&twodoe class 90a1 &2 simple comparative experiments statistical plots sampling and sampling...

&Two DOE Class 90a 1

&2 Simple Comparative Experiments

Statistical Plots

Sampling and Sampling Distributions

Hypothesis Testing

Confidence Interval


點圖 (Dot Diagram)

~


直方圖 (Histogram)


盒形圖 (Box Plot)


時間序列圖 (Time Series Plot)


期望值與變異數之公式母體平均數 () = 隨機變數之期望值 E(X) 母體變異數 (2) = 隨機變數之變異數 V(X)

2

)(

,

,)()(

xEXV

discretexp(x)

continuousdxxxfXE

xAll


期望值與變異數之公式

)()()(

)()()(

)()()( ,0),( ,,

),(

),,(2)()()(

)()(

)()()(

2

1

2

1

2121

21212121

221121

212121

12

1

2121

xExE

xxE

xExExxE

xVxVxxVxxCovxx

xxExxCov

xxCovxVxVxxV

xVaaxV

xbExaEbxaxE

且則獨立若其中


Sample and Sampling


點估計 (Point Estimation) 以抽樣得來之樣本資料 , 依循某一公式計算出單一數

值 , 來估計母體參數 , 稱為點估計 . 好的點估計公式之條件 :

不偏性最小變異

常用之點估計 : 母體平均數 ()

母體變異數 ()

n

XX i

1

1

2

2

n

XXS

n

ii


Central Limit Theorem


假設檢定 (Hypothesis Testing) “A person is innocent until proven guilty beyond a

reasonable doubt.” 在沒有充分證據證明其犯罪之前 , 任何人皆是清白的 .

假設檢定H0: = 50 cm/sH1: 50 cm/s

Null Hypothesis (H0) Vs. Alternative Hypothesis (H1) One-sided and two-sided Hypotheses A statistical hypothesis is a statement about the pa

rameters of one or more populations.


About Testing

Critical Region Acceptance Region Critical Values


Errors in Hypothesis Testing

檢定結果可能為

Type I Error(): Reject H0 while H0 is true. Type II Error(): Fail to reject H0 while H0 is false.


The Defendant isThe Jury finds the

person Innocent Guilty

Innocent Type II Error

Guilty Type I Error

)(:

)(:

1

0

GuiltyH

InnocentH

有罪無辜


Making Conclusions

We always know the risk of rejecting H0, i.e., the significant level or the risk.

We therefore do not know the probability of committing a type II error ().

Two ways of making conclusion:1. Reject H0

2. Fail to reject H0, (Do not say accept H0)

or there is not enough evidence to reject H0.


Significant Level () = P(type I error) = P(reject H0 while H0 is true)

n = 10, = 2.5/n = 0.79


The Power of a Statistical Test

Power = 1 - Power = the sensitivity of a statistical test


1. From the problem context, identify the parameter of interest.2. State the null hypothesis, H0.3. Specify an appropriate alternative hypothesis, H1.4. Choose a significance level a.5. State an appropriate test statistic.6. State the rejection region for the statistic.7. Compute any necessary sample quantities, substitute these into the equation for the test statistic, and compute that value.8. Decide whether or not H0 should be rejected and report that in the problem context.

General Procedure for Hypothesis Testing


Inference on the Mean of a Population-Variance Known

H0: = 0

H1: 0 , where 0 is a specified constant. Sample mean is the unbiased point estimator for

population mean.

1,0~

then),( trueis H if Therefore,

.,~ then , varianceand mean with

ondistributi a fromdrawn samples are ,,, If

00

00

22

21

Nn

XZ

nNX

XXX n


Example 8-2Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 cm/s. We know that the standard deviation of burning rate is 2 cm/s. The experimenter decides to specify a type I error probability or significance level of α = 0.05. He selects a random sample of n = 25 and obtains a sample average of the burning rate of x = 51.3 cm/s. What conclusions should be drawn?


1. The parameter of interest is , the meaning burning rate.2. H0: = 50 cm/s3. H1: 50 cm/s4. = 0.055. The test statistics is:

6. Reject H0 if Z0 > 1.96 or Z0 < -1.96 (because Z= Z0.025 = 1.96)7. Computations:

8. Conclusions: Since Z0 = 3.25 > 1.96, we reject H0: = 50 at the 0.05 level of significance. We conclude that the mean burning rate differs from 50 cm/s, based on a sample of 25 measurements. In fact, there is string evidence that the mean burning rate exceeds 50 cm/s.

n

xZ

/0

0

25.325/2

503.510

Z


P-Values in Hypothesis Tests

Where Z0 is the test statistic, and (z) is the standard normal cumulative function.


