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Ch 11 實習 (2)Ch 11 實習 (2)
Jia-Ying Chen2
A Two - Tail Test
Example 11.2 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 long-distance bills for all AT&T residential customers are $17.09 and $3.87 respectively.
Jia-Ying Chen3
A Two - Tail Test
Example 11.2 - continued 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 Chen4
Solution Is the mean different from 17.09?
H0: = 17.09
09.17:H1
– Define the rejection region
A Two - Tail Test
/ 2 / 2z z or z z
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17.09
We want this erroneous rejection of H0 to be a rare event, say 5% chance.
x x
If H0 is true ( =17.09), can still fall far above or far below 17.09, in which case we erroneously reject H0 in favor of H1
x
)09.17(
20.025 20.025
Solution - continued
A Two – Tail Test
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20.025
17.09
0
x x20.025
20.025 20.025
19.110087.3
09.1755.17
n
xz
-z= -1.96 z= 1.96
Rejection region
Solution - continued
A Two – Tail Test
55.17x
From the sample we have:
17.55
Jia-Ying Chen7
20.025 20.025
19.110087.3
09.1755.17
n
xz
-z= -1.96 z= 1.96
There is insufficient evidence to infer that there is a difference between the bills of AT&T and the competitor.
-1.19
Also, by the p value approach:The p-value = P(Z< -1.19)+P(Z >1.19) = 2(.1173) = .2346 > .05
1.190
A Two – Tail Test
Example 1
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 Chen8
Solution
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 Chen9
Jia-Ying Chen10
Example 2
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 Chen11
Solution
Testing hypotheses and intervals estimators
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 ( is not equal to the hypothesized value).
Jia-Ying Chen12
Drawbacks
Two-tail interval estimators may not provide the right answer to the question posed in one-tail hypothesis tests.
The interval estimator does not yield a p-value.
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Example 3
Using a confidence interval when conducting a two-tail test for m, we do not reject H0 if the hypothesized value for m:
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.
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To calculate Type II error we need to… express the rejection region directly, in terms of the
parameter hypothesized (not standardized). specify the alternative value under H1.
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
單尾 雙尾
Let us revisit Example 11.1
Calculation of the Probability of a Type II Error
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Express the rejection region directly, not in standardized terms
34.175xL
=.05
= 170
Calculation of the Probability of a Type II Error
Let us revisit Example 11.1 The rejection region was with = .05.175.34x
Do not reject H0
180
H1: = 180
H0: = 170
Specify the alternative value
under H1.
Let the alternative value be = 180 (rather than just >170)
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34.175xL
=.05
= 170
Calculation of the Probability of a Type II Error
34.175x 180
H1: = 180
H0: = 170
A Type II error occurs when a false H0 is not rejected.
A false H0……is not rejected
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34.175xL = 170
Calculation of the Probability of a Type II Error
180
H1: = 180
H0: = 170
)180thatgiven34.175x(P
)falseisHthatgiven34.175x(P 0
0764.)40065
18034.175z(P
Jia-Ying Chen19
Example 4
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
Rejection region: z>zα
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Example 5
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
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Solution
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Decreasing the significance level increases the value of and vice versa
Effects on of changing
= 170 180
2 >2 <
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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 increasing , and/or increasing the sample size.
Judging the Test
Jia-Ying Chen25
Increasing the sample size reduces
Judging the Test
By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, decreases.
Lx
nzxthus,
nx
z:callRe LL
Jia-Ying Chen26 Lx 180= 170
Judging the Test
Lx
Note what happens when n increases:
Lx LxLx Lx
does not change,but becomes smaller
Increasing the sample size reduces
nzxthus,
nx
z:callRe LL
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Power of a test The power of a test is defined as 1 - It represents the probability of rejecting the null
hypothesis when it is false.
Judging the Test
Jia-Ying Chen28
Example 6
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 Chen29
Example 7
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 Chen30
Solution
H0: μ=3 H1: μ>3
Rejection region:z>zα
β = P( < 3.52 given that μ = 3.5) = P(z < 0.06) = 0.5239
Power = 1 - β = 0.4761
x
0.05
31.645 3.52
1
10
xz x