Ch 11 實習 (1)Ch 11 實習 (1)

Jia-Ying Chen2

Introduction

所謂假設檢定，係指在尚未蒐集樣本資料、進行推論之前，就事先對母體的某種特徵性質作一合理的假設敘述，再利用隨機抽出的樣本及抽樣分配，配合機率原理，以判斷此項假設是否為真

若抽出的樣本資料與所陳述的假設很不一致，檢定的結果必然認為這個假設不對，則否定或拒絕（ reject）這個假設

Jia-Ying Chen3

Introduction

若抽出的樣本資料與所陳述的假設不會很不一致，檢定的結果就沒有充分理由斷定這個假設不對，但也不認為這個假設是對的

假設檢定的主要精神在於尋找證據來拒絕 H0，所以假設檢定只能檢定 H0是否顯著錯誤，而不能判斷其絕對正確

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Developing Null and Alternative Hypotheses

Hypothesis testing can be used to determine whether a statement about the value of a population parameter should or should not be rejected.

The null hypothesis, denoted by H0 , is a tentative assumption about a population parameter.

The alternative hypothesis, denoted by H1, is the opposite of what is stated in the null hypothesis.

Hypothesis testing is similar to a criminal trial. The hypotheses are:

H0: The defendant is innocent

H1: The defendant is guilty

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A Summary of Forms for Null and Alternative Hypotheses about a Population Mean

The equality part of the hypotheses always appears in the null hypothesis.

In general, a hypothesis test about the value of a population mean must take one of the following three forms (where 0 is the hypothesized value of the population mean).

H0: = 0 H0: = 0 H0: = 0

H1: < 0 H1: > 0 H1: 0

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Concepts of Hypothesis Testing

The critical concepts of hypothesis testing.Example:

An operation manager needs to determine if the mean demand during lead time is greater than 350.

If so, changes in the ordering policy are needed. There are two hypotheses about a population

mean: H0: The null hypothesis μ = 350

H1: The alternative hypothesis μ > 350

This is what you want to prove

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Example 1

A spouse suspects that the average amount of money spent on Christmas gifts for immediate family members is above $1,200. The correct set of hypotheses is:

a. H0: μ = 1200 vs. H1: μ < 1200

b. H0: μ > 1200 vs. H1: μ = 1200

c. H0: μ = 1200 vs. H1: μ > 1200

d. H0: μ < 1200 vs. H1: μ = 1200

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Concepts of Hypothesis Testing

= 350

Assume the null hypothesis is true (= 350).

Sample from the demand population, and build a statistic related to the parameter hypothesized (the sample mean).

Pose the question: How probable is it to obtain a sample mean at least as extreme as the one observed from the sample, if H0 is correct?

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Since the is much larger than 350, the mean is likely to be greater than 350. Reject the null hypothesis.

x

355x

Concepts of Hypothesis Testing

= 350

Assume the null hypothesis is true (= 350).

450x

In this case the mean is not likely to be greater than 350. Do not reject the null hypothesis.

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Types of Errors

Two types of errors may occur when deciding whether to reject H0 based on the statistic value.

Type I error: Reject H0 when it is true.

Type II error: Do not reject H0 when it is false. Example continued

Type I error: Reject H0 ( = 350) in favor of H1 ( > 350) when the real value of is 350.

Type II error: Believe that H0 is correct ( = 350) when the real value of is greater than 350.

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Type I and Type II Errors

Since hypothesis tests are based on sample data, we must allow for the possibility of errors.

The person conducting the hypothesis test specifies the maximum allowable probability of making aType I error, denoted by and called the level of significance.

Generally, we cannot control for the probability of making a Type II error, denoted by .

Statistician avoids the risk of making a Type II error by using “do not reject H0” and not “accept H0”.

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(

( )

Null True Null False

Fail toreject null

CorrectDecision

Type II error)

Reject null Type I error

Correct Decision(Power)

Decision Table for Hypothesis Testing

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Example 2

Which of the following conclusions is not an appropriate conclusion from a hypothesis test?

a. Reject H0. Sufficient evidence to support H1.

b. Fail to reject H0. Insufficient evidence to support H1.

c. Accept H0. Sufficient evidence to support H0.

d. All of these choices are true.

