Ch 11 實習 (2)
A Two - Tail Test
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Example 11.2
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2
AT&T has been challenged by competitors
who argued that their rates resulted in lower
bills.
A statistics practitioner determines that the
mean and standard deviation of monthly longdistance bills for all AT&T residential customers
are $17.09 and $3.87 respectively.
Jia-Ying Chen
A Two - Tail Test
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Example 11.2 - continued
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3
A random sample of 100 customers is selected
and customers’ bills recalculated using a
leading competitor’s rates (see Xm11-02).
Assuming the standard deviation is the same
(3.87), can we infer that there is a difference
between AT&T’s bills and the competitor’s bills
(on the average)?
Jia-Ying Chen
A Two - Tail Test
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Solution

Is the mean different from 17.09?
H0: m = 17.09
H1 : m  17.09
– Define the rejection region
z   z / 2 or z  z / 2
4
Jia-Ying Chen
A Two – Tail Test
Solution - continued
/2 = 0.025
x
/2 = 0.025
17.09
If H0 is true (m =17.09), x can still fall far
above or far below 17.09, in which case
we erroneously reject H0 in favor of H1
(m  17.09)
5
x
We want this erroneous
rejection of H0 to be a
rare event, say 5%
chance.
Jia-Ying Chen
A Two – Tail Test
Solution - continued
z=
/2 = 0.025
xm

=
n
17.55  17.09
= 1.19
3.87 100
17.55
x
17.09
x
From the sample we have:
/2 = 0.025
/2 = 0.025
/2 = 0.025
x = 17.55
-z/2 = -1.96
6
0 z/2 = 1.96
Rejection region
Jia-Ying Chen
A Two – Tail Test
There is insufficient evidence to infer that there is a
difference between the bills of AT&T and the competitor.
Also, by the p value approach:
The p-value = P(Z< -1.19)+P(Z >1.19)
= 2(.1173) = .2346 > .05
/2 = 0.025
z=
7
xm

n
/2 = 0.025
-1.19 0 1.19
=
17.55  17.09
= 1.19
-z/2 = -1.96
z/2 = 1.96
3.87 100
Jia-Ying Chen
Example 1
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8
A machine that produces ball bearings is set
so that the average diameter is 0.5 inch. A
sample of 10 ball bearings was measured
with the results shown here. Assuming that
the standard deviation is 0.05 inch, can we
conclude that at the 5% significance level that
the mean diameter is not 0.5 inch?
0.48 0.50 0.49 0.52 0.53 0.48 0.49 0.47
0.46 0.51
Jia-Ying Chen
Solution
9
Jia-Ying Chen
Calculation of the Probability of a
Type II Error
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To calculate Type II error we need to…
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型二誤差的定義是，H1 正確卻無法拒絕H0
在什麼規則下你無法拒絕H0
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10
express the rejection region directly, in terms of the
parameter hypothesized (not standardized).
specify the alternative value under H1.
單尾
雙尾
Let us revisit Example 11.1
Jia-Ying Chen
Calculation of the Probability of a
Type II Error
Express the rejection

region directly, not in
standardized terms
Let us revisit Example 11.1
 The rejection region was x  175.34 with  = .05.

Let the alternative value be m = 180 (rather
than just m>170) H : m = 170
0
H1: m = 180
Do not reject H0
=.05
m= 170
xL =
Specify the
alternative value
under H1.
m=180
175.34
11
Jia-Ying Chen
Calculation of the Probability of a
Type II Error

A Type II error occurs when a false H0 is
not rejected.
H0: m = 170
A false H0…
…is not rejected
H1: m = 180
x  175.34
m= 170
xL =
=.05
m=180
175.34
12
Jia-Ying Chen
Calculation of the Probability of a
Type II Error
 = P( x  175.34 given that H 0 is false)
= P( x  175.34 given that m = 180)
= P( z 
175.34  180
65
400
) = .0764
H0: m = 170
H1: m = 180
m= 170
xL =
m=180
175.34
13
Jia-Ying Chen
Example 2
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14
A statistics practitioner wants to test the
following hypotheses with σ=20 and n=100:
H0: μ=100
H1: μ>100
Using α=0.1 find the probability of a Type II
error when μ=102
Jia-Ying Chen
Solution
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15
Rejection region: z>zα
Jia-Ying Chen
Example 3
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16
Calculate the probability of a Type II error for
the following test of hypothesis, given that
μ=203.
H0: μ=200
H1: μ≠200
α=0.05, σ=10, n=100
Jia-Ying Chen
Solution
17
Jia-Ying Chen
Effects on  of changing 

Decreasing the significance level ,
increases the value of , and vice versa.
2 < 1
m= 170
18
2 > 1
m=180
Jia-Ying Chen
Judging the Test
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
A hypothesis test is effectively defined by
the significance level  and by the sample
size n.
If the probability of a Type II error  is
judged to be too large, we can reduce it by
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19
increasing , and/or
increasing the sample size.
Jia-Ying Chen
Judging the Test

Increasing the sample size reduces 
xL  m

Recall : z  =
, thus xL = m  z 
 n
n
By increasing the sample size the
standard deviation of the sampling
distribution of the mean decreases.
Thus, x Ldecreases.
20
Jia-Ying Chen
Judging the Test

Increasing the sample size reduces 
xL  m

Recall : z  =
, thus xL = m  z 
 n
n
Note what happens when n increases:
 does not change,
but  becomes smaller
21
m= 170
xxxLxLLxLxLL
m=180
Jia-Ying Chen
Judging the Test
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Power of a test
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22
The power of a test is defined as 1 - .
It represents the probability of rejecting the null
hypothesis when it is false.
Jia-Ying Chen
Example 4
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23
For a given sample size n, if the level of
significance α is decreased, the power of the
test will:
a.increase.
b.decrease.
c.remain the same.
d.Not enough information to tell.
Jia-Ying Chen
Example 5
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24
During the last energy crisis, a government official
claimed that the average car owner refills the tank
when there is more than 3 gallons left. To check the
claim, 10 cars were surveyed as they entered a gas
station. The amount of gas remaining before refill
was measured and recorded as follows (in gallons):
3, 5, 3, 2, 3, 3, 2, 6, 4, and 1. Assume that the
amount of gas remaining in tanks is normally
distributed with a standard deviation of 1 gallon.
Compute the probability of a Type II error and the
power of the test if the true average amount of gas
remaining in tanks is 3.5 gallons. (α=0.05)
Jia-Ying Chen
Solution
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Rejection region:z>zα
x 3
 z0.05 = 1.645 = x  3.52
1
10
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25
β = P( x < 3.52 given that μ = 3.5) = P(z < 0.06) =
0.5239
Power = 1 - β = 0.4761
Jia-Ying Chen