Download Two-Sample Inference Procedures

Two-Sample
Inference
Procedures with
Means
Remember:
   
x y

x y
x
y
  
2
2
x
y
We will be
interested
in the
difference
of means,
so we will
use this to
find
standard
error.
Suppose we have a population of
adult men with a mean height of
71 inches and standard deviation
of 2.6 inches. We also have a population of
adult women with a mean height of 65 inches
and standard deviation of 2.3 inches. Assume
heights are normally distributed.
Describe the distribution of the difference in
heights between males and females (malefemale).
Normal distribution with
x-y =6 inches & x-y =3.471 inches
Female
65
Male
71
Difference = male - female
6
 = 3.471
a) What is the probability that the
height of a randomly selected man is
at most 5 inches taller than the
height of a randomly selected
woman?
P((xM-xF) < 5) = normalcdf(-∞,5,6,3.471) = .3866
b) What is the 70th percentile for the
difference (male-female) in heights
of a randomly selected man &
woman?
(xM-xF) = invNorm(.7,6,3.471) = 7.82
a) What is the probability that the
mean height of 30 men is at most 5
inches taller than the mean height of
30 women?
P((xm – xw)< 5) = .0573
b) What is the 70th percentile for the
difference (male-female) in mean
heights of 30 men and 30 women?
6.332 inches
Two-Sample Procedures
When we
with means compare,
what are we
interested
in?
• The goal of these inference
procedures is to compare the
responses to two treatments or to
compare the characteristics of two
populations.
• We have INDEPENDENT samples
from each treatment or population
Assumptions:
• Have two SRS’s from the populations or
two randomly assigned treatment groups
• Samples are independent
• Both distributions are approximately
normally
– Have large sample sizes
– Graph BOTH sets of data
•
’s unknown
Hypothesis Statements:
H0: 1 =
- 2 = 0
Ha:
Ha:
H
Haa::
1<- 22 < 0
1>- 22 > 0
11 -≠ 22 ≠ 0
Be sure
to define
BOTH 1
and 2!
Formulas
Since in real-life, we
will NOT know both ’s,
we will do t-procedures.
Hypothesis Test:
Test statistic 
Since we usually
assume H0 is true,
statistic
parameter
then this equals 0 –
can usually
SDsoofwestatistic
leave it out
 x  x      
t
1
2
1
2
2
1
2
1
2
s s

n n
2
Degrees of Freedom
Option 1: use the smaller of the two
values n1 – 1 and n2 – 1
This will produce conservative
results – higher p-values & lower
confidence.
Calculator
Option 2: approximation used bydoes this
automatically!
technology
s s 
2
2
1
2
1
2
2
  
n n 

df 
1 s 
1 s
  

n  1 n  n  1 n
1
2
2
1
2
1
2
2



If there were such a thing as a
personality test, do you think
that the guys’ personalities
would be different from the
girls?
VS
Dr. Phil’s Survey
Two competing headache remedies claim to give fastacting relief. An experiment was performed to
compare the mean lengths of time required for bodily
absorption of brand A and brand B. Assume the
absorption time is normally distributed. Twelve people
were randomly selected and given an oral dosage of
brand A. Another 12 were randomly selected and given
an equal dosage of brand B. The length of time in
minutes for the drugs to reach a specified level in the
blood was recorded. The results follow:
mean
SD
n
Brand A
20.1
8.7
12
Brand B
18.9
7.5
12
Is there sufficient evidence that these drugs differ in
the speed at which they enter the blood stream?
Have 2 independent randomly assigned treatments
State assumptions!
Given the absorption rate is normally
distributed
’s unknown
H0: A= B
Hypotheses & define variables!
Where A is the true mean absorption time
for Brand A & B is the true mean
absorption time for Brand B
Ha:A= B
x1  x2
20.1  18.9
t

