Transcript Chapter 22 Confidence Intervals and Hypothesis Tests for the

Chapter 22 Comparing
2 Proportions
© 2006 W.H. Freeman and Company
Objectives (Chapter 22)
Comparing two proportions (comparing
ABILITY in two different contexts)
•
Comparing two independent samples
•
Large-sample CI for two proportions
•
Test of statistical significance
Point Estimator: pˆ 1  pˆ 2
• Two random samples are selected from two
populations.
• The number of successes in each sample is
recorded.
• The sample proportions are computed.
Sample 1
Sample size n1
Number of successes x1
Sample proportion
pˆ 1 =
x1
n1
Sample 2
Sample size n2
Number of successes x2
Sample proportion
pˆ 2 =
x2
n2
3
Comparing two independent samples
We often need to estimate the difference p1 – p2 between two unknown
population proportions based on independent samples. We can compute the
difference between the two sample proportions and compare it to the
corresponding, approximately normal sampling distribution model for pˆ 1  pˆ 2
Large-sample CI for two proportions
For two independent samples of sizes n1 and n2 with
sample proportion of successes pˆ1 and pˆ 2
respectively, an approximate level C confidence
interval for p1 – p2 is
( pˆ 1  pˆ 2 )  z *
pˆ 1 (1  pˆ 1 )
n1

pˆ 2 (1  pˆ 2 )
n2
w here z* is the appropriate value from
the z-table that depends on the
confidence level C
C is the area under the standard normal curve between −z* and z*.
Use this method when
npˆ 1  10, n (1  pˆ 1 )  10, npˆ 2  10, n (1  pˆ 2 )  10
Icing the Kicker: 95% Confidence Interval
Football coaches often employ the “icing the kicker” strategy. To ice the kicker the
opposing coach calls for a timeout just before the kicker attempts a field goal, hoping
that the delay interrupts the kicker’s concentration and causes him to miss the kick.
Standard error of the difference p1− p2:
SE =
SE =
pˆ 1 (1  pˆ 1 )
n1
0 . 797 ( 0 . 203 )


pˆ 2 (1  pˆ 2 )
n2
0 . 773 ( 0 . 227 )
= 0 . 0344
pˆ
Made FG
n
Timeout (icing) p1
157
197
79.7%
No Timeout p2
377
488
77.3%
The confidence
197
488
95% CI
(0.797  0.773)  1.96(0.0343)
interval
is ( pˆ 1  pˆ 2 )  z * SE
= 0.024  .0672
So the 95% CI is 0.024 ± 0.0672 = (0.0432, 0.0912)
We are 95% confident that the interval 4.32% to 9.12% captures the true difference in the
ABILITY of kickers to make a field goal when iced and their ABILITY to make a field goal when
not iced. Because 0 is in the interval, we do not have convincing evidence that there is a
significant difference in the ABILITY of kickers to make field goals when iced and when
not iced.
Example: 95% confidence interval for p1 – p2
The age at which a woman gives birth to her first child may be an important factor in the
risk of later developing breast cancer. An international study conducted by WHO selected
women with at least one birth and recorded if they had breast cancer or not and whether
they had their first child before their 30th birthday or after.
Cancer
Sample
Size
Age at
683
First Birth
> 30
3220
Age at
1498
First Birth
<= 30
10,245
The parameter to be estimated is p1 – p2.
p1 = cancer rate when age at 1st birth >30
p2 = cancer rate when age at 1st birth <=30
21.2%
pˆ 1
( pˆ 1  pˆ 2 )  1 . 96
pˆ 1 (1  pˆ 1 )

pˆ 2 (1  pˆ 2 )
n1
n2
14.6%
pˆ 2
We estimate that the cancer rate when
age at first birth > 30 is between .05
and .082 higher than when age <= 30.
(.212  .146 )  1.96
.212 (.788)
3220

.146 (.854 )
10, 245
.066  1.96 (.008) or .066  .016
(.05, .082 )
7
Beware!! Common Mistake !!!
A common mistake is to calculate a one-sample confidence interval for
p1, a one-sample confidence interval for p2,and to then conclude that
p1 and p2 are equal if the confidence intervals overlap.
This is WRONG because the variability in the sampling distribution for
pˆ 1  pˆ 2
from two independent samples is more complex and must take
into account variability coming from both samples. Hence the more
complex formula for the standard error.
SE =
pˆ 1 (1  pˆ 1 )
n1

