【正文】 表AM911AM2PM123123123位置板間空銲橋銲其他空銲橋銲其他空銲橋銲其他空銲橋銲其他空銲橋銲其他空銲橋銲其他空銲橋銲其他空銲橋銲其他空銲橋銲其他中右ABCDE中左ABCDE右邊ABCDE左邊ABCDE)1. 綜合判斷defect比率31llppm與往日紀(jì)錄相近，所以數(shù)據(jù)可信。可配因子A B CD F G H列 行1 2 34 5 6 71 1 11 1 1 121 1 12 2 2 231 2 21 1 2 241 2 22 2 1 152 1 21 2 1 262 1 22 1 2 172 2 11 2 2 182 2 12 1 1 2成分 AA B B A BC C C A B CYATES演算法,求偏差平方和【例2】如【例1】求因子之效果差及各種偏差平方和代號(hào)數(shù)據(jù)(1)(2)(3) 效果差(3)2/8名稱(1)25162898CTc3131224/8SCb564864/8SBbc86200SBCa318416/8SAac330636/8SACab402864/8SABabc222416/8SABC【練習(xí)1】：試計(jì)算如下之因子之效果差及各種偏差平方和A1A2B1B2B1B2C1D12423D23128C2D165686D2810910解ABCDDATA1234S11111251939165CT1112d3142012663SD1121c65195115SC1122cd8151076869SCD1211b4431989SB1212bd11529675SBD1221cd5113975SCD1222bcd1096717877SBCD2111a219187SA2112ad22108873SAD2121ac6311177SAC2122acd95857487SACD2211ab301187SAB2212abd8387473SABD2221abc8653773SABC2222abcd1076818491SABCD散佈圖定公差通常產(chǎn)品之規(guī)格包括兩種數(shù)值，一為數(shù)值表示，另一為範(fàn)圍表示，數(shù)值部分稱為規(guī)格值，範(fàn)圍表示稱為公差，製程條件也有此兩種數(shù)值，因此當(dāng)製程條件決定時(shí)(田口稱為參數(shù)設(shè)計(jì))除了數(shù)值部份，亦應(yīng)檢討重要條件之公差範(fàn)圍(田口稱為公差設(shè)計(jì))，使製程規(guī)格明確，以方便製程管制。這種完全組合之實(shí)驗(yàn)，我們稱為完全配置，這種配置主要目的在求得各個(gè)因子效果之大小，及因子組合後之組合效果(又稱交互作用)之大小，一般說來，效果大小來自各水準(zhǔn)間之差異，差異愈大，表示效果大，因此有如下之問題必須解決(1)有多少效果必須計(jì)算(2)因子效果如何計(jì)算，包括主因子效果，交互作用(3)如何比較效果，下表表示因子數(shù)與效果數(shù)。2. 之分配，n[()/s]2為趨近分配，此處值如下表所示，從表中可知，d2=(1+1/4n)，當(dāng)組數(shù)夠大時(shí)，=d23. 當(dāng)組數(shù)k=1，即為一般所稱之R，在不致誤解下，有時(shí)亦表示成，同理有時(shí)以d2表示。C(L3)－7。C(L2)－5。C(L1) O。例:HOUREMETER(Component serach)An hour meter , built by an electronics pany , had a 2025 percent defect rate because several of the units could not meet the customer39。判斷法：如果A因子有重要影響，則應(yīng)該有AL在RH群中，會(huì)有顯著的差別出現(xiàn)，同時(shí)AH在RL群中會(huì)有顯著的差別出現(xiàn)，也因此我們有如下之判斷方法。s test for two independent samples.（5） 判斷所選擇因子中有影響的大要因存在，可進(jìn)行步驟2（6）如果 判斷所選擇的因子中無影響大要因存在，回到步驟1步驟2：消去(elimination)逐一確認(rèn)要因中是那一個(gè)重要(Red X or Pink X)，同時(shí)去除不重要因子（消去法）作法：從第一個(gè)變數(shù)稱為A開始，選擇好的條件AH，AL，其餘的變數(shù)組合為差的組合稱為RL，RH。同值時(shí)之計(jì)算，有兩種可能發(fā)生1 當(dāng)兩組之最大值等於聯(lián)合最大值，或者兩組之最小值等於聯(lián)合最小值，在這些情形其中任何一個(gè),T值為0.2 如果一組樣本包含聯(lián)合最大值和另一組有聯(lián)合最小值，若發(fā)生另組數(shù)值與聯(lián)合最小相同時(shí)或若發(fā)生另組數(shù)值與聯(lián)合最大相同時(shí)，此時(shí)個(gè)數(shù)以1/2個(gè)計(jì)算組件及變數(shù)搜尋(Component amp。步驟終止。s test to cover all cases in which the sample sizes n. and n1，n2 are both 20 or less. l The range method can also be used when the sample size exceeds 20. With two samples each of size 24, for example, each sample may be divided at random into two groups of size 12. The range is found for each group, and the average of the four ranges is taken. Lord (3) gives the necessary tables. This device keeps the efficiency of the range test high for samples greater than 20, though the calculation takes a little longer. l To summarize, the range test is convenient for normal samples if a 5% to 10% loss in information can be tolerated. It is often used when many routine tests of significance or calculations of confidence limits have to be made. SHAININ使用Tukey Quick Test之步驟1 找出聯(lián)合的那些二組樣本裡的最大和最小值，稱為聯(lián)合最大，及聯(lián)合最小2 如果聯(lián)合最大值和聯(lián)合最小值兩者在相同組的樣本發(fā)生,我們沒有足夠證據(jù)證實(shí)二組樣本是不同的。s t, used in example for these data, gave closely similar results for the significance level and the confidence limits. l For two independent samples of unequal sizes。Cmedian－37－5range57D=32，=(5+7)/2=6，D: =32:6=:1，The test for a significant and repeatable difference between the good units and bad units is determined by the formula : D: :1，The Red X and Pink X are among the causes being considered and there is good repeatability in the disassembly / reassembly process .Lord39。CResults after 2nd disassembly/ reassembly(H3):－37。CResults after lst disassembly/ reassembly(H2):－35。s reliability requirement of perfect operation at 40 C .The worst units could only reach 0 C before malfunction .T