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Horng-Chyi HorngStatistics II41 Inference on the Mean of a Population - Variance Known H 0 :  =  0 H 0 :  =  0 H 1 :    0, where  0 is a specified.

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Presentation on theme: "Horng-Chyi HorngStatistics II41 Inference on the Mean of a Population - Variance Known H 0 :  =  0 H 0 :  =  0 H 1 :    0, where  0 is a specified."— Presentation transcript:

1 Horng-Chyi HorngStatistics II41 Inference on the Mean of a Population - Variance Known H 0 :  =  0 H 0 :  =  0 H 1 :    0, where  0 is a specified constant. n Sample mean is the unbiased point estimator for population mean. &4-4 (&8-2)

2 Horng-Chyi HorngStatistics II42 The Reasoning n For H 0 to be true, the value of Z 0 can not be too large or too small. n Recall that 68.3% of Z 0 should fall within (-1, +1) 95.4% of Z 0 should fall within (-2, +2) 99.7% of Z 0 should fall within (-3, +3) What values of Z 0 should we reject H 0 ? (based on  value) What values of Z 0 should we reject H 0 ? (based on  value) What values of Z 0 should we conclude that there is not enough evidence to reject H 0 ? What values of Z 0 should we conclude that there is not enough evidence to reject H 0 ?

3 Horng-Chyi HorngStatistics II43

4 Horng-Chyi HorngStatistics II44 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 = 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?

5 Horng-Chyi HorngStatistics II45

6 Horng-Chyi HorngStatistics II46 Hypothesis Testing on  - Variance Known

7 Horng-Chyi HorngStatistics II47 P-Values in Hypothesis Tests(I) Where Z 0 is the test statistic, and  (z) is the standard normal cumulative function. Where Z 0 is the test statistic, and  (z) is the standard normal cumulative function. In example 8-2, Z 0 = 3.25, P-Value = 2[1-  (3.25)] = 0.0012 In example 8-2, Z 0 = 3.25, P-Value = 2[1-  (3.25)] = 0.0012

8 Horng-Chyi HorngStatistics II48 P-Values of Hypothesis Testing on  - Variance Known

9 Horng-Chyi HorngStatistics II49 P-Values in Hypothesis Tests(II)  -value is the maximum type I error allowed, while P- value is the real type I error calculated from the sample.  -value is the maximum type I error allowed, while P- value is the real type I error calculated from the sample.  -value is preset, while P-value is calculated from the sample.  -value is preset, while P-value is calculated from the sample. When P-value is less than  -value, we can safely make the conclusion “Reject H 0 ”. By doing so, the error we are subjected to (P-value) is less than the maximum error allowed (  -value). When P-value is less than  -value, we can safely make the conclusion “Reject H 0 ”. By doing so, the error we are subjected to (P-value) is less than the maximum error allowed (  -value).

10 Horng-Chyi HorngStatistics II50 Type II Error - Fail to reject H 0 while H 0 is false

11 Horng-Chyi HorngStatistics II51 How to calculate Type II Error? (I) (H 0 :  =  0 Vs. H 1 :    0 ) Under the circumstance of type II error, H 0 is false. Supposed that the true value of the mean is  =  0 + , where  > 0. The distribution of Z 0 is: Under the circumstance of type II error, H 0 is false. Supposed that the true value of the mean is  =  0 + , where  > 0. The distribution of Z 0 is:

12 Horng-Chyi HorngStatistics II52 How to calculate Type II Error? (II) - refer to section &4.3 (&8.1) n Type II error occurred when (fail to reject H 0 while H 0 is false) n Therefore,

13 Horng-Chyi HorngStatistics II53 The Sample Size (I) Given values of  and , find the required sample size n to achieve a particular level of .. Given values of  and , find the required sample size n to achieve a particular level of ..

