工學(xué)grr分析ppt課件-資料下載頁
2025-01-19 09:17本頁面
【正文】 THOD 1: ? 選取 20件樣品 ,2名檢驗人員 ,每人進行兩次測詴 . 若所有判定均相同 ,則接受 ,否則 ,應(yīng)改善或再評價 五 、 研究特征 GRR的方式 MEASUREMENT SYSTEM / GAUGE CAPABILITY ATTRIBUTE CALCULATION WORKSHEETPART NAME: PART NUMBER:CHARACTERISTIC: GAGE NAME:GAUGE NUMBER: GAGE TYPE:OPERATOR:SAMPLE 1st 2nd 1st 2ndNUMBER TRIAL TRIAL TRIAL TRIAL 1 GO GO GO GO2 NO GO GO NO GO NO GO3 GO GO GO GO4 GO GO GO GO5 NO GO NO GO NO GO NO GO6 GO GO GO GO7 NO GO NO GO NO GO NO GO8 GO GO GO GO9 GO GO GO GO10 NO GO NO GO NO GO NO GO11 GO GO GO GO12 GO GO GO GO13 NO GO NO GO NO GO NO GO14 GO GO GO GO15 GO GO NO GO GO16 GO GO GO GO17 NO GO NO GO NO GO NO GO18 GO GO GO GO19 GO GO GO GO20 GO GO GO GOB:A:? METHOD 2 : ? 選取 100件樣品 ,3名檢驗人員 ,每人進行二次測詴 . 依據(jù)所有判定結(jié)果計算 ,若正確判定之機率 70%, 則接受 ,否則 ,應(yīng)改善或再評價 五 、 研究特征 GRR的方式 方法 2:統(tǒng)計表單淺析 所有的檢查者對該樣品的判定是否一致 . DATA ENTRY WORKSHEETDATE: When study took placeAttribute Legend5 (used in putations) NAME: Study CoOrdinator1 pass ITEM: Name and Number (Identification) All operators2 fail LOCATION: Where Study took place agree within and All Operatorsbetween each agree withOther standardKnown Population Y/N Y/NSample Attribute Try 1 Try 2 Try 1 Try 2 Try 1 Try 2 Agree Agree1 pass pass pass pass pass fail fail N N2 pass pass pass pass pass fail fail N N3 fail fail fail fail pass fail fail N N4 fail fail fail fail fail fail fail Y YOperator 1 Operator 2 Operator 3所有的檢查者對樣品的判定與標準是否一致 . 100% OPERATOR SCORE (1) % % %% SCORE VS. ATTRIBUTE (2) % % %SCREEN % EFFECTIVE SCORE (3) %SCREEN % EFFECTIVE SCORE vs. ATTRIBUTE (4) %? METHOD 3 : ? 選取 30件樣品 ,2名檢驗人員 ,每人進行三次測詴 . 依據(jù)所有判定結(jié)果計算 ,若正確判定之機率 80%, 則接受 ,否則 ,應(yīng)改善或再評價 五 、 研究特征 GRR的方式 OPERATOR:SAMPLE 1st 2nd 3rd 1st 2nd 3rdNUMBER TRIAL TRIAL TRIAL TRIAL TRIAL TRIAL 1 1 1 1 1 1 1 1 12 1 1 1 1 1 1 1 13 1 1 1 1 0 0 0 14 1 1 1 1 1 1 1 15 0 0 0 1 0 0 0 16 1 1 1 1 1 1 1 17 1 1 1 1 1 1 1 18 1 1 1 1 0 1 1 09 0 0 0 1 1 1 1 110 0 1 1 0 0 0 0 111 0 0 0 1 1 0 1 012 1 1 1 1 1 1 1 113 1 1 1 1 1 1 1 114 0 1 1 0 0 0 0 115 1 1 1 1 1 1 1 116 1 1 1 1 1 0 1 017 0 0 0 1 1 1 1 118 1 1 1 1 1 1 1 119 1 1 1 1 1 1 1 120 0 0 0 1 1 1 1 121 1 1 1 1 0 0 0 122 0 0 0 1 1 0 1 023 1 1 1 1 1 1 1 124 1 1 1 1 1 1 1 125 1 1 1 1 1 1 1 126 1 1 0 0 1 1 1 127 0 0 0 1 0 0 0 128 1 0 1 0 1 1 1 129 1 1 1 1 1 1 1 130 1 1 1 1 0 1 1 0TOTALS 21 22 22 26 22 21 24 2526 25AD BDSUM(A)= 65 SUM(B)= 67Number of Samples:( S )=30 Number of Trials:( T )= 3A: B:DIFF A DIFF A方法 3:結(jié)果判定 ? 同一操作者間的一致性 : ? 一致性 TEST(Z) = [AD(S*)]/ (S**) ? (Z), 表示錯誤的判定 20%, REJECT THE GAUGE (Z)=, ? 表示正確的的判定 80%, ACCEPT THE GAUGE ? 兩個操作者間的一致性 : ? TEST(B) = ? ? TEST(B), REJECT THE GAUGE ? TEST(B)=, ACCEPT THE GAUGE S*T*(A+B)(A2+B2) (S*T*2)*(A2+B2)S*T*(A+B)2 T H A N K S !
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