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Example 3

Suppose we wish to test H0: m = 45 vs. H1: m > 45. What will result if we conclude that the mean is greater than 45 when the actual mean is 50?

a.We have made a Type I error.

b.We have made a Type II error.

c.We have made both a Type I error and a Type II error.

d.We have made the correct decision.

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Controlling the probability of conducting a type I error

Recall: H0: = 350 and H1: > 350.

H0 is rejected if is sufficiently large

Thus, a type I error is made if when = 350.

By properly selecting the critical value we can limit the probability of conducting a type I error to an acceptable level.

xvaluecriticalx

Critical value

x= 350

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Testing the Population Mean When the Population Standard Deviation is Known

Example A new billing system for a department store will be

cost- effective only if the mean monthly account is more than $170.

A sample of 400 accounts has a mean of $178. If accounts are approximately normally distributed

with = $65, can we conclude that the new system will be cost effective?

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Example 11.1 – Solution The population of interest is the credit accounts at

the store. We want to know whether the mean account for

all customers is greater than $170.

H1 : > 170

The null hypothesis must specify a single value of the parameter

H0 : = 170

Testing the Population Mean ( is Known)

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Approaches to Testing

There are two approaches to test whether the sample mean supports the alternative hypothesis (H1) The rejection region method is mandatory for

manual testing (but can be used when testing is supported by a statistical software)

The p-value method which is mostly used when a statistical software is available.

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The rejection region is a range of values such that if the test statistic falls into that range, the null hypothesis is rejected in favor of the alternative hypothesis.

The rejection region is a range of values such that if the test statistic falls into that range, the null hypothesis is rejected in favor of the alternative hypothesis.

The Rejection Region Method

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Example 11.1 – solution continued

• Define a critical value for that is just large enough to reject the null hypothesis.

xLx

• Reject the null hypothesis if

Lx xLx x

The Rejection Region Method for a Right - Tail Test

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Allow the probability of committing a Type I error be (also called the significance level).

Find the value of the sample mean that is just large enough so that the actual probability of committing a Type I error does not exceed Watch…

Determining the Critical Value for the Rejection Region

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P(commit a Type I error) = P(reject H0 given that H0 is true)Lx

170x x

= P( given that H0 is true)Lx x

40065

170xz L

Example – solution continued

… is allowed to be

( )Since P Z Z we have:

Determining the Critical Value – for a Right – Tail Test

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Determining the Critical Value – for a Right – Tail Test

.34.17540065

645.1170x

.645.1z,05.0selectweIf

.40065

z170x

L

05.

L

40065

170xz L

= 0.05

170x Lx

Example – solution continued

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Determining the Critical value for a Right - Tail Test

Re

175.34

ject the null hypothesis if

x Re

175.34

ject the null hypothesis if

x

ConclusionSince the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence to infer that the mean monthly balance is greater than $170 at the 5% significance level.

ConclusionSince the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence to infer that the mean monthly balance is greater than $170 at the 5% significance level.

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Example 4

Suppose that 10 observations are drawn from a normal population whose variance is 64. The observations are: 58, 62, 45, 50, 59, 65, 39, 40, 41, and 52. Test at the 10% level of significance to determine if there is enough evidence to conclude that the population mean is greater than 45.

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Solution

Reject the null hypothesis if

Since sample mean=51.1

Reject ,Yes, there is enough statistical evidence at the 10% significance level to conclude that the population mean is greater than 45.

8 845 1.28 48.24

10 10LX X z

45:0 H 45:1 H

0H

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Instead of using the statistic , we can use the standardized value z.

Then, the rejection region becomes

x

n

xz

z zOne tail test

The standardized test statistic

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Example - continued We redo this example using the standardized

test statistic.

Recall: H0: = 170

H1: > 170 Test statistic:

Rejection region: z > z.051.645.

46.240065

170178

n

xz

The standardized test statistic

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Example - continued

The standardized test statistic

Re

1.645

ject the null hypothesis if

Z Re

1.645

ject the null hypothesis if

Z

ConclusionSince Z = 2.46 > 1.645, reject the null hypothesis in favor of the alternative hypothesis.

