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Inference on the Mean of a Population &4-4 (&8-2) -Variance Known H0: m = m0 H1: m m0 , where m0 is a specified constant. Sample mean is the unbiased point estimator for population mean. If X 1 , X 2 , , X n are samples drawn from a distributi on 2 with mean m and variance , then X ~ N m , . n 2 Therefore, if H 0 is true ( m m 0 ), then X m0 Z0 ~ N 0,1 n Horng-Chyi Horng Statistics II 41 The Reasoning For H0 to be true, the value of Z0 can not be too large or too small. Recall that 68.3% of Z0 should fall within (-1, +1) 95.4% of Z0 should fall within (-2, +2) 99.7% of Z0 should fall within (-3, +3) What values of Z0 should we reject H0? (based on a value) What values of Z0 should we conclude that there is not enough evidence to reject H0? Horng-Chyi Horng Statistics II 42 Horng-Chyi Horng Statistics II 43 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? Horng-Chyi Horng Statistics II 44 Horng-Chyi Horng Statistics II 45 Hypothesis Testing on m - Variance Known H1 Test Statistic m m0 m > m0 Z0 X m0 n m < m0 Horng-Chyi Horng Reject H0 if Z0 > za or Z0 < - za Z0 > za Z0 < -za Statistics II 46 P-Values in Hypothesis Tests(I) Where Z0 is the test statistic, and (z) is the standard normal cumulative function. In example 8-2, Z0 = 3.25, P-Value = 2[1-(3.25)] = 0.0012 Horng-Chyi Horng Statistics II 47 P-Values of Hypothesis Testing on m - Variance Known H1 m m0 m > m0 Test Statistic X m0 Z0 n m < m0 Horng-Chyi Horng P-Value P = 2 * P(Z |Z0|) P = P(Z Z0 ) P = P(Z Z0 ) Statistics II 48 P-Values in Hypothesis Tests(II) a-value is the maximum type I error allowed, while Pvalue is the real type I error calculated from the sample. a-value is preset, while P-value is calculated from the sample. When P-value is less than a-value, we can safely make the conclusion “Reject H0”. By doing so, the error we are subjected to (P-value) is less than the maximum error allowed (a-value). Horng-Chyi Horng Statistics II 49 Type II Error - Fail to reject H0 while H0 is false Horng-Chyi Horng Statistics II 50 How to calculate Type II Error? (I) (H0: m = m0 Vs. H1: m m0) Under the circumstance of type II error, H0 is false. Supposed that the true value of the mean is m = m0 + d, where d > 0. The distribution of Z0 is: X m 0 X m 0 d d Z0 n n n N 0, 1 d n d Therefore, Z 0 ~ N , 1 n Horng-Chyi Horng Statistics II 51 How to calculate Type II Error? (II) - refer to section &4.3 (&8.1) Type II error occurred when (fail to reject H0 while H0 is false) d Za Z 0 Za where Z 0 ~ N , 1 2 2 / n Therefore, P Za / 2 Z 0 Za / 2 d d d Z Z Z a /2 0 a /2 / n / n / n P 1 1 1 d d P Za / 2 Z Za / 2 / n / n d d Za / 2 Za / 2 / n / n Horng-Chyi Horng Statistics II 52 The Sample Size (I) Given values of a and d, find the required sample size n to achieve a particular level of .. d d Since Za / 2 Za / 2 / n / n d Za / 2 when d 0 / n Let Z Then, Z Za / 2 Horng-Chyi Horng d / n 2 Za / 2 Z 2 n d 2 Statistics II whe re d m m 0 53 The Sample Size (II) Two-sided Hypothesis Testing One-sided Hypothesis Testing Horng-Chyi Horng Statistics II 54 Example 8-3 Horng-Chyi Horng Statistics II 55 The Operating Characteristic Curves - Normal test (z-test) Use to performing sample size or type II error calculations. The parameter d is defined as: d | m m0 | |d | so that it can be used for allproblems regardless of the values of m0 and . Chart VI a,b,c,d are for Z-test. Horng-Chyi Horng Statistics II 56 Example 8-5 Horng-Chyi Horng Statistics II 57 Horng-Chyi Horng Statistics II 58 Horng-Chyi Horng Statistics II 59 Large Sample Test If n 30, then the sample variance s2 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. Horng-Chyi Horng Statistics II 60 Large Sample Hypothesis Testing on m - Variance Unknown but n 30 H1 Test Statistic m m0 m > m0 Z0 X m0 s n m < m0 Horng-Chyi Horng Reject H0 if Z0 > za or Z0 < - za Z0 > za Z0 < -za Statistics II 61 Statistical Vs. Practical Significance Practical Significance = 50.5-50 = 0.5 Statistical Significance P-Value for each sample size n. Horng-Chyi Horng Statistics II 62 Notes be careful when interpreting the results from hypothesis testing when the sample size is large, because any small departure from the hypothesized value m0 will probably be detected, even when the difference is of little or no practical significance. In general, two types of conclusion can be drawn: 1. At a = 0.**, we have enough evidence to reject H0. 2. At a = 0.