【正文】 % S De v i a t i o n , S t d 2 . 2 8 3 61 3 . 7 2 . 1 ) 2 . 7 3 . 4 0 . 8 3 ( 1 . 7 X M e a n ,0 . 0 5 l e v e l , Co n f i d e n c e6 n s i z e , S a m p l e2) / (???????????????????????aa 6Using student t distribution 0 . 9 9 7 2 . 2 8 3 ( 0 . 3 8 8) 2 . 5 7 1 2 . 2 8 3 nS t X i n t erv al c o n f i d e n c e 95% S De v i at i o n , S t d 2 . 2 8 3 61 3 . 7 2 . 1 ) 2 . 7 3 . 4 0 . 8 3 ( 1 . 7 X M e a n ,1) 6 ( i . e . 5 f r e e d o m , of De g r e e0 . 0 5 l ev el , Co n f i d en c e6 n s i z e , S a m p l e2) / (????????????????????????aua 6Although the confidence level is the same, using t distribution will result in a larger interval value, because: ? 標(biāo)準(zhǔn)偏差 , S for small sample size is probably not accurate ? 標(biāo)準(zhǔn)偏差 , S for small sample 。 then 流程 is not centered Cpk = Cp 。Abcot Example (con’t) At x = , z = (x m) / ? = ( 5) / = At x = , z = (x m) / ? = ( 5) / = From standardised normal distribution table, at z = P(?) = 1 = at z = P(?) = Hence, P( ? x ? ) = = (approx equal to by binomial distribution) ： ； :； ： Abcot Introduction to Sampling What is population in statistic ? A population in statistic refers to all items that have been chosen for study. What is a sample in statistic ? A sample in statistic refers to a portion chosen from a population, by which the data obtain can be used to infer on the actual performance of the population Population Sample 2 Sample 6 Sample 8 Sample 1 Sample 3 Sample 7 Sample 4 Sample 5 Abcot Symbols for population 和 sample For population, Population mean = m Population size = N Population 標(biāo)準(zhǔn)偏差 = ? For sample, Sample mean = X Sample size = n Sample 標(biāo)準(zhǔn)偏差 = S Sampling distribution a distribution of sample means If you take 10 samples out of the same populations, you will most likely end up with 10 different sample means 和 sample 標(biāo)準(zhǔn)偏差 s. A Sampling distribution describes the probability of all possible means of the samples taken from the same population. Population mean, m = 150 Sampling distribution mean, X Sample 2, mean X2 Sample 1, mean X1 Sample 3, mean X3 Sample n, mean Xn Collection of sample means Abcot m = 150 Population distribution with ? = 25 When sample size increases, the standard error (or the std deviation of sampling distribution) will get smaller. Sample distribution with std error ?x much less than 25 (when n = 30) Sample distribution with std error ?x 25 (when n = 5) Sampling distribution (con’t) Like all normal curve, the sampling distribution can be described by its mean, x 和 its 標(biāo)準(zhǔn)偏差 ?x (which is also known as standard error of the mean). As such, the standard error measures the extent to which we expect the means from different samples to vary due to chance error in sampling 流程 . Standard error = Population 標(biāo)準(zhǔn)偏差 / square root of sample size ?x = ? / n Abcot Central Limit Theorem 1. The mean x of the sampling distribution will approximately equal to the population mean regardless of the sample size. The larger the sample size, the closer the sample mean is towards the population mean. 2. The sampling distribution of the mean will approach normality regardless of the actual population distribution. 3. It assures us that the sampling distribution of the mean approaches normal as the sample size increases. It allow us to use sample statistics to make inferences about population parameters without knowing anything about the actual population distribution, other than what we can obtain from the sample. m = 150 Population distribution x = 150 Sampling distribution (n = 5) x = 150 Sampling distribution (n = 20) Abcot Example of sampling distribution The population distribution of annual ine of engineers is skewed negatively. This distribution has a mean of $48,000, 和 a 標(biāo)準(zhǔn)偏差 of $5000. If we draw a sample of 100 engineers, what is the probability that their average annual ine is $48700 和 more. m = $48K Population distribution ? = $5000 x = $48K Sampling distribution (n = 100) ?x = ? $ Since population mean is equal to sampling distribution mean (. central limit theorem), hence X = m = 48000 Sampling distribution mean = Standard error = ?x = ? / n = 5000 / 100 = 5000 / 10 = 500 Abcot Example of sampling distribution (con’t) x = $48K Sampling distribution (n = 100) ?x = 500 $ Therefore, mean = 48000 sigma = 500 X = 50000 Z = (48700 48000) / 500 = 700 / 500 = From the standardized normal distribution table, P(X ? $48700) = Therefore。 3? variation is called natural tolerance Area under a Normal Distribution Abcot 流程 capability potential, Cp Based on the assumptions that : 1. 流程 is normal Normal Distribution 6? 5? 4? 3? 2? 1? 0 1? 2? 3? 4? 5? 6? Lower Spec Limit LSL Upper Spec Limit USL Specification Center 2. It is a 2sided specification 3. 流程 mean is centered to the device specification Spread in specification Natural tolerance CP = ? USL LSL 6? ? 8? 6? = Abcot 流程 Capability Index, Cpk 1. Based on the assumption that the 流程 is normal 和 in control 2. An index that pare the 流程 center with specification center Normal Distribution 6? 5? 4? 3? 2? 1? 0 1? 2? 3? 4? 5? 6? Lower Spec Limit LSL Upper Spec Limit USL Specification Center Therefore when , Cpk C