pˆ 2 (1  pˆ 2 )
n2
INCORRECT Two single-sample 95% confidence intervals:
The confidence interval for the rightie BA and the confidence
interval for the leftie BA overlap, suggesting no significant
difference between Ryan Howard’s ABILITY to hit righthanded pitchers and his ABILITY to hit left-handed pitchers.
Hits AB
Rightie interval: (0.274, 0.366)
phat(BA)
Rightie 126
394 .320
Leftie 50
222 .225
Leftie interval: (0.170, 0.280)
C O R R E C T T he 2-sam ple 95% confidence interval of the form
( p R  p L )  1.96
p R (1  p R )
nR

p L (1  p L )
nL
for the difference p R  p L betw een the A B IL IT IE S
is (.023, .167). Interval is entirely positive, su ggestin g sign ifican t d ifferen ce
betw een H ow ard's A B IL IT IE S to hit righties and lef ties
(evidence that p R is larger than p L ).
0 .023
.095
.167
Reason for Contradictory Result
It's alw ays true that
a  b 
pˆ 1 (1  pˆ 1 )
n1
a 

b . S pecifically,
pˆ 2 (1  pˆ 2 )

pˆ 1 (1  pˆ 1 )
n2
n1

pˆ 2 (1  pˆ 2 )
n2
SE ( pˆ 1  pˆ 2 )  SE ( pˆ 1 )  SE ( pˆ 2 )
10
Hypothesis Tests for p1  p2
If the null hypothesis is true, then we can rely on the properties of the
sampling distribution of pˆ1  pˆ 2 to estimate the probability of selecting 2
samples with proportions pˆ1 and pˆ 2
S am pling distribution of pˆ 1  pˆ 2
H 0 : p1  p 2 = 0
Ha
( th at is, p1 = p 2 = p )
w hen H 0 : p1  p 2 = 0 is true.
 0

: p1  p 2   0
 0

 1
1 
p (1  p ) 


 n2 n2 
O u r b est estim ate o f p is pˆ ,
th e p o o led sam p le p ro p o rtio n
pˆ =
to tal su ccesses
=
to tal o b servatio n s
z =
co u n t 1  co u n t 2
n1  n 2
pˆ 1  pˆ 2
 1
1 
pˆ (1  pˆ ) 


n2 
 n2
This test is appropriate when
npˆ 1  10, n (1  pˆ 1 )  10, npˆ 2  10, n (1  pˆ 2 )  10
=0
Do NFL Teams With Domes Have an Unfair Advantage?
Do NFL teams that play their home games in a dome have an
advantage over teams that do not play their home games in a dome?
Home Home
Wins Games
Home
field is
dome
50
Home
field is
not dome
103
79
The parameter to be estimated is p1 – p2.
p1 = home win rate for dome teams
p2 = home win rate for non-dome teams
.633
pˆ 1
187
.551
pˆ 2
Do not reject H0 :p1 – p2 = 0. There is
no significant difference between the
home win rate for dome teams and the
home win rate for non-dome teams.
50  103
: p  p = 0 pˆ =
= . 575
0 1
2
79  187
H :p  p 0
A 1
2
H
. 575 (1 . 575 ) . 575 (1 . 575 )

= . 066
79
187
(. 633  . 551 )  0
:z =
= 1 . 24
. 066
SE ( pˆ  pˆ ) =
1
2
Test Statistic
P  va lu e :
P ( z  1 .2 4 ) = .1 0 7 5
12
Gastric Freezing
Gastric freezing was once a treatment for ulcers. Patients would
swallow a deflated balloon with tubes, and a cold liquid would be
pumped for an hour to cool the stomach and reduce acid production,
thus relieving ulcer pain. The treatment was shown to be safe,
significantly reducing ulcer pain, and was widely used for years.
A randomized comparative experiment later compared the outcome of gastric freezing
with that of a placebo: 28 of the 82 patients subjected to gastric freezing improved,
while 30 of the 78 in the control group improved.
H0: pgf - pplacebo = 0
Ha: pgf - pplacebo > 0
pˆ gf =
28
pgf = proportion that receive relief from gastric freezing
pplacebo = proportion that receive relief using a placebo
= .3 4 1
P  value = P ( z   0.499) = .69
82
pˆ placebo =
pˆ pooled =
30
= .3 8 5
78
28  30
82  78
= 0 .3 6 2 5
z=
pˆ gf  pˆ placebo
 1
1 
pˆ (1  pˆ )  

 n1 n 2 
=
0.341  0.385
1 
 1
0.363 * 0.637 


 82 78 
=
 0.044
=  0.499
0.231 * 0.025
Conclusion: The gastric freezing was no better than a placebo (P-value 0.69), and this
treatment was abandoned. ALWAYS USE A CONTROL!