Announcements
• Exams are graded. I’ll hand them back at
end of class. Solution will be on the web.
Many people did better than I think they’d
guess…
• Easy optional homework due Thursday.
See web. (This is to get you to open the
book and take a look at chapter 8.)
Can only add to your homework score
Example
• A toxocologist* conducts an experiment to
investigate the relation between PCB (industrial
chemical and pollutant) exposure and estrogen
production in breast cancer cells.
• She conducts the following
“concentration / response” experiment.
*Following is from the research of Professor
Kathleen Arcaro (department of Veterinary and
Animal Sciences, Umass)
Picture of a plate
There are 48 “wells”
on each plate. Breast
cancer cells are grown in
each 32 of them (dark
circles).
24 of those
wells contain a
certain amount of PCB
(one of several types of
PCBs).
White circles are not part
of the experiment.
After growth, a
machine determines
the amount of estrogen
produced in each well.
Left two columns
are “controls”
(no PCB in these wells)
Experiment is repeated on 30 plates.
One way to think about the data:
Each plate provides one number
(we’ll see a better way to analyze the
data later):
Plate Number
1
2
3
4
5
etc
X (Overall Mean)
S (Overall Std Dev)
Average
Average Estrogen
Produced by
minus Control Cell Estrogen
production
cells exposed to PCBs
X1 = 1146.83
X2 = -95.67
X3 = -60.67
X4 = 160.17
X5 = 509.17
etc
516.24
1078.67
Summary of
all 30 plates.
Less estrogen in
PCB exposed cells.
12
More estrogen in
PCB exposed cells (relative to controls).
0
2
4
6
8
10
Each plate contributes
one data point to this
histogram.
-500
0
500
1000
1500
Estrogen Production: PCB Exposed Cells - Control Cells
This is an estimate of the distribution of X, not the distribution of X.
2000
Statistical Question:
• A x > 0 says that more estrogen is produced
by cells that are exposed to PCBs on
average.
• Observed x = 516.24, but there’s a fair
amount of “noise” in the data (s = 1078.67).
• Is this x statistically significantly positive?
(in other words: If the experiment were
repeated, how certain should we be that we’ll
see x>0 again?)
To answer the question, we need to know the
probability distribution of x
In general, the central limit theorem says that the
sample mean has a normal distribution (when
n>30 or so…)
X ~ N(m,s2/n)
We don’t know m or s, but we do have estimates:
x = 516.24, s = 1078.67
This gives us a good guess at the distribution of
X (i.e. A model for how X would behave if we
took another sample):
X ~ N(516.24, 1078.672/30)
0.0020
0.0010
0.0005
0.0
normal density
0.0015
Distribution of X-Bar
N(516.24,1078.672/30)
0
500
1000
X-bar
Informally, since 0 is in the “tail” of the distribution of X,
then seeing an x < 0 is unlikely
(i.e. these PCBs cause estrogen production on average)
More formally:
Confidence intervals for the mean
• A 1-a% confidence interval for the mean:
x +/- za/2s/sqrt(n)
x and s are the sample mean and sample
std deviation from n observations
za/2 = number such that Pr(Z<za/2) = 1-a/2
Rough Interpretation: interval covers
“inner” 1-a% of the best guess at x’s
distribution
95% of the
area under
the curve
Distribution of X
0.0010
0.0005
0.0
normal density
0.0015
0.0020
Observed x
0
130.44
500
920.44 1000
X-bar
95% confidence interval (a = 0.05)
516.24 +/- 1.96 (1078.67) / 5.48
516.24 +/- 385.80 = (130.44 , 902.44)
x +/- za/2s/sqrt(n)
•
95% confidence interval for
mean estrogen production in cells + PCB – mean
production in control cells
•
is (130.44 , 902.44).
Since 0 is not in this interval, there’s
“statistically significant” evidence (at 95%
confidence) that mean estrogen
production is higher for cells plus PCB.
