6-15 The length of industrial filters is a quality characteristic of interest. T

Question

6-15 The length of industrial filters is a quality characteristic of interest. Thirty samples, each of size 5, are chosen from the process. The data yield an average length of 110 mm, with the process standard deviation estimated to be 4 mm. a. Find the warning limits for a control chart for the average length. D. Find the 3 control limits. What is the probability of a type I error? c. If the process mean shifts to 112 mm, what are the chances of detecting this shift by the third sample drawn after the shift? d. What is the chance of detecting the shift for the first time on the second sample point drawn after the shift? e. What is the ARL for a shift in the process mean to 112 mm? How many samples, on average, would it take to detect a change in the process mean to 116 mm?

Explanation / Answer

HERE SAMPLE SIZE = 30 X 5 =150 ,S.D. OF SAMPLE = 4 ,MEAN LENGTH = 110 MM

HENCE STANDARD ERROR = s / n - 1 = 4 / 150 -1 = 0.328

HENCE CONFIDENCE INTERVAL FOR POPULATION MEAN = 110 ± S.E X Z ( FOR ALARMING SITUATION

Z = 3 OTHER WISE Z= 1.96)

PUTTING VALUES CONFIDENCE INTERVAL = 110 ± 0.328 X 3 = 110± 0.984

= 109.016 TO 110.984 i.e length between this limit

FOR 3 S.D. CONFIDENCE LIMIT I.E, 99.73% CONFIDENCE Z =3 ON WHICH CALCULATION IS ALREADY

DONE. NORMALLY Z = 1.96 FOR 95 % CONFIDENCE

TYPE ONE ERROR PROBABILITY IS GIVEN AS = REJECTING H0 / H1 IS TRUE

(C) IF MEAN IS SHIFTED TO 112 MM.AND CHECKED BY ON THE BASIS OF 3X5 = 15 SAMPLES

THEN S,E,= 4/ 15 -1 = 1.07 AND`t` DISTRIBUTION WILL BE USED AS n < 30

AND t = (112-110)/1.07 =1.87 by table chance can be found

FOR SHIFTING MEAN 112 TO 116 AGAIN S.E, = 4/ 10-1 =1.27 1S SAMPLE SIZE 10

AND `t` value =(116-112)/1.27 =3.14 and chance from t table

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