14 Horng-Chyi HorngStatistics II54 The Sample Size (II) n Two-sided Hypothesis Testing n One-sided Hypothesis Testing

15 Horng-Chyi HorngStatistics II55 Example 8-3

16 Horng-Chyi HorngStatistics II56 The Operating Characteristic Curves - Normal test (z-test) n Use to performing sample size or type II error calculations. n The parameter d is defined as: so that it can be used for all problems regardless of the values of  0 and . n Chart VI a,b,c,d are for Z-test.

17 Horng-Chyi HorngStatistics II57 Example 8-5

18 Horng-Chyi HorngStatistics II58

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20 Horng-Chyi HorngStatistics II60 Large Sample Test If n  30, then the sample variance s 2 will be close to  2 for most samples. If n  30, then the sample variance s 2 will be close to  2 for most samples. Therefore, if population variance  2 is unknown but n  30, we can substitute  with s in the test procedure with little harmful effect. Therefore, if population variance  2 is unknown but n  30, we can substitute  with s in the test procedure with little harmful effect.

21 Horng-Chyi HorngStatistics II61 Large Sample Hypothesis Testing on  - Variance Unknown but n  30

22 Horng-Chyi HorngStatistics II62 Statistical Vs. Practical Significance n Practical Significance = 50.5-50 = 0.5 n Statistical Significance P-Value for each sample size n.

23 Horng-Chyi HorngStatistics II63 Notes be careful when interpreting the results from hypothesis testing when the sample size is large, because any small departure from the hypothesized value  0 will probably be detected, even when the difference is of little or no practical significance. be careful when interpreting the results from hypothesis testing when the sample size is large, because any small departure from the hypothesized value  0 will probably be detected, even when the difference is of little or no practical significance. n In general, two types of conclusion can be drawn: 1. At  = 0.**, we have enough evidence to reject H 0. 2. At  = 0.**, we do not have enough evidence to reject H 0.

24 Horng-Chyi HorngStatistics II64 Confidence Interval on the Mean (I) n Point Vs. Interval Estimation n The general form of interval estimate is L    U in which we always attach a possible error  such that P(L   U) = 1-  That is, we have 1-  confidence that the true value of  will fall within [L, U]. n Interval Estimate is also called Confidence Interval (C.I.).

25 Horng-Chyi HorngStatistics II65 Confidence Interval on the Mean (II) n L is called the lower-confidence limit and U is the upper-confidence limit. n Two-sided C.I. Vs. One-sided C.I.

26 Horng-Chyi HorngStatistics II66 Construction of the C.I. n From Central Limit Theory, n Use standardization and the properties of Z,

27 Horng-Chyi HorngStatistics II67 Formula for C.I. on the Mean with Variance Known n Used when 1. Variance known 2. n  30, use s to estimate .

28 Horng-Chyi HorngStatistics II68 Example 8-6 Consider the rocket propellant problem in Example 8-2. Find a 95% C.I. on the mean burning rate? 95% C.I =>  = 0.05, z  /2 = z 0.025 = 1.96 z  /2 = z 0.025 = 1.96

29 Horng-Chyi HorngStatistics II69 Notes - C.I. n Relationship between Hypothesis Testing and C.I.s Confidence level (1-  ) and precision of estimation (C.I. * 1/2) Confidence level (1-  ) and precision of estimation (C.I. * 1/2) n Sample size and C.I.s

30 Horng-Chyi HorngStatistics II70 Choice of Sample Size to Achieve Precision of Estimation

31 Horng-Chyi HorngStatistics II71 Example 8-7

32 Horng-Chyi HorngStatistics II72 One-Sided C.I.s on the Mean

33 Horng-Chyi HorngStatistics II73 Inference on the Mean of a Population - Variance Unknown H 0 :  =  0 H 0 :  =  0 H 1 :    0, where  0 is a specified constant. Variance unknown, therefore, use s instead of  in the test statistic. Variance unknown, therefore, use s instead of  in the test statistic. &4-5 (&8-3) n If n is large enough (  30), we can use the test procedure in &4-4 (&8-2). However, n is usually small. In this case, T0 will not follow the standard normal distribution.