Meaning of 1-a% confidence
interval:
We can be “1-a% confident” that the true
mean is in this interval…
Note that the
interval is the random
thing. The true mean
is a fixed number.
Order these from narrowest to widest:
90% confidence interval
99% confidence interval
95% confidence interval
80% confidence interval
Ordered from narrowest to widest:
Narrowest:
Widest:
80% confidence interval
90% confidence interval
95% confidence interval
99% confidence interval
Common sense: we can say with high confidence that the
mean is somewhere within an extremely wide interval…
Math: z0.005 is bigger than z0.1 since the subscript
describes a “tail” probability…
za/2 (quick reference)
• za/2 is a number such that
Pr(-za/2<Z<za/2)=1-a where Z~N(0,1).
• Examples:
Area 1-a
1-a
a/2
za/2
90%
5%
1.645
95%
2.5%
1.96
99%
0.5%
2.58
Note also:
Pr(|Z|<za/2) = 1-a
-za/2
area a/2
0
za/2
area a/2
95% of the
area under
the curve
Distribution of X
0.0010
0.0005
0.0
normal density
0.0015
0.0020
Observed x
0
130.44
500
920.44 1000
X-bar
95% confidence interval (a = 0.05)
516.24 +/- 1.96 (1078.67) / 5.48
516.24 +/- 385.80 = (130.44 , 902.44)
x +/- za/2s/sqrt(n)
Distribution of X-Bar
when mean
Is lower
limit
If true mean is at lower confidence
limit and the experiment is repeated,
the probability of seeing the observed
x or something larger is 0.025 (=a/2)
0.0010
This interpretation is more precise
because the true mean is not random,
the data are…
0.0005
Area 0.025 (=a/2)
0.0
normal density
0.0015
0.0020
A more precise interpretation of the
95% confidence interval
0
130.44
516.24
920.44 1000
If true mean is at upper confidence
limit, then if the experiment is repeated,
the probability of seeing the observed
x or something smaller is 0.025 (=a/2)
Distribution of X-Bar
when mean
Is upper
limit
0.0010
0.0005
Area 0.025 (=a/2)
0.0
normal density
0.0015
0.0020
Equivalently…
0
516.24
920.44 1000
Calculation:
Pr( X > observed x when lower limit is true mean)
=Pr(X > x)
Observed number
Random variable
=Pr( (X – lower limit)/(s/sqrt(n))
> (x – lower limit)/(s/sqrt(n))
=Pr(Z > [x – (x – za/2s/sqrt(n))]/[s/sqrt(n)] )
=Pr(Z > za/2) = a/2 by definition…
CONFIDENCE INTERVALS
FOR ESTIMATES OF PROPORTIONS
• Now that Pedro is health,, what are your expectations for the
Red Sox this season?
Will make the playoffs 59%
90 wins 16%
80 wins 14%
70 wins 11%
• Number polled (when I took poll) = 81
What’s a 90% confidence interval for proportion of people who think
they’ll make the playoffs?
Is 59% from 81 people significantly greater than ½ (at 90%
confidence level?)
Review: The CLT for proportions
Let Xi
= 1 if person i thinks the Sox will
make playoffs.
= 0 otherwise
where i=1,…,81 (indexes people polled).
Mean of Xi = p
Variance of Xi = p(1-p)
Xi~Bin(n=1,p)
p = (X1+…+X81)/81
By central limit theorem,
P ~ N(p,p(1-p)/n)
Confidence Intervals for Proportions
• 90% CI for proportion:
p +/- z0.10/2sqrt[p*(1-p)/n]
0.59 +/- 1.64 sqrt(0.59*0.41/81)
0.59 +/- 0.0895
[0.505, 0.6795]
(uses that p ~ N(p,p(1-p)/n)
Confidence Intervals for Proportions
• 90% CI for proportion:
p +/- z0.10/2sqrt[p*(1-p)/n]
0.59 +/- 1.64 sqrt(0.59*0.41/81)
0.59 +/- 0.0895
[0.505, 0.6795]
(uses that p ~ N(p,p(1-p)/n)
In general, a 1-a level confidence interval is:
estimator +/- za/2std error(estimator)
p=0.59
Area=0.05
2
4
Area=0.05
Lower limit
0.505
Upper limit
0.6795
0
normal density
6
Distribution of P
when true p = 0.59.
