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Ch' 8 Estimation Ch' 9 Hypothesis Testing Confidence Intervals

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Sample members are chosen in pairs, one member from each population. ... Duncan. New. Rogaine. d. name. 95% Confidence Interval. 417. 95% Confidence Interval. n=6 ... – PowerPoint PPT presentation

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Title: Ch' 8 Estimation Ch' 9 Hypothesis Testing Confidence Intervals


1
Ch. 8 EstimationCh. 9 Hypothesis
TestingConfidence Intervals Hypothesis
Tests differences between means of normal
populations1. Matched pairs/ dependent
samples2. Independent samples
2
Matched pairs/ dependent samples
  • Sample members are chosen in pairs,
  • one member from each population.
  • Members of a pair should be
  • very similar to one another
  • OR
  • the same member before and after
  • a specific event.

3
Paired t-test
  • We have n matched pairs of observations,
  • (x1 ,y1 ), (x2 ,y2 ), , (xn ,yn )
  • from populations with mean mx and my .
  • Consider the n differences di xi yi
  • d sample mean of differences
  • sd sample standard deviation of differences
  • md mx - my

_
4
Paired t-test
A drug company wants to compare the effect of a
new drug for treating male-pattern baldness with
Rogaine. The new drug is applied to a 4
square inch area on the left side of the scalp of
six bald men, and Rogaine is applied to an equal
area on the right. Find the 95 confidence
interval.
5
Confidence Interval for matched pairs
L
,
U
d
t
s



_
a
/
2
d
di Xi new - yi Rogaine
for i 1, . . . , n.
We need normality for small sample problems.
6
95 Confidence Interval
df n 1
7
95 Confidence Interval
New
Rogaine
d
name
202
185
17
Joe
314
251
63
Fred
84
25
59
Harry
578
412
166
Chester
874
414
460
Max
602
444
158
Duncan
8
95 Confidence Interval
_
d 153.833
n6 dfn-1 df5
9
95 Confidence Interval
df n?1 5
L
,
U
d
t
s



_
/
2
a
d

_

L
,
U
153
.
833
2
.
571
65
.
80
(
)

(-15.27 , 322.94)
10
Difference between pop. means Independent
samples
Use the differences between sample means as the
base point for inference.
_
_
_
_
E(X-Y) E(X) E(Y) mX - my
11
_
_
If X and Y are normally distributed, so is (X
Y)
_
_
All linear combinations of jointly normally
distributed random variables are always normally
distributed.
12
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15
In general,
_
_
_
_
_
_
Var(X-Y) Var(X) Var(Y) 2 Cov(X,Y)
However, random sampling ? independence so
covariance 0
_
_
_
_
Var(X-Y) Var(X) Var(Y)
16
For independent X and Y
_
_
_
_
Var(X-Y) Var(X) Var(Y) s2 s2
___
___
x
y
ny
nx
________
sx-y s2 s2
__
__
v
nx
ny
17
Difference between pop. means Independent
samples, s known
C
.
V
.
,
C
.
V
.

L
U
(
)
z
m
-
m
s

_

2
/
2
x
x
1
0
a
-
1
2
18
Difference between pop. means Independent
samples, s known
U.S. companies claim that the Japanese are
dumping steel in the U.S. by selling it at a
lower average price than the average price in
Japan. The price of Japanese steel is normally
distributed and has a known variance of 180 yen
in Japan and 160 yen in U.S. at current rate of
exchange. A sample of 10 sales in the U.S. had
an avg. price of 200 yen while 20 sales in Japan
averaged 210 yen. At the 1 level of
significance, test the null hypothesis that Japan
is dumping steel in the U.S.
19
H

m

m


0
J
a
p
a
n
U
.
S
.
H

m
gt
m


1
J
a
p
a
n
U
.
S
.
H

0
m
-
m



0
J
a
p
a
n
U
.
S
.
H

0
m
-
m
gt


1
J
a
p
a
n
U
.
S
.
raw score test statistic
average price difference
X
X
-

J
a
p
a
n
U
S
20
Under the effective null hypothesis
mean
variance
21
A sample of 10 sales in the U.S. show an
average price of 200 yen while a sample of 20
sales in Japan average 210 yen.
raw score space
? 0.01
standardized space
? 0.01
z
22
First, carry out the test in standardized space
? 0.01
z
For??? 0.01, the normal table yields Z? 2.33
Z 2 falls to the left of Z? 2.33 . Do not
reject the null hypothesis
23
Next, carry out the test in raw score space
? 0.01
24
Testing in raw score space (continued)
? 0.01
10 lt 11.65 raw score test statistic value lt
critical value (C.V.) Do not reject the null
hypothesis.
25
Now, carry out the test using the p-value
Normal table shows P (0 lt z lt 2) 0.4772 so
for a one-tail test the p-value 0.0228 .
Z 2 is the standardized test statistic value.
The p-value is the probability of observing what
you observe or something more extreme.
Since p-value gt ? , then do not reject null
hypothesis.
26
Difference between pop. meansIndependent
samples, s unknown (use s)
Domino Pizza is choosing between the Ford Escort
and Honda Civic. If the difference for
time-in-repair after 2 years is more than 15 days
in favor of Honda, Honda will be cheaper. At
a0.1, test the null hypothesis that the
time-in-repair difference is less than or equal
to 15 days. Ford nf 100, sf 8.6, xf
26 Honda nh 100, sh 7.2, xh 10
_
_
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28
1.122
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