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Ch 11 實習 (2). A Two - Tail Test. Example 11.2 AT&T has been challenged by competitors who argued that their rates resulted in lower bills. - PowerPoint PPT Presentation
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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
Jia-Ying Chen5
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
Jia-Ying Chen6
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
Jia-Ying Chen8
Example 1
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 Chen9
Solution
Jia-Ying Chen10
To calculate Type II error we need to… express the rejection region directly, in terms of the paramet
er 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
Jia-Ying Chen11
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)
Jia-Ying Chen12
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
Jia-Ying Chen13
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 Chen14
Example 2
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 Chen15
Solution
Rejection region: z>zα
Jia-Ying Chen16
Example 3
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 Chen17
Solution
Jia-Ying Chen18
Decreasing the significance level increases the value of and vice versa
Effects on of changing
= 170 180
2 >2 <
Jia-Ying Chen19
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 Chen20
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 Chen21 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
Jia-Ying Chen22
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 Chen23
Example 4
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 Chen24
Example 5
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 Chen25
Solution
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