 .361& calculations
Formula
s12 s22
8.7 2 7.52


n1 n2
12
12
Conclusion in context
p  value  .7210 df  21.53 α  .05
Since p-value > a, I fail to reject H0. There is not
sufficient evidence to suggest that these drugs differ in
the speed at which they enter the blood stream.
Suppose that the sample mean of Brand
B is 16.5, then is Brand B faster?
t
x1  x2
s12 s22

n1 n2

20.1  16.5
8.7 2 7.52

12
12
 1.085
p  value  .2896 df  21.53 α  .05
No, I would still fail to reject the null
hypothesis.
A modification has been made to the process
for producing a certain type of time-zero film
(film that begins to develop as soon as the
picture is taken). Because the modification
involves extra cost, it will be incorporated only
if sample data indicate that the modification
decreases true average development time by
more than 1 second. Should the company
incorporate the modification?
Original 8.6 5.1 4.5 5.4
Modified 5.5 4.0 3.8 6.0
6.3 6.6
5.8 4.9
5.7 8.5
7.0 5.7
Assume we have 2 independent SRS of film
Both distributions are approximately normal due to
approximately symmetrical boxplots
’s unknown
H0: O- M = 1
Where O is the true mean developing time
for original film & M is the true mean
developing time for modified film
Ha:O- M > 1
t

x1  x2   1  2  6.3375  5.3375  1


0
s
s

n1 n2
2
1
2
2
1.5146
1.0636

8
8
2
p  value  .5 df  7   .05
Since p-value > , I fail to reject H0. There is not
sufficient evidence to suggest that the company
incorporate the modification.
2
Confidence
Called
intervals:
standard
error
CI  statistic  critical value SD of statistic
s
s
x  x   t *

n n
1
2
2
1
2
1
2
2

Two competing headache remedies claim to give fastacting relief. An experiment was performed to
compare the mean lengths of time required for bodily
absorption of brand A and brand B. Assume the
absorption time is normally distributed. Twelve people
were randomly selected and given an oral dosage of
brand A. Another 12 were randomly selected and given
an equal dosage of brand B. The length of time in
minutes for the drugs to reach a specified level in the
blood was recorded. The results follow:
mean
SD
n
Brand A
20.1
8.7
12
Brand B
18.9
7.5
12
Find a 95% confidence interval difference in
mean lengths of time required for bodily
absorption of each brand.
Assumptions:
State assumptions!
Thinkrandomly
“Price assigned
is Right”!
Have 2 independent
treatments
Given the absorption rate is normally distributed
’s unknown Closest without going
over
& calculations
s12 s22 Formula
x1  x2   t *


df  21.53
n1 n2
2
2
8.7 7.5
20.1  18.9  2.080

 (5.685,8.085)
12
12
From calculator df =
Conclusion in context
We are 95% confident that the true difference in mean
21.53, use t* for df =
lengths of time required for bodily absorption of each
21 & 95% confidence
brand is between –5.685 minutes and 8.085 minutes.
level
In an attempt to determine if two competing
brands of cold medicine contain, on the average,
the same amount of acetaminophen, twelve
different tablets from each of the two
competing brands were randomly selected and
tested for the amount of acetaminophen each
contains. The results (in milligrams) follow.
Brand A
517, 495, 503, 491
503, 493, 505, 495
498, 481, 499, 494
Brand B
493, 508, 513, 521
541, 533, 500, 515
536, 498, 515, 515
Compute a 95% confidence interval for the mean difference in
amount of acetaminophen in Brand A and Brand B.
Input Brand A data into L1 and Brand B data into
L2.
I am computing a 2-sample T interval for means at
a 95% confidence level.
Confidence level: (-28.48, -7.189) with 17 df
I am 95% confident that the true difference in
the mean amount of acetaminophen in Brand A is
between 28.5 and 7.2 mg lower than in Brand B.
Note: confidence interval
statements
• Matched pairs – refer to
“mean difference”
• Two-Sample – refer to
“difference of means”
Pooled procedures:
• Used for two populations with the
same variance
• When you pool, you average the
two-sample variances to estimate
the common population variance.
• DO NOT use on AP Exam!!!!!
We do NOT know the variances of the population,
so ALWAYS tell the calculator NO for pooling!
Robustness:
• Two-sample procedures are more
robust than one-sample procedures
• BEST to have equal sample sizes! (but
not necessary)