Comparing Two Population Treatments

1
Comparing Two Population Treatments
Types of Problems Does the group receiving blood
pressure medication have lower blood pressure
than the control group receiving a placebo? Is
the mean grade point of students who have
part-time jobs in high school less than students
who dont have jobs? Do students in engineering
pay more than other students for text books in
College?
2
Two Means Unknown s
Population 1
Population 2
µ1 s1
µ2 s2
Sampling Dist.
Sampling Dist.
Sampling Dist.
A T-Distribution Applies
3
Two Means
4
Independent Samples
One Sample Independent Knowing the outcome of
one selection, does not change the probability of
the outcome of another selection.Example
Suppose probability of selecting an Obama
supporter at random is Seattle is 70. Suppose
that the probability of a spouse supports Obama
is 85 if the partner supports Obama. If the
spouse and partner are both selected for the
sample, the sample is not independent. Two
Samples Independent Knowing the outcome of a
selection from one sample does not change the
probability of the outcome from a selection from
the other sample. Example Suppose samples of
men and women are taken to evaluate the frequency
of arthritis. Suppose the probability of having
arthritis is 5. Suppose that the probability of
having arthritis is 23 if a family member is
known to have the disease. Two samples would not
be independent if members of one sample had
family members in the other sample.
5
Brain Size Problem
Solving with Add-in ProgramsFind
PRGM7STDERROR52-MEANSSx1117.5n1152Sx2111
.3n2139 SE 13.4146 df288.636
6
Brain Size Problem (Continued)
t-sampling distribution
p-value
0
PRGM8TDISTLOWER BOUND -EE99UPPER BOUND
-45.10MEAN 0SE 13.4146df288.636P-VALUE
(ONE TAIL) 4.39E-4 0.000439t-3.3602Since
P-value is less than .05, we reject the null
hypothesis and conclude children with ADHD have
smaller brains.
7
Brain Size Problem (Continued)
Solve with Black Box ProgramSTAT
TESTS42-SampTTestInput Statsx(bar)1
1059.4Sx1 117.5n1 152x(bar)2 1104.5Sx2
111.3n2 139µ1ltµ2Pooled NoCalculate-
Entert-3.3602p(p-value)4.39E-4
0.000439Since P-value is less that .05, we
reject the null hypothesis
8
Brain Size ProblemConfidence Interval
Find the 95 confidence interval for the
difference in brain sizes, with ADHD without
ADHD.Solve using TI-83 Add-in Programs
We are 95 confident that the brain size of
children with ADHD is between 71.5 and 18.7
milliliters less than brains of children without
ADHD.
9
Brain Size ProblemConfidence Interval (Cont.)
Find the 95 confidence interval for the
difference in brain sizes using the TI-83 Black
Box Program. STAT TESTS02-SampTIntervalInput
Statsx(bar)11059.4Sx1 117.5n1
152x(bar)21104.5Sx2 111.3n2
139C-Level.95Pooled NoCalculate
Enter(-71.5, -18.7)
10
Problems
11
Problems
12
Problems
13
Paired Samples
Benefits of Ultrasound StudyStudy is done to
measure the effectiveness of an ultrasound and
stretch therapy for knee injuries.
We have two sets of data, but we cannot do a
2-Sample T-Test because the data is not
independent. The same subjects relate to both
the before and after treatments. When sets of
data have a natural relationship, we can use a
paired sample T-Test, which is usually a better
study procedure because it better isolates
exogenous variables.
14
Ultra Sound Study Cont.

• For a paired sample test, we calculate the
  difference set, Pre-treatment Post-treatment.
  We then do a 1-Sample T-Test or a 1-Sample
  T-Interval on the difference set.
• Hypothesis TestHo µd 0 (There is no
  improvement)HA µdlt0 (There is improvement,
  treatment works)
• Conditions for a Paired Sample Test
• 1) The data in not independent. The data is
  naturally related. Each set has the same
  number of samples.
• The paired data samples are a random sample
• The paired data samples are independent of each
  other
• The number of paired samples is large enough
  (n30 or sample is normal)

