Transcript Two-Sample Inference Procedures14

Two-Sample
Inference
Procedures with
Means
Unit 13, Notes Page 208
Two-Sample Procedures
When we
with means compare,
•
what are we
The goal of these inferenceinterested
procedures is to compare the in?
responses to two treatments or
to compare the characteristics
of two populations.
• We have INDEPENDENT samples
from each treatment or
population
We will be
interested
in the
difference
of means,
so we will
use this to
find
standard
error.
Remember:
   
x y

x y
x
y
  
2
2
x
y
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
Purpose of Two-Sample
Procedures with Means
• 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 known/unknown
Formulas
Since in real-life, we
will NOT know both ’s,
we will do t-procedures.
Formula for CI:
Called
standard
error
CI  statistic  critical value SD of statistic
s
s
x  x   t *

n n
1
2
2
1
2
1
2
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



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)
Example: Two competing headache remedies claim to
give fast-acting 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
Describe the shape & standard error for sampling
distribution of the differences in the mean speed of
absorption. (answer on next screen)
Describe the sampling distribution of the differences
in the mean speed of absorption.
Normal distribution with S.E. = 3.316
Find a 95% confidence interval difference in
mean lengths of time required for bodily
absorption of each brand. (answer on next screen)
Assumptions:
State assumptions!
Think randomly
“Price isassigned
Right”!treatments
Have 2 independent
Given the absorption rate is normally distributed
’s unknown Closest without going
Formula & calculations
s12 s22 over
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
Calculator: STAT, TEST, 2-SampTInt
Conclusion
(Stats) in context
We are 95% confident that the true difference in mean
C-Level: .95, Pooled NO, (arrow down), Calculate.
lengths of time required for bodily absorption of each
Calculator gives: df = 21.53, and interval answer.
brand is between –5.685 minutes and 8.085 minutes.
Note: confidence interval
statements
• Matched pairs – refer to
“mean difference”
• Two-Sample – refer to
“difference of means”
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!
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
The length of time in minutes for the drugs
to reach a specified level in the blood was
recorded. The results follow:
Brand A
Brand B
mean
20.1
18.9
SD
8.7
7.5
n
12
12
Example: 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
Formula & Calculations
x1  x2
20Use:
.1  18
.
9
STAT, TEST,
t

.361 2-SampTTest
s12 s22
8.7 2 7.52 Pooled NO


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.
Example 3 - 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
Ha:O- M > 1
t
Where O is the true mean developing time
for original film & M is the true mean
developing time for modified film

x1  x2   1  2  6.3375  5.3375  1


0
s12 s22

n1 n2
1.51462 1.06362

8
8
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.
Homework:
Page 213