Transcript A Two - Tail Test

Ch 11 實習 (2)
A Two - Tail Test
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Example 11.2
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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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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
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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
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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)
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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
3.87
= 1.19
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
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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
3.87
= 1.19
-z/2 = -1.96
z/2 = 1.96
100
Jia-Ying Chen
Example 1
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A random sample of 100 observations from a
normal population whose standard deviation
is 50 produced a mean of 75. Does this
statistic provide sufficient evidence at the 5%
level of significance to infer that the
population mean is not 80?
Jia-Ying Chen
Solution
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H0: μ=80 vs. H1: μ ≠80
Rejection region: |z| > z0.025=1.96
Test statistic: z = (75-80)/(50/10)=-1.0
Conclusion: Don’t reject . No sufficient
evidence at the 5% level of significance to
infer that the population mean is not 80.
Jia-Ying Chen
Example 2
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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
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Solution
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Jia-Ying Chen
Testing hypotheses and intervals estimators
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Interval estimators can be used to test hypotheses.
Calculate the 1 -  confidence level interval
estimator, then
if the hypothesized parameter value falls within the
interval, do not reject the null hypothesis
if the hypothesized parameter value falls outside the
interval, conclude that the null hypothesis can be
rejected (m is not equal to the hypothesized value).
Jia-Ying Chen
Drawbacks
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Two-tail interval estimators may not provide the right
answer to the question posed in one-tail hypothesis
tests.
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The interval estimator does not yield a p-value.
Jia-Ying Chen
Example 3
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Using a confidence interval when conducting a twotail test for m, we do not reject H0 if the
hypothesized value for m:
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a. is to the left of the lower confidence limit (LCL).
b. is to the right of the upper confidence limit (UCL).
c. falls between the LCL and UCL.
d. falls in the rejection region.
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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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
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Calculation of the Probability of a
Type II Error
Express the rejection
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region directly, not in
standardized terms
Let us revisit Example 11.1
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The rejection region was x  175.34
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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
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Jia-Ying Chen
Calculation of the Probability of a
Type II Error
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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
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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
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Jia-Ying Chen
Example 4
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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
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Solution
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Rejection region: z>zα
Jia-Ying Chen
Example 5
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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
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Effects on  of changing 
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Decreasing the significance level ,
increases the value of , and vice versa.
2 < 1
m= 170
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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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increasing , and/or
increasing the sample size.
Jia-Ying Chen
Judging the Test
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Increasing the sample size reduces 
xL  m
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Re call : z  =
, thus x L = m  z 
 n
n
By increasing the sample size the
standard deviation of the sampling
distribution of the mean decreases.
Thus, x Ldecreases.
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Jia-Ying Chen
Judging the Test
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Increasing the sample size reduces 
xL  m
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Re call : z  =
, thus x L = m  z 
 n
n
Note what happens when n increases:
 does not change,
but  becomes smaller
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m= 170
xxxLLxLxLxLL
m=180
Jia-Ying Chen
Judging the Test
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Power of a test
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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 6
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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 7
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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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H0: μ=3 H1: μ>3
Rejection region:z>zα
x 3
 z0.05 = 1.645 = x  3.52
1
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β = P( x < 3.52 given that μ = 3.5) = P(z < 0.06) =
0.5239
Power = 1 - β = 0.4761
Jia-Ying Chen