ConclusionSince Z = 2.46 > 1.645, reject the null hypothesis in favor of the alternative hypothesis.

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Example 5

In testing the hypotheses H0:μ=50 vs. H1:μ>50, the following information are known: n = 64, sample mean= 53.5, and σ = 10. The standardized test statistic z equals?

Solution ：53.5 50

2.810 / 64

z

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The p-value provides information about the amount of statistical evidence that supports the alternative hypothesis.

– The p-value of a test is the probability of observing a test statistic at least as extreme as the one computed, given that the null hypothesis is true.

–虛無假設成立，獲得檢定統計量及更極端數值之機率。 –Let us demonstrate the concept on Example

P-value Method

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178 170( )

65 400

( 2.4615) .0069

P z

P z

170x

178x

The probability of observing a test statistic at least as extreme as 178, given that = 170 is…

The p-value

P-value Method

( 178 170)P x when

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Because the probability that the sample mean will assume a value of more than 178 when = 170 is so small (.0069), there are reasons to believe that > 170.

178x

170:H x0 170:H x1

…it becomes more probable under H1, when 170x

Note how the event is rare under H0

when but...178x

,170x

Interpreting the p-value

Jia-Ying Chen34 178x

170:H x0 170:H x1

We can conclude that the smaller the p-value the more statistical evidence exists to support the alternative hypothesis.

We can conclude that the smaller the p-value the more statistical evidence exists to support the alternative hypothesis.

Interpreting the p-value

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The p-value can be used when making decisions based on rejection region methods as follows:Define the hypotheses to test, and the required significance

level α.Perform the sampling procedure, calculate the test statistic

and the p-value associated with it.Compare the p-value to α. Reject the null hypothesis only if

p-value < α; otherwise, do not reject the null hypothesis.

The p-value

34.175xL

= 0.05

170x

178x

The p-value and the Rejection Region Methods

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If we reject the null hypothesis, we conclude that there is enough evidence to infer that the alternative hypothesis is true.

If we do not reject the null hypothesis, we conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true.

If we reject the null hypothesis, we conclude that there is enough evidence to infer that the alternative hypothesis is true.

If we do not reject the null hypothesis, we conclude that there is not enough statistical evidence to infer that the alternative hypothesis is true. The alternative hypothesis

is the more importantone. It represents whatwe are investigating.

The alternative hypothesisis the more importantone. It represents whatwe are investigating.

Conclusions of a Test of Hypothesis

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1. Establish hypotheses: state the null and alternative hypotheses.

2. Determine the appropriate statistical test and sampling distribution.

3. Specify the Type I error rate (4. State the decision rule.

5. Gather sample data.

6. Calculate the value of the test statistic.

7. State the statistical conclusion.

8. Make a managerial decision.

Steps in Hypothesis Testing

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Example 6

If a hypothesis is not rejected at the 0.10 level of significance, it:

a.must be rejected at the 0.05 level.

b.may be rejected at the 0.05 level.

c.will not be rejected at the 0.05 level.

d.must be rejected at the 0.025 level.

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Example 7

Calculate the value of the test statistic, set up the rejection region, determine the p-value, interpret the result, and draw the sampling distribution.

1.0,3.14,25,2

15:

15:H

1

0

xn

H

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Solution

Rejection region z<z0.1=-1.28

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Example 8

A business student claims that on average an MBA students is required to prepare more than five cases per week. To examine the claim, a statistics professor asks a random sample of 10 MBA students to report the number of cases they prepare weekly. The results are exhibited here. Can the professor conclude at the 5% significance level that the claim is true, assuming that the number of cases is normally distributed with a standard deviation of 1.5?

2 7 4 8 9 5 11 3 7 4

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Solution

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Example 9

A random sample of 12 second-year university students enrolled in a business statistics course was drawn. At the course’s completion, each student was asked how many hours he or she spent doing homework in statistics. The data are listed here. It is known that the population standard deviation is σ=8. The instructor has recommended that students devote 3 hours per week for the duration of the 12-week semester, for a total of 36 hours. Test to determine whether there is evidence that the average student spent less than the recommend amount of time. Compute the p-value of the test.

31 40 26 30 36 38 29 40 38 30 35 38

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Solution