**, we do not have enough evidence to reject H0. Horng-Chyi Horng Statistics II 63 Confidence Interval on the Mean (I) Point Vs. Interval Estimation The general form of interval estimate is L m U in which we always attach a possible error a such that P(L m U) = 1-a That is, we have 1-a confidence that the true value of m will fall within [L, U]. Interval Estimate is also called Confidence Interval (C.I.). Horng-Chyi Horng Statistics II 64 Confidence Interval on the Mean (II) L is called the lower-confidence limit and U is the upper-confidence limit. Two-sided C.I. Vs. One-sided C.I. Horng-Chyi Horng Statistics II 65 Construction of the C.I. From Central Limit Theory, 2 If X ~ m , and n 25, X ~ N m , . n 2 Use standardization and the properties of Z, X m Z and P za / 2 Z za / 2 1 a n X m P za / 2 za / 2 1 a n P X za / 2 / n m X za / 2 / n 1 a Horng-Chyi Horng Statistics II 66 Formula for C.I. on the Mean with Variance Known Used when 1. Variance known 2. n 30, use s to estimate . Horng-Chyi Horng Statistics II 67 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 => a = 0.05, za/2 = z0.025 = 1.96 Horng-Chyi Horng Statistics II 68 Notes - C.I. Relationship between Hypothesis Testing and C.I.s Confidence level (1-a) and precision of estimation (C.I. * 1/2) Sample size and C.I.s Horng-Chyi Horng Statistics II 69 Choice of Sample Size to Achieve Precision of Estimation Horng-Chyi Horng Statistics II 70 Example 8-7 Horng-Chyi Horng Statistics II 71 One-Sided C.I.s on the Mean Horng-Chyi Horng Statistics II 72 Inference on the Mean of a Population &4-5 (&8-3) -Variance Unknown H0: m = m0 H1: m m0 , where m0 is a specified constant. Variance unknown, therefore, use s instead of in the test statistic. 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. Horng-Chyi Horng Statistics II 73 Inference on the Mean of a Population -Variance Unknown Let X1, X2, …, Xn be a random sample for a normal distribution with unknown mean m and unknown variance 2. The quantity has a t distribution with n - 1 degrees of freedom. Horng-Chyi Horng Statistics II 74 pdf of t distributi on : k 1 / 2 1 k 1 / 2 k k / 2 x 2 / k 1 k is the number of degrees of freedom. f x Horng-Chyi Horng Statistics II 75 The Reasoning For H0 to be true, the value of T0 can not be too large or too small. What values of T0 should we reject H0? (based on a value) What values of T0 should we conclude that there is not enough evidence to reject H0? Although when n 30, we can use Z0 in section &8-2 to perform the testing instead. We prefer using T0 to more accurately reflect the real testing result if t-table is available. Horng-Chyi Horng Statistics II 76 Horng-Chyi Horng Statistics II 77 Example 8-8 Horng-Chyi Horng Statistics II 78 Horng-Chyi Horng Statistics II 79 Testing for Normality (Example 8-8) - t-test assumes that the data are a random sample from a normal population (1) Box Plot Plot Horng-Chyi Horng (2) Normality Probability Statistics II 80 Hypothesis Testing on m - Variance Unknown H1 m m0 m > m0 Test Statistic X m0 T0 s n m < m0 Horng-Chyi Horng Reject H0 if T0 > ta or T0 < - ta T0 > ta T0 < -ta Statistics II 81 Finding P-Values Steps: 1. Find the degrees of freedom (k = n-1)in the t-table. 2. Compare T0 to the values in that row and find the closest one. 3. Look the a value associated with the one you pick. The p-value of your test is equal to this a value. In example 8-8, T0 = 4.90, k = n-1 = 21, P-Value < 0.0005 because the t value associated with (k = 21, a = 0.0005) is 3.819. Horng-Chyi Horng Statistics II 82 P-Values of Hypothesis Testing on m - Variance Unknown H1 m m0 m > m0 Test Statistic X m0 T0 s n m < m0 Horng-Chyi Horng P-Value P = 2 * P(tn-1 |T0|) P = P(tn-1 T0 ) P = P(tn-1 T0 ) Statistics II 83 The Operating Characteristic Curves - t-test Use to performing sample size or type II error calculations. The parameter d is defined as: d | m m0 | |d | so that it can be used for all problems regardless of the values of m0 and . Chart VI e,f,g,h are used in t-test. (pp. A14-A15) Horng-Chyi Horng Statistics II 84 Example 8-9 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 H0 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 H0: m = 10 if the true mean exceeds this by 1.0 MPa (reject H0 while H0 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 Horng-Chyi Horng Statistics II 85 Horng-Chyi Horng Statistics II 86 Construction of the C.I. on the Mean - Variance Unknown T X m In general, the distribution of is t with n-1 d.f. Use the properties of t with n-1 d.f., s n P ta / 2,n 1 T ta / 2,n 1 1 a X m P ta / 2,n 1 ta / 2,n 1 1 a s n P X ta / 2,n 1s / n m X ta / 2,n 1s / n 1 a Horng-Chyi Horng Statistics II 87 Formula for C.I. on the Mean with Variance Unknown Horng-Chyi Horng Statistics II 88 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, ta/2,n-1 = t0.025,21 = 2.080 X ta / 2,n1s / n m X ta / 2,n1s / n 13.71 - 2.080 (3.55) / 22 m 13.71 + 2.080 (3.55) / 22 13.71 - 1.57 m 13.71 + 1.57 12.14 m 15.28 The 95% C.I. On the mean is [12.14, 15.28] Horng-Chyi Horng Statistics II 89 Final Note for the Inference on the Mean Variance Sample Size Population Type Testing Method Z Z Chebyshev’s Inequality T or Z T Non-parametric Known N > 30 N < 30 All Normal Non-normal Unknown N > 30 N < 30 All Normal Non-normal Horng-Chyi Horng Statistics II 90