Ch 11 實習 (2)Ch 11 實習 (2)

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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.

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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)?

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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

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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

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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

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Solution

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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

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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

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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

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Solution

Rejection region: z>zα

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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

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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

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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

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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

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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.

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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)

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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