0.4
0.5
0.6
P-hat
0.7
0.8
p=0.59
2
4
Approximate
Area=0.05
Lower limit
0.505
0
normal density
6
Distribution of P
when true p = 0.505.
0.4
0.5
0.6
P-hat
0.7
0.8
p=0.59
Distribution of P
when true p = 0.505.
2
4
Approximate
Area=0.05
True p is in
mean and
variance
Lower limit
0.505
0
normal density
6
Approximate because
Distn of P is
N(true p, (true p)(1- true p)/n)
0.4
0.5
0.6
P-hat
0.7
0.8
If the true p is 0.505, how “rare” is
seeing the poll result of 59%?
• “Rarity” = Pr( P > 0.59 when true p = 0.505)
=Pr[ (P-0.505)/sqrt(.505*.495/81) >
(0.59-0.505)/sqrt(.505*.495/81)
=Pr( Z > 0.085/0.06) = Pr(Z>1.53) = 0.0630
p=0.59
2
4
Area=0.0630
Lower limit
0.505
0
normal density
6
Distribution of P
when true p = 0.505.
0.4
0.5
0.6
P-hat
0.7
0.8
If the true p is 0.5, how “rare” is
seeing and poll result of 59%?
• “Rarity” =
Pr( P > 0.59 when true p = 0.5)
=Pr[ (P-0.5)/sqrt(.5*.5/81) >
(0.59-0.5)/sqrt(.5*.5/81)
=Pr( Z > 0.09/0.06) = Pr(Z>1.62) = 0.0526
(This is the p-value!)
p=0.59
2
4
Area=0.0526
0.5
0
normal density
6
Distribution of P
when true p = 0.5
0.4
0.5
0.6
P-hat
0.7
0.8
Differences between means
• Can people discern a difference between Advil and
generic ibuprofen on average?
• Experiment:
“Double Blind Study”
– Randomly separate 80 people into two groups of 40 each.
– Give one group Advil and generic ibuprofen to the other
group.
– “Blinding”: person handing out pills and people being tested
don’t know which pill they’re getting (pills were dyed).
– People report their opinion of how well the pill works (on a
scale of 1 to 10).
– When we did the experiment, 2 people from the Advil group
dropped out.
When might this matter?
Results:
Generic Group
0
0
2
2
4
4
6
6
8
Advil Group
2
4
6
8
10
2
4
6
score
score
X Advil = 6.35
S Advil = 2.32
n Advil = 38
X Generic = 6.03
S Generic = 2.47
n Generic = 40
Is X Advil – X Generic = 0.32. Is this
“statistically significantly” greater than zero?
8
10
In General:
1-a confidence interval is
“Estimate +/- za/2 se(estimate)”
Variance of X Advil – X Generic
is s2Advil /n Advil + s2Generic /n Generic
Variance of X Advil – X Generic
is s2Advil /n Advil + s2Generic /n Generic
1-a confidence interval is
“Estimate +/- za/2 se(estimate)”
When nAdvil and nGeneric are both >30ish,
a 1-a confidence interval is
XAdvil–XGeneric+/-za/2 sqrt(s2Advil/nAdvil + s2Generic /nGeneric)
Not: sAdvil/sqrt(nAdvil) + sGeneric/sqrt(nGeneric)
95% CI for Advil / Generic Example
0.32 +/- 1.96 sqrt(5.38/38 + 6.10/40)
= [-0.74,1.38]
Can’t claim that people can discern a difference at 95% confidence.
Is there a level of confidence at which we could make the claim?
Suppose n = 1,000,000. Do you think we could claim a significant
difference then?