34 Horng-Chyi HorngStatistics II74 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 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.

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36 Horng-Chyi HorngStatistics II76 The Reasoning n For H 0 to be true, the value of T 0 can not be too large or too small. What values of T 0 should we reject H 0 ? (based on  value) What values of T 0 should we reject H 0 ? (based on  value) What values of T 0 should we conclude that there is not enough evidence to reject H 0 ? What values of T 0 should we conclude that there is not enough evidence to reject H 0 ? n Although when n  30, we can use Z 0 in section &8-2 to perform the testing instead. We prefer using T 0 to more accurately reflect the real testing result if t-table is available.

37 Horng-Chyi HorngStatistics II77

38 Horng-Chyi HorngStatistics II78 Example 8-8

39 Horng-Chyi HorngStatistics II79

40 Horng-Chyi HorngStatistics II80 Testing for Normality (Example 8-8) - t-test assumes that the data are a random sample from a normal population (1) Box Plot(2) Normality Probability Plot

41 Horng-Chyi HorngStatistics II81 Hypothesis Testing on  - Variance Unknown

42 Horng-Chyi HorngStatistics II82 Finding P-Values n Steps: 1. Find the degrees of freedom (k = n-1)in the t-table. 2. Compare T 0 to the values in that row and find the closest one. 3. Look the  value associated with the one you pick. The p-value of your test is equal to this  value. In example 8-8, T 0 = 4.90, k = n-1 = 21, P-Value < 0.0005 because the t value associated with (k = 21,  = 0.0005) is 3.819. In example 8-8, T 0 = 4.90, k = n-1 = 21, P-Value < 0.0005 because the t value associated with (k = 21,  = 0.0005) is 3.819.

43 Horng-Chyi HorngStatistics II83 P-Values of Hypothesis Testing on  - Variance Unknown

44 Horng-Chyi HorngStatistics II84 The Operating Characteristic Curves - t-test n Use to performing sample size or type II error calculations. n The parameter d is defined as: so that it can be used for all problems regardless of the values of  0 and . n Chart VI e,f,g,h are used in t-test. (pp. A14-A15)

45 Horng-Chyi HorngStatistics II85 Example 8-9 n In example 8-8, if the mean load at failure differs from 10 MPa by as much as 1 MPa, is the sample size n = 22 adequate to ensure that H 0 will be rejected with probability at least 0.8? s = 3.55, therefore, d = 1.0/3.55 = 0.28. Appendix Chart VI g, for d = 0.28, n = 22 =>  = 0.68 The probability of rejecting H 0 :  = 10 if the true mean exceeds this by 1.0 MPa (reject H 0 while H 0 is false) is approximately 1 -  = 0.32, which is too small. Therefore n = 22 is not enough. At the same chart, d = 0.28,  = 0.2 (1-  =0.8) => n = 75

46 Horng-Chyi HorngStatistics II86

47 Horng-Chyi HorngStatistics II87 Construction of the C.I. on the Mean - Variance Unknown n In general, the distribution of is t with n-1 d.f. n Use the properties of t with n-1 d.f.,

48 Horng-Chyi HorngStatistics II88 Formula for C.I. on the Mean with Variance Unknown

49 Horng-Chyi HorngStatistics II89 Example 8-10 Reconsider the tensile adhesive problem in Example 8-8. Find a 95% C.I. on the mean? N = 22, sample mean = 13.71, s = 3.55, t  /2,n-1 = t 0.025,21 = 2.080 N = 22, sample mean = 13.71, s = 3.55, t  /2,n-1 = t 0.025,21 = 2.080 13.71 - 2.080 (3.55) /  22    13.71 + 2.080 (3.55) /  22 13.71 - 1.57    13.71 + 1.57 12.14    15.28 The 95% C.I. On the mean is [12.14, 15.28]

50 Horng-Chyi HorngStatistics II90 Final Note for the Inference on the Mean


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