Download EGR252S14_Chapter9_Lecture2_v9th_ed 922

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Estimating the Difference Between
Two Means
Given two independent random samples, a
point estimate the difference between μ1 and
μ2 is given by the statistic
x1  x 2
We can build a confidence interval for μ1 - μ2
(given σ12 and σ22 known) as follows:
( x 1  x 2 )  z / 2
 12  22
 12  22

 1  2  ( x 1  x 2 )  z / 2

n1 n2
n1 n2
Note: Use this calculation for large n as well
(n>30).
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 1
Example: Estimating Difference Btwn
Two Means
A study was conducted and two types of
engines, A and B, were compared. Fifty
experiments were performed using engine A
and 75 using B. The average gas mileage
for A was 36 mpg, and 42 mpg for B. Find a
96% confidence interval on B-A where A
and B are population mean gas mileages for
engines A and B. Assume population
standard deviations are A = 6 and B = 8.
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 2
Example 9.10 Page 286
Find a 96% Confidence Interval
 xbarA = 36 mpg σA = 6 nA = 50
 xbarB = 42 mpg σB = 8 nB = 75
 α=0.04
α/2 =0.02 Z0.02 = 2.055
Calculations:
6 – 2.055 sqrt(64/75 + 36/50) < (μB - μA) < 6 + 2.055 sqrt(64/75 + 36/50)
Results:
3.4224 < (μB - μA) < 8.5776
96% CI is (3.4224, 8.5776)
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 3
Differences Between Two Means:
Variances Unknown
Case 1: σ12 and σ22 unknown but “equal”
Pages 287 and 288
( x1  x 2 )  t ( / 2,n1  n2 2 )S p
1 1
1 1
  1  2  ( x1  x 2 )  t ( / 2,n1  n2 2 )S p

n1 n2
n1 n2
2
2
(
n

1
)
S

(
n

1
)
S
1
2
2
Where, Sp2  1
n1  n2  2
Note v = n1 + n2 - 2
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 4
Differences Between Two Means:
Variances Unknown (Page 290)
Case 2: σ12 and σ22 unknown and not equal
( x 1  x 2 )  t / 2,
Where,
s12 s22
s12 s22

 1  2  ( x 1  x 2 )  t / 2,

n1 n2
n1 n2
(S12 / n1  S22 / n2 )2

 S2 / n 2   S2 / n 2 
 1 1  2 2 
 n1  1   n2  1 




WOW!
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 5
Estimating μ1 – μ2
Example (σ12 and σ22 known) :
A farm equipment manufacturer wants to compare the average
daily downtime of two sheet-metal stamping machines located in
two different factories. Investigation of company records for 100
randomly selected days on each of the two machines gave the
following results:
x1 = 12 minutes
x2 = 10 minutes
s12 = 12
s22 = 8
n1 = n2 = 100
Construct a 95% C.I. for μ1 – μ2
Since n is large, we can estimate σi2 with si2
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 6
Solution
( x 1  x 2 )  z / 2
 95% CI
 12  22
 12  22

 1  2  ( x 1  x 2 )  z / 2

n1 n2
n1 n2
α/2 = ?
Z.025 , Z.975 = 1.96
 (12-10) + 1.96*sqrt(12/100 + 8/100) = 2 + 0.8765
 1.1235 < μ1 – μ2 < 2.8765
 Interpretation: If CI contains 0, then μ1 – μ2 may be either
positive or negative (can’t say that one population mean
is larger than the other); however, since the CI for μ1 – μ2 is
positive, we conclude μ1 must be larger than μ2 .
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 7
μ1 – μ2 : σi2 Unknown
Example (σ12 and σ22 unknown but “equal”):
Suppose the farm equipment manufacturer was
unable to gather data for 100 days. Using the data
they were able to gather, they would still like to
compare the downtime for the two machines. The
data they gathered is shown below. Assume
population standard deviations are equal (12 = 22).
x1 = 12 minutes
s12 = 12
n1 = 18
x2 = 10 minutes
s22 = 8
n2 = 14
Construct a 95% C.I. for μ1 – μ2
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 8
Solution
Governing Equations:
( x 1  x 2 )  t / 2,n1 n2 2Sp
S
Calculations:
t0.025,30 = 2.042
2
p
1
1
1
1

 1  2  ( x 1  x 2 )  t / 2,n1 n2 2Sp

n1 n2
n1 n2
( n1  1) S12  (n2  1) S 22


n1  n2  2
sp2 = ((17*12)+(13*8))/30 = 10.267
sp = 3.204
2 + 2.042*3.204*sqrt(1/18 + 1/14) = 2 + 2.3314
-0.3314 < μ1 – μ2 < 4.3314
Interpretation:
Since this CI contains 0, we can’t conclude μ1 is significantly different
from μ2 .
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 9
Paired Observations
Suppose we are evaluating observations that
are not independent …
For example, suppose a teacher wants to
compare results of a pretest and posttest
administered to the same group of students.
Paired-observation or Paired-sample test …
Example: murder rates in two consecutive years
for several US cities. Construct a 90% confidence
interval around the difference in consecutive
years.
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 10
Calculation of CI for Paired Data
Example 9.13 We have 20 pairs of values. We calculate the difference for
each pair. We calculate the sample standard deviation for the difference
values. The appropriate equations are:
μd = μ1 – μ2
sd 
2
(
d

d
)
 i
n 1
Based on the data in Table 9.1 Dbar = -0.87
We determine that a (1-0.05)100% CI for μd is:
sd
d  t / 2,n 1(
)
n
Sd = 2.9773
n=20
-2.2634 < μd < 0.5234
Interpretation: If CI contains 0, then μ1 – μ2 may be
either positive or negative (can’t say that one is larger
than the other). Since this CI contains 0, we conclude
there is no significant difference between the mean
TCDD levels in the fat tissue.
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 11
C.I. for Proportions
 The proportion, P, in a binomial experiment may be

estimated by
X
P
n
where X is the number of successes in n trials.
 For a sample, the point estimate of the parameter is

x
p
n
 The mean for the sample proportion is    p
p
and the sample variance  2  pq

n
p
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 12
C.I. for Proportions
An approximate (1-α)100% confidence
interval for p is:

p  z / 2
 
pq
n
Large-sample C.I. for p1 – p2 is:


( p1  p2 )  z / 2




p1 q1 p2 q2

n1
n2
Interpretation: If the CI contains 0 …
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 13
Interpretation of the Confidence
Interval Significance
1. If the C.I. for p1 – p2 = (-0.0017, 0.0217), is
there reason to believe there is a significant
decrease in the proportion defectives using
the new process?
2. What if the interval were (+0.002, +0.022)?
3. What if the interval were (-0.900, -0.700)?
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 14
Determining Sample Sizes for Developing
Confidence Intervals
 Requires specification of an error amount е
 Requires specification of a confidence level
 Examples in text

Example 9.3 Page 273
•

Single sample estimate of mean
Example 9.15 Page 299
•
Single sample estimate of proportion
EGR 252 Ch. 9 Lecture2 9th ed.
MDH 2015
Slide 15
Related documents