15
Ultra Sound Study Cont.
Calculating the P-value using TI-83 Add-in
Program Enter the Pre-treatment data in L1 and
the Post-treatment data in L2PRGMPARDSAMP2NO
(Lists are not independent, they are paired)LIST
1 2nd L1LIST 2 2nd L21YES (We need to
check the probability plot to see if the sample
difference set is normal. It is close
enough.ENTER (Cancels display of
Plot)2T-Test-1 (The Alt. Hypothesis is
µdlt0)P-value is .0207. Since this is less than
5, we reject the null hypothesis and accept the
alternate hypothesis that the treatment
works. Find the 90 confidence interval for the
ud.Enter (Returns to program menu)1C
INTERVALC LEVEL .9090 confidence level is
(-6.00, -.85)ENTER (Returns of program menu)3
QUIT (Closes program)
16
Problems
17
Problems

• Is there convincing evidence that the mean
  cost-to charge ratio for Oregon hospitals is
  lower for outpatient care? Write the appropriate
  hypothesis, find the p-value and state your
  conclusion. (a.05)
• Find the 90 confidence interval for the
  differences in the 2002 inpatient ratio and the
  outpatient ratio.

18
Problems
19
Difference between 2 Population Proportions
20
Two Proportions Unknown p
Population 1
Population 2
p1 s1 n 1
p2 s2 n1
Sampling Dist. p2
Sampling Dist. p1
Sampling Dist. p1 p2
The Normal Model Applies
21
Two ProportionsConditions for Normal Model
The sampling distribution for the difference of
two sample proportions is a normal model
if Both samples are random Samples are
independent of each other Each sample has ten
successes and 10 failures
22
Difference between 2 Population Proportions
For a hypothesis test, the null hypothesis
isHo p1 p2. If the two proportions are
equal, then the standard deviation of the
proportions must be also equal. Since p1 seldom
equals p2, the SE(p1-p2) does not reflect this
fact. To get around this problem for hypothesis
tests, we calculate the weighted average of p1
and p2, called pcombined pc, and substitute is
value for both p1 and p2 in the standard error
calculation.
23
Wart Problem (1)
Normal Sampling Distribution
0
p1-p2 60/100 88/104 -.2462
24
Wart Problem (2)
Solve using TI-83 Add-in Programs PRGM7STDERROR
32-PROP COMBINDp160/100n1100p288/104n210
4SE(PCombined).0625 PRGMNORMDIST1LOWER
BOUND- 2ND EE99UPPER BOUND-.2462MEAN0STNDRD
DEVIATN.0625PROBABILITY 4.0894E-5
.00004089Z ( -.2462-0)/.0625 -3.9440 Since
the P-value is less than 5, we reject the null
hypothesis and conclude that the duct tape
treatment works best.
25
Wart Problem (2)
Solving the Problem using TI-83 Black Box
Program STAT TESTSB 6-2PropZTestx160
(number of success in p1 as integer)n1100x288
(number of success in p2 as integer)n2104p1ltp2
p4.1041E-5 .00004104 z-3.9383
26
Wart Problem (3)
Find the 90 confidence interval for the
difference in the true proportions.Solve using
TI-83 Add-in Programs C.I. p1-p2 Z
SE(p1-p2) PRGM CRITVAL1NORMAL
DISTRIBUTIONCONFIDENCE LEVEL .90CR VALUE
1.6449PRGM STDERROR2 2-PROPp160/100n1100
p288/104n2104SE(p1-p2).0604 C.I. p1-p2
Z SE(p1-p2)C.I. .60/100 88/104 1.6449
.0604C.I. (-.346, -.147)
27
Wart Problem (4)
Find the 90 confidence interval for
the difference in the true proportions.Solve
using TI-83 Black Box Program.Stat
TestsB2-PropZIntx160 (The number of successes
in p1)n1100x288 (The number of successes in
p2) N2104C-Level .90C.I. (-.346, -.147) We
are 90 confident that the difference in the true
proportions, p1-p2, is in the above interval. We
can also conclude that we are 90 sure that the
success rate of the duct tape treatment is better
than Liquid nitrogen freezing.
28
Problems
29
Problems
30
Problems