x>"4IPM.Microsoft Mail.Note1 H#  #& )6p Suggested version of prc241.htmq m jx@Ȏ)T>? SMTP:PTOBIAS@AUSTIN.RR.COM @N  ydBuBbr[!€  *&oLZFu " rcpg1252 `n033Och set0 } vwk d4 `cP  obja@ph\'   A,I'vle h  i of t nfinc tv@ g  ;!=nps.4 Iy`pp`,@ aorc2@41.htmCu @ s a 7e.Pa.u P#  F F  F FRt FT9.0  F : F< F= F R FS FydBuBbr[!ydBuBbr[!8+*VPSTPRX.DLLNITA7nC:\Documents and Settings\Paul\Local Settings\Application Data\Microsoft\Outlook\outlook.pst475^beALAN,IVEATTACHEDAVERSIONOFTHECONFIDENCEINTERVALPAGETHATHASBOTHTHEOL;DANDTHENEWMETHODSIFYOUAPPROVE,CHE A2In< 7.2.4.1. Confidence intervals Next Page Previous Page Home Tools & Aids Search Handbook
7. Product and Process Comparisons
7.2. Comparisons based on data from one process
7.2.4. Does the proportion of defectives meet requirements?

7.2.4.1.

Confidence intervals
Confidence intervals using the method of Agresti and Coull The method recommended by Agresti and Coull (1998) and also by Brown, Cai and DasGupta (2001) is to use the form of the confidence interval that corresponds to the hypothesis test given in Section 7.2.4. That is, solve for the two values of p0 (say, pupper and plower) that result from setting z = z(alpha/2 and solving for p0 = pupper, and then setting z = -z(alpha/2 and solving for p0 = plower. (Here, as in Section 7.2.4, z(alpha/2 denotes the variate value from the standard normal distribution such that the area to the right of the value is alpha/2.) Although solving for the two values of p0 might sound complicated, the appropriate expressions can be obtained by straightforward but slightly tedious algebra. Such algebraic manipulation isn't necessary, however, as the appropriate expressions are given in various sources. Specifically, we have
Formulas for the confidence intervals
    U. L. = {phat + z(alpha/2)^(2)/(2n) + z(alpha/2/(sqrt[(phat*(1-phat))/n + z(alpha/2)^(2)/(4n^2)]}/ {1 + z(alpha/2)^(2)/n}

    L. L. = {phat + z(alpha/2)^(2)/(2n) - z(alpha/2/(sqrt[(phat*(1-phat))/n + z(alpha/2)^(2)/(4n^2)]}/ {1 + z(alpha/2)^(2)/n}

Procedure does not strongly depend on values of p and n This approach can be substantiated on the grounds that it is the exact algebraic counterpart to the (large-sample) hypothesis test given in section 7.2.4 and is also supported by the research of Agresti and Coull. One advantage of this procedure is that its worth does not strongly depend upon the value of n and/or p, and indeed was recommended by Agesti and Coull for virtually all combinations of n and p.
Another advantage is that the lower limit cannot be negative Another advantage is that the lower limit cannot be negative. That is not true for the confidence expression most frequently used:
    phat +/- z(alpha/2)*SQRT[(phat*(1-phat))/n]
A confidence limit approach that produces a lower limit which is an impossible value for the parameter for which the interval is constructed is an inferior approach. This also applies to limits for the control charts that are discussed in Chapter 6.
One-sided confidence intervals A one-sided confidence interval can also be constructed simply by replacing each z(alpha/2) by z(alpha) in the expression for the lower or upper limit, whichever is desired. The 95% one-sided interval for p for the example in the preceding section is:
Example
    p >= lower limit

    p >= {phat + z(alpha)^(2)/(2n) - z(alpha)*sqrt[(phat*(1-phat))/n + z(alpha)^(2)/(4n^2)]}/ {1 + z(alpha)^(2)/n}

    p >= {0.13 + 1.645^2/(2*200) - 1.645*sqrt[(0.13*0.8713)/200 + 1.645^2/(4*200^2)]}/ {1 + z(alpha)^(2)/n}

    p >= 0.09577

Conclusion from the example Since the lower bound does not exceed 0.10, in which case it would exceed the hypothesized value, the null hypothesis that the proportion defective is at most .10, which was given in the preceding section, would not be rejected if we used the confidence interval to test the hypothesis. Of course a confidence interval has value in its own right and does not have to be used for hypothesis testing.
Constrution of exact two-sided confidence intervals based on the binomial distribution Exact Intervals for Small Numbers of Failures and/or Small Sample Sizes

If the the number of failures is very small or if the sample size N is very small, symmetical confidence limits that are approximated using the normal distribution may not be accurate enough for some applications. An exact method based on the binomial distribution is shown next.To construct a two-sided confidence interval at the 100(1 - )% confidence level for the true proportion defective p where Nd defects are found in a sample of size N follow the steps below.

