410 Ch. 8 Estimation Ch. 9 Hypothesis Testing
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Transcript of 410 Ch. 8 Estimation Ch. 9 Hypothesis Testing
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Ch. 8 EstimationCh. 9 Hypothesis Testing
Confidence Intervals & Hypothesis Tests:
differences between means
of normal populations
1. Matched pairs/ dependent samples
2. Independent samples
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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 anotherOR
the same member “before” and “after” a specific event.
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Paired t-test
We have n matched pairs of observations,
(x1 ,y1 ), (x2 ,y2 ), … , (xn ,yn )
from populations with mean µx and µy .
Consider the n differences di = xi – yi
d = sample mean of differences
sd = sample standard deviation of differences
µd = µx - µy
_
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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.
Paired t-test
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L,U = d tα/2sd
Confidence Intervalfor matched pairs:
di = Xi new - yi Rogainefor i = 1, . . . , n.
+_
We need normality for small sample problems.
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sd =sd
n
sd =di − d ( )2
i=1
n∑
n −1
df = n – 1
95% Confidence Interval
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202
314
84
578
874
602
185
251
25
412
414
444
17
63
59
166
460
158
Joe
Fred
Harry
Chester
Max
Duncan
New Rogaine dname95% Confidence Interval
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sd = 161.165
sd = 65.80
95% Confidence Interval
n=6df=n-1df=5
d = 153.833_
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L,U = d tα/2sd
L,U = 153.833 2.571 65.80( )
95% Confidence Intervaldf = n−1 = 5
+_
+_
= (-15.27 , 322.94)
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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) = µX - µy
_ _ _ _
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All linear combinationsof jointly normally distributedrandom variables are always
normally distributed.
If X and Y are normally distributed,
so is (X – Y)
_ _
_ _
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µ1µ 2 x 1x 2
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x 1 − x 2µ1 − µ 2
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x 1 − x 2µ1 − µ2( )0acceptance region rejection
regionrejection region
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In general,
However,random sampling independence
so covariance = 0
Var(X-Y) = Var(X) + Var(Y) – 2 Cov(X,Y)
Var(X-Y) = Var(X) + Var(Y)_ _ _ _
_ _ _ _ _ _
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For independent X and Y:
Var(X-Y) = Var(X) + Var(Y)
= σ2 + σ2
_ _
___ ___
_ _
x y
nx ny
σx-y = σ2 + σ2__ __nx ny
________
√
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C.V.L
,C.V.U =
µ1 − µ2( )
0zα/2σx
1−x 2
+_
Difference between pop. means:Independent samples, σ known
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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.
Difference between pop. means:Independent samples, σ known
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H0: µJapan ≤ µU.S.
H1: µ Japan > µU.S.
H0: µJapan − µU.S. ≤ 0
H1: µJapan − µU.S. > 0
raw score test statistic:
X Japan − X
US
average pricedifference
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Under the “effective” null hypothesis:
E X
Japan− X
US( )= µJapan
− µUS
= 0
X Jap an −X US
2
σ =X Japan
2
σ +X US
2
σ
mean:
variance:
σ
X Japan − X US
2 = Japan
2σn
Japan
+ US
2σn
US
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X Japan
− X US
A sample of 10 sales in the U.S. showan average price of 200 yen while a sampleof 20 sales in Japan average 210 yen.
α = 0.01
0 C.V.
raw score space:
0 Zα
α = 0.01
z
standardized space:
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α = 0.01
z
First, carry out the test in standardized space:
For α = 0.01, the normal table yields Zα = 2.33
z =x
Japan− x
US( )− µJapan
− µUS( )
σJapan
2
nJapan
+ σUS
2
nUS
z =210 − 200( )− 0( )
180
20+160
10
= 2
Z = 2falls to the left of
Zα = 2.33 .
Do not rejectthe null hypothesis
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X Japan
− X US
α = 0.01
0 C.V.
Next, carry out the test in raw score space:
C.V .= (µ
Japan− µ
US) + Zα
σJapan
2
nJapan
+σ
US
2
nUS
C.V .= (0) + 2.33
180
20+
160
10= 11.65
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X Japan
− X US
α = 0.01
Testing in raw score space (continued):
X Japan
− X US
=210 −200 = 10
10 < 11.65raw score test statistic value < critical value (C.V.)
Do not reject the null hypothesis.
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= 2.33
α = 0.01
= 2.33Z = 2
p-value = 0.0228
Since p-value > α , then do not reject null hypothesis.
Now, carry out the test using the p-value:Normal table shows P (0 < z < 2) = 0.4772 so for a one-tail test the p-value = 0.0228 .
Z = 2 is thestandardizedtest statisticvalue.
The p-value is the probabilityof observing what youobserve or somethingmore extreme.
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Difference between pop. means:Independent samples, σ 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 α=0.1, 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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H0 :µFord – µHonda ≤ 15
H1:µFord – µHonda > 15
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s x 1 − x 2=
8.62
100+
7.22
100
sx 1 − x 2= s1
2
n1+ s2
2
n2
= 1.122
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H0 :µFord – µHonda ≤ 15
H1:µFord – µHonda > 15
C.V. = 15 + 2.33 1.122( )C.V. = 17.614
C.V. = µF −µH( )0 + tαsx 1−x 2
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acceptance region rejection region
15
17.614
x F − x H
16