&TwoDOE Class 90a1 &2 Simple Comparative Experiments Statistical Plots Sampling and Sampling Distributions Hypothesis Testing Confidence Interval

&TwoDOE Class 90a2 點圖 (Dot Diagram) ~

&TwoDOE Class 90a3 直方圖 (Histogram)

&TwoDOE Class 90a4 盒形圖 (Box Plot)

&TwoDOE Class 90a5 時間序列圖 (Time Series Plot)

&TwoDOE Class 90a6 期望值與變異數之公式 母體平均數 ( m  ) = 隨機變數之期望值 E(X) 母體變異數 ( s  2 ) = 隨機變數之變異數 V(X)

&TwoDOE Class 90a7 期望值與變異數之公式

&TwoDOE Class 90a8 Sample and Sampling

&TwoDOE Class 90a9 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 ( m ) 母體變異數 ( s 2 )

&TwoDOE Class 90a10 Central Limit Theorem

&TwoDOE Class 90a11 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/s H1:   50 cm/s Null Hypothesis (H 0 ) Vs. Alternative Hypothesis (H 1 ) One-sided and two-sided Hypotheses A statistical hypothesis is a statement about the parameters of one or more populations.

&TwoDOE Class 90a12 About Testing Critical Region Acceptance Region Critical Values

&TwoDOE Class 90a13 Errors in Hypothesis Testing 檢定結果可能為 Type I Error(  ): Reject H 0 while H 0 is true. Type II Error(  ): Fail to reject H 0 while H 0 is false.

&TwoDOE Class 90a14

&TwoDOE Class 90a15 Making Conclusions We always know the risk of rejecting H 0, 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 H 0 2. Fail to reject H 0, (Do not say accept H 0 ) or there is not enough evidence to reject H 0.

&TwoDOE Class 90a16 Significant Level (  )  = P(type I error) = P(reject H 0 while H 0 is true) n = 10,  = 2.5  /  n = 0.79

&TwoDOE Class 90a17

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&TwoDOE Class 90a20

&TwoDOE Class 90a21 The Power of a Statistical Test Power = 1 -  Power = the sensitivity of a statistical test

&TwoDOE Class 90a22 1. From the problem context, identify the parameter of interest. 2. State the null hypothesis, H Specify an appropriate alternative hypothesis, H 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 H 0 should be rejected and report that in the problem context. General Procedure for Hypothesis Testing

&TwoDOE Class 90a23 Inference on the Mean of a Population - Variance Known H 0 :  =  0 H 1 :    0, where  0 is a specified constant. Sample mean is the unbiased point estimator for population mean.

&TwoDOE Class 90a24 Example 8-2 Aircrew 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 = 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?

&TwoDOE Class 90a25 1.The parameter of interest is , the meaning burning rate. 2.H 0 :  = 50 cm/s 3.H 1 :   50 cm/s  = The test statistics is: 6.Reject H0 if Z 0 > 1.96 or Z 0 < (because Z  = Z = 1.96) 7.Computations: 8.Conclusions: Since Z 0 = 3.25 > 1.96, we reject H 0 :  = 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.

&TwoDOE Class 90a26 P-Values in Hypothesis Tests Where Z 0 is the test statistic, and  (z) is the standard normal cumulative function.

&TwoDOE Class 90a27 The Sample Size (I) Given values of  and , find the required sample size n to achieve a particular level of ..

&TwoDOE Class 90a28 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。

&TwoDOE Class 90a29

&TwoDOE Class 90a30 Construction of the C.I. From Central Limit Theory, Use standardization and the properties of Z,

&TwoDOE Class 90a31 Inference on the Mean of a Population - Variance Unknown Let X 1, X 2, …, X n 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.

&TwoDOE Class 90a32 Inference on the Mean of a Population - Variance Unknown H 0 :  =  0 H 1 :    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, T 0 will not follow the standard normal distribution.

&TwoDOE Class 90a33 Inference for the Difference in Means -Two Normal Distributions and Variance Unknown Why?

&TwoDOE Class 90a34

&TwoDOE Class 90a35 is distributed approximately as t with degrees of freedom given by

&TwoDOE Class 90a36 C.I. on the Difference in Means

&TwoDOE Class 90a37 C.I. on the Difference in Means

&TwoDOE Class 90a38 Paired t-Test When the observations on the two populations of interest are collected in pairs. Let (X 11, X 21 ), (X 12, X 22 ), …, (X 1n, X 2n ) be a set of n paired observations, in which X 1j ~(  1,  1 2 ) and X 2j ~(  2,  2 2 ) and D j = X 1j – X 2j, j = 1, 2, …, n. Then, to test H 0 :  1 =  2 is the same as performing a one- sample t-test H 0 :  D = 0 since  D = E(X 1 -X 2 ) = E(X 1 )-E(X 2 ) =  1 -  2

&TwoDOE Class 90a39

&TwoDOE Class 90a40 Inference on the Variance of a Normal Population (I) H 0 :  2 =  0  H 1 :  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,

&TwoDOE Class 90a41 Inference on the Variance of a Normal Population (II) Let X 1, X 2, …, X n be a random sample for a normal distribution with unknown mean  and unknown variance  2. To test the hypothesis H 0 :  2 =  0  H 1 :  2   0 , where  0  is a specified constant. We use the statistic If H 0 is true, then the statistic has a chi-square distribution with n-1 d.f..

&TwoDOE Class 90a42

&TwoDOE Class 90a43 The Reasoning For H 0 to be true, the value of  0 2 can not be too large or too small. What values of  0 2 should we reject H 0 ? (based on  value) What values of  0 2 should we conclude that there is not enough evidence to reject H 0 ?

&TwoDOE Class 90a44

&TwoDOE Class 90a45 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 s 2 = (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.

&TwoDOE Class 90a46 1. The parameter of interest is the population variance  H 0 :  2 = H 1 :  2   = The test statistics is 6. Reject H 0 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.

&TwoDOE Class 90a47 Hypothesis Testing on Variance - Normal Population

&TwoDOE Class 90a48 The Test Procedure for Two Variances Comparison

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&TwoDOE Class 90a51

&TwoDOE Class 90a52 Hypothesis Testing on the Ratio of Two Variances