The Sample Size (I)

Given values of and , find the required sample size n to achieve a particular level of ..

02

222/

2/

2/

2/2/

re whe

/ Then,

Let

0 when /

// Since

ZZn

nZZ

Z

nZ

nZ

nZ


The Operating Characteristic Curves- Normal test (z-test)

Use to performing sample size or type II error calculations.

The parameter d is defined as:

so that it can be used for all problems regardless of the values of 0 and .

課本 41 頁之公式為兩平均數差之假設檢定所需之樣本數公式。

|||| 0

d


Construction of the C.I. From Central Limit Theory,

. ,~ ,25 and ,~ 22

nNXnXIf Use standardization and the properties of Z,

1//

1

1 and

2/2/

2/2/

2/2/

nzXnzXP

zn

XzP

zZzPn

XZ


Inference on the Mean of a Population-Variance Unknown

Let X1, X2, …, Xn be a random sample for a normal distribution with unknown mean and unknown variance 2. The quantity

has a t distribution with n - 1 degrees of freedom.


Inference on the Mean of a Population-Variance Unknown

H0: = 0

H1: 0 , where 0 is a specified constant. Variance unknown, therefore, use s instead of

in the test statistic.

If n is large enough ( 30), we can use Z-test. However, n is usually small. In this case, T0 will not follow the standard normal distribution.


Inference for the Difference in Means-Two Normal Distributions and Variance Unknown

Why?


have, we still and S with S and σplacing σ 22

21

22

21Re

is distributed approximately as t with degrees of freedom given by


C.I. on the Difference in Means


Paired t-Test

When the observations on the two populations of interest are collected in pairs.

Let (X11, X21), (X12, X22), …, (X1n, X2n) be a set of n paired observations, in which X1j~(1, 1

2) and X2j~(2, 22)

and Dj = X1j – X2j, j = 1, 2, …, n. Then, to test H0: 1= 2 is the same as performing a one-sample t-test H0: D = 0 since D = E(X1-X2) = E(X1)-E(X2) = 1 - 2


Inference on the Variance of a Normal Population (I) H0: 2 = 0

H1: 2 0 , where 0

is a specified constant.

Sampling from a normal distribution with unknown mean and unknown variance 2, the quantity

has a Chi-square distribution with n-1 degrees of freedom. That is,

2

22 1

Sn

212

22 ~

1

n

Sn


Inference on the Variance of a Normal Population (II) Let X1, X2, …, Xn be a random sample for a normal dist

ribution with unknown mean and unknown variance 2. To test the hypothesis

H0: 2 = 0

H1: 2 0 , where 0

is a specified constant.We use the statistic

If H0 is true, then the statistic has a chi-square distribution with n-1 d.f..


k

k

k

xexk

xf xkk

2

addition,In

freedom. of degrees ofnumber theis

0 2/2

1

:ondistributi square-chi of PDF

2

2/12/2/


The Reasoning

For H0 to be true, the value of 0

2 can not be

too large or too small.

What values of 0

2 should we reject H0? (based

on value) What values of

02 should we conclude that

there is not enough evidence to reject H0?


Example 8-11

An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 20 bottles results in a sample variance of fill volume of s2 = 0.0153 (fluid ounces)2. If the variance of fill volume exceeds 0.01 (fluid ounces)2, an unacceptable proportion of bottles will be underfilled and overfilled. Is there evidence in the sample data to suggest that the manufacturer has a problem with underfilled and overfilled bottles? Use = 0.05, and assume that fill volume has a normal distribution.


1. The parameter of interest is the population variance 2.2. H0: 2 = 0.01

3. H1: 2 0.01

4. = 0.055. The test statistics is

6. Reject H0 if

7. Computations:

8. Conclusions: Since , we conclude that there is no strong evidence that the variance of fill volume exceeds 0.01 (fluid ounces)2.

14.3007.29 219,05.0

20

07.29

01.0

0153.01920

14.30219,05.0

20

20

220

1

sn


Hypothesis Testing on Variance - Normal Population

H 1 Te s t S ta tis t ic R e je c t H 0 i f

0

2 21,2/1

20

21,2/

20 or nn

> 0

2 21,

20 n

< 0

2

20

220

1

Sn

21,1

20 n


The Test Procedure for Two The Test Procedure for Two Variances ComparisonVariances Comparison


Hypothesis Testing on the Ratio of Two Variances

&twodoe class 90a1 &2 simple comparative experiments statistical plots sampling and sampling...

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