  1. Solve the equation 


    2. SUM[k=0 to Nd][(N k)*p(U)**k*(1-p(U))**(N-k)] \= alpha/2

    for pU to obtain the upper 100(1-)% limit for p. 

  2. Next solve the equation 


    3. SUM[k=0 to Nd-1][(N k)*p(L)**k*(1-p(L))**(N-k)] \= 1 - alpha/2

    for pL to obtain the lower 100(1 - alpha)% limit for p.
Note The interval {pL, pU} is an exact 100(1 - alpha)% confidence interval for p. However, it is not symmetric about the observed proportion defective,phat = Nd/N.
Example of calculation of upper limit for binomial confidence intervals using EXCEL The equations above that determine pL and pU can easily be solved using functions built into EXCEL. Take as an example the situation where twenty units are sampled from a continuous production line and four items are found to be defective. The proportion defective is estimated to bephat = 4/20 = 0.20. The calculation of a 90% confidence interval for the true proportion defective, p, is demonstrated using EXCEL spreadsheets.
Upper confidence limit from EXCEL To solve for pU:
  1. Open an EXCEL spreadsheet and put the starting value of 0.5 in the A1 cell. 
  2. Put =BINOMDIST(Nd, N, A1, TRUE) in B1, where Nd = 4 and N = 20. 
  3. Open the Tools menu and click on GOAL SEEK. The GOAL SEEK box requires 3 entries. 
    • B1 in the "Set Cell" box 
    • alpha/2 = 0.05 in the "To Value" box 
    • A1 in the "By Changing Cell" box. 
    The picture below shows the steps in the procedure.
Excel spread sheet showing initial values
Final step
  1. Click OK in the GOAL SEEK box. The number in A1 will change from 0.5 to PU. The picture below shows the final result.
Excel spread sheet showing the computation of P(U)
Example of calculation of lower limit for binomial confidence limits using EXCEL The calculation of the lower limit is similar. To solve for pL:
  1. Open an EXCEL spreadsheet and put the starting value of 0.5 in the A1 cell. 
  2. Put =BINOMDIST(Nd -1, N, A1, TRUE) in B1, where Nd -1 = 3 and N = 20. 
  3. Open the Tools menu and click on GOAL SEEK. The GOAL SEEK box requires 3 entries. 
    • B1 in the "Set Cell" box 
    • 1 - alpha/2 = 1 - 0.05 = 0.95 in the "To Value" box 
    • A1 in the "By Changing Cell" box. 
    The picture below shows the steps in the procedure.
Excel spread sheet showing the initial values
Final step
  1. Click OK in the GOAL SEEK box. The number in A1 will change from 0.5 to pL. The picture below shows the final result.
Excel spread sheet showing the comptation of p(L)
Interpretation of result A 90% confidence interval for the proportion defective, p, is {0.071, 0.400}. Whether or not the interval is truly "exact" depends on the software. Notice in the screens above that GOAL SEEK is not able to find upper and lower limits that correspond to exact 0.05 and 0.95 confidence levels; the calculations are correct to two significant digits which is probably sufficient for confidence intervals. The calculations using a package called SEMSTAT agree with the EXCEL results to two significant digits.
Calculations using SEMSTAT The downloadable software package SEMSTAT contains a menu item "Hypothesis Testing and Confidence Intervals." Selecting this item brings up another menu that contains "Confidence Limits on Binomial Parameter." This option can be used to calculate binomial confidence limits as shown in the screen shot below.
SEMSTAT output for exact confidence interval

Calculations using Dataplot This computation can also be performed using the following Dataplot program. 
. Initalize
let p = 0.5
let nd = 4
let n = 20
. Define the functions
let function fu = bincdf(4,p,20) - 0.05
let function fl = bincdf(3,p,20) - 0.95
. Calculate the roots
let pu = roots fu wrt p  for p = .01 .99
let pl = roots fl wrt p  for p = .01 .99
. print the results
let pu1 = pu(1)
let pl1 = pl(1)
print "PU = ^pu1"
print "PL = ^pl1"
Dataplot generated the following results. 
 PU = 0.401029
 PL = 0.071354
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