi-Square with 3 DF
P( X <= x)
x
0.0250
0.2158
0.9750
9.3484
Estimation 2: 55
5. Compute Limits:
Lower limit of
the 95% CI:
Upper limit of
the (1– a) CI:
(n - 1) s
2
 n2-1,1-a / 2
3(3.67)

 1.18
9.348
(n - 1) s 2

2
n -1,a / 2
3(3.67)

 50.97
.216
The 95% CI for s2 = (1.18, 50.97)
Estimation 2: 56
6. To compute a confidence interval for the standard
deviation, s
•
always compute a CI for the variance
•
then take the square root of the upper and
lower limits:
95% CI for s: ( 1.18, 50.97 ) = (1.09, 7.14)
Point estimate for s = 3.67 = 1.92
Does this precision instrument meet its ‘guarantee’
to accuracy within  2 units?
Estimation 2: 57
Note that the confidence intervals for s
•
are wide
•
are not symmetric about the point estimate
LL
s
1.09 1.92
UL
7.14
•
Only with very large n
•
will you find relatively narrow confidence
interval estimates for the variance and standard
deviation.
Estimation 2: 58
Confidence Interval calculation for the ratio of two
variances – introducing the F-distribution
We are often interested in comparing the variances of
2 groups.
• This may be the primary question of interest:
I have a new measurement procedure – are the
results more variable than the standard
procedure?
• Comparison of variances may also be a
preliminary analysis to determine whether it is
appropriate to compute a pooled variance
estimate or not, when the goal is comparing the
mean levels of two groups.
Estimation 2: 59
For comparing variances, we use a RATIO rather than
a difference.
• We look at the ratio of variances:
sx2/sy2
• If this ratio is 1  the variances are the same
• If it is far from 1  the variances differ.
In order to
• make probability statements about ratios of
variances
•
to compute confidence intervals
we need to introduce another distribution, known as
the
F-distribution
Estimation 2: 60
A Definition of the F Distribution
IF
x1, … xnx are each independent Normal (x, sx2)
and y1, … yny are each independent Normal (y, sy2)
and if we calculate sample variances in the usual way
nx
1
2
sx2 
(
x
x
)

i
nx - 1 i 1
n
y
1
2
2
sy 
( yi - y )

ny - 1 i 1
Estimation 2: 61
THEN
s /s
~ Fnx -1;ny -1
s /s
2
x
2
y
2
x
2
y
The ratio follows an F-distribution with two degree of
freedom specifications – for the numerator and for
the denominator.
numerator df = nx – 1
denominator df = ny – 1
Estimation 2: 62
Percentiles of the F-distribution are tabulated, as in
the Appendix of Rosner, Table 9, pages 762-764.
Using the Table:
• Each row defines a different percentile distribution
for a given denominator df.
• Each column defines a different numerator df.
• The body of the table gives values of the Fdistribution.
• Only the upper-tail percentiles (.90, …, .999) of the
distribution are tabulated.
Estimation 2: 63
df for
denominator
Freedom
p
…
20
.90
.95
…
30
.90
.95
df for numerator
…
7
8
12
24

2.04
2.51
…
…
…
2.00
2.45
1.89
2.28
1.77
2.08
1.61
1.84
Example: Find the 95th percentile of an F-distribution with
df=12,20? (num,den is the standard order)
1. Under the denominator df col, find the row for df=20
2. Under p, the percentile, find the row for p=.95
3. Find the column headed by numerator df=12.
4. Read the value at their intersection: F12,20; .95 = 2.28 .
Estimation 2: 64
For lower-tail percentiles (.005, …, .10) which are not
tabulated, we use the fact that:
The percentiles of an F with
numerator df=a
denominator df=b
are related to
The percentiles of an F with
numerator df=b
denominator df=a
as:
Fa ,b;(a ) 
1
Fb,a;(1-a )
Estimation 2: 65
Example: What is the 5th percentile of an Fdistribution with df=12,20?
F20,12;(.05) 
1
F12,20;(.95)
1

 .439
2.28
We have already looked
up this value.
Lower and upper tail percentiles can be computed
directly using Minitab – no need to invert.
Estimation 2: 66
Use Minitab: Calc  Probability Distributions  F
Inverse
Cumulative Prob
Numerator
and
Denominator
df
Desired Percentile
Estimation 2: 67
Inverse Cumulative Distribution Function
F distribution with 12 DF in numerator and 20
DF in denominator
P( X <= x)
0.9500
x
2.2776
Estimation 2: 68
Confidence Interval for the ratio of 2 Variances: sx2/sy2
IF
x1, … xnx are each independent Normal (x, sx2)
and y1, … yny are each independent Normal (y, sy2)
A point estimate of sx2/sy2 is
sx2/sy2
Estimation 2: 69
A (1-a) Confidence Interval Estimate has:

 sx2  
1

LL :  2  
 s  F

y
n
1,
n
1;(1
a
/
2)
  x y


 sx2  
1

UL :  2  
 s  F

y
n
1,
n
1;(
a
/
2)
  x y

 sx2 
  2  Fny -1,nx -1;(1-a / 2)
s 
 y


Estimation 2: 70
Example:
Pelicans were exposed to DDT. Is there a difference
in variability of residue found in juveniles and
nestling birds?
Juvenile Pelicans
Nestling Pelicans
n1 = 10
n2 = 13
s1 = .017
s2 = .006
Compute a 95% Confidence Interval for the ratio of
true variances, s12/s22 .
Estimation 2: 71
1. Point estimate:
s12/s22 = (.017)2 / (.006)2 = 8.03
2. Percentiles of F:
F9,12;.975 = 3.44
F9,12;.025 = .286
3. Confidence Limits:

 sx2  
1
 1 
  8.03 
LL :  2  
 2.33

 s  F

 3.44 
 y   nx -1,ny -1;(1-a / 2) 
Estimation 2: 72

 sx2  
1
 1 
  8.03 
UL :  2  
 31.07

 s  F

 .286 
 y   nx -1,ny -1;(a / 2) 
A 95% CI for the variance ratio is (2.33, 31.07) .
Notice that the width of the sampling distribution of
variance ratios is very broad. Since 1 is not in the
interval, it appears that the variances of the groups
are different, with juvenile pelicans having a greater
variability in DDT residue.
Note: always work with variances, not standard
deviations.
Estimation 2: 73