Lecture #39 (tape #39) Vishesh Kumar TA. Dec. 6, 2002

Transcription

1 Lecture #39 (tape #39) Vishesh Kumar TA Dec. 6, 2002

2 Control Chart for Number of Defectives Sometimes, more convenient to make chart by plotting d rather than p. In particular, plotting d appropriate if - n is constant - p is very small Referred to as np chart. Basically same as p chart except for scaling factor, n.

3 np chart requires one less calculation since p = d/n need to be calculated for each sample Recall that µ d = np' σ d 2 = np' ( 1 p' ) or σ d = np' ( 1 p' ) Form of the control limits (use p to estimate p ): np ± 3 np( 1 p)

4 Revisit the instrument panel assembly example. p = = np = 100( 0.079) = 7.9 np ± 3 np( 1 p) 7.9 ± 3 7.9( ) UCL np = LCL np = so use a lower limit of 0.0

5 Problems with np Chart If sample size varies, control limits will vary from sample to sample If sample size varies, centerline will vary as well from sample to sample This makes chart interpretation difficult d value (easier calculations) easy to communicate -- but what does a value of d = 12 mean?? Doesn t mean much unless you also know sample size.

6 Variable-Sample-Size p Charts Speaking of dealing with varying sample sizes, how do we manage this for our p Chart???? Good news, p will not change from sample to sample, but control limits depend on n. How to handle: Separate limits for each subgroup Limits based on n-bar Standardized p-chart

7 Manufacture of a Package Tray Wood fiber/polymer composite extruded into thin sheets for use in hatchbacks. Sheets heated then molded into shape of tray. Carpet applied to sheet w/ adhesive. Carpeted tray trimmed to size trays per shift. Defects observed: cloth coverage on hinge, carpet coverage problem, bleed-through, soiled carpet, improper flex on hinge, (chips, scratches, abrasions on back surface), wrinkles

8 Data for Package Tray Example Sample n d p LCL p UCL p : : : : : : : :

9 p Chart - Individual Limits sum of the n s = 7433 sum of the d s = 582 k k p = d i n i i = 1 i = 1 = 582 / 7433 = p ± 3 p ( 1 p) n For sample 1, ± : 0.026, 0.131

10 p Chart --- n-bar Limits Plot the p s and p. Put limits on the chart based on n. n = k n i i = 1 k = 7433 / 30 = approx. 248 p ± 3 p ( 1 p) ± 3 n ( ) 248 Limits are & Pts near the limits must be interpreted separately.

11 p Chart --- Standardized Limits Calc. standardized p value for each sample Sample p p - pbar n σ p Z=(p - pbar)/σ p

12 Attribute Control Charts So far, p chart variable sample size p-charts - individual control limits - n-bar limits - standardized chart Now we want to construct a chart for defects, a c-chart

13 c - Control Chart c Sample No.

14 c - Control Chart To find limits for the c Chart we need to more about how defects arise in a probabilistic sense. Defectives and fraction defective -- binomial distn. 4 defects on the part small enough area: B or W

15 Defects We decompose the surface into a very large number of sections that are either B or W. Let s say N=1,000, on the average 5 black spots, p = (big n, little p, np from 1 to 20) P(4) = ( ) ( ) 4 = We could use the binomial distn. to work with defects But, the math can be messy.

16 Poisson Distribution Pc ( ) = λ c e λ c! λ is the mean number of defects per sample (c, µ c ) # defects in the sample = c (sometimes x) So, for c = 5, P(c=4) = 5 c e c! = (very close to binomial calc.)

17 Sample Size The mean number of defects, c, on one of our surfaces (say 12" x 12") is 5. IMPORTANT!!!! If we were to sample surfaces like this for the purpose of constructing a control chart OPPORTUNITY SPACE FOR OCCURRENCE OF DEFECTS MUST BE KEPT CONSTANT!! Surface area must be the same (all 12" x 12")

18 What is c if surface is only 1/4 the size?? What if surface is twice the size?? c =2 (# of breaks in a piece of wire that is 20 yds long) What is probability of 2 breaks in a wire that is 10 yds long? What is probability of 2 or more breaks in a wire that is 10 yds long?

19 Poisson (c = λ = 6) Probability # of defects (c or x)

20 Poisson (c = λ = 3) Probability # of defects (c or x)

21 µ c = λ = c' Mean and Variance σ c 2 = λ = c' In principle, µ c ± 3σ c or c' ± 3 c' We don t know c so we must estimate it with c. c = k c i i = 1 k