530/R-09-007a

      ERRATA SHEET —MARCH 2009 UNIFIED GUIDANCE
                                August 9, 2010

      The following are corrections made to a number of Example calculations and
equations in the March 2009 Unified Guidance [EPA 530/R-09-007]:

Chapter 10, Example 10-1: In the initial table of Nickel concentrations, Tears' has
been changed to 'Wells' to maintain consistency with  succeeding examples. At the
bottom of page 10-11, the following sentence was added (to allow for pooling):
"Assume that the individual well data sets can be shown to arise from a single common
population."

Chapter 10, Example 10-4 Calculations for the Multiple Group Shapiro-Wilk
Test (full revised example text in parentheses):

      "The previous examples in this chapter pooled the data of Example 10-1 into a
single group before testing for normality. This time, treat each well separately and
compute the Shapiro-Wilk multiple group test of normality at the a = .05 level.

      SOLUTION
Step 1.  The nickel data in  Example 10-1 come  from K = 4 wells  with  m = 5
        observations per well. Using equation  [10.10], the SWi individual well test
        statistics are calculated as:

            Welll:      SWi = 0.7577
            Well 2:      SW2 = 0.7396
            Well 3:      SW3 = 0.7065
            Well 4:      SW4 = 0.8149
Step 2.  Since «i = 5 for each  well, use Table 10-7 of Appendix D to find £ = .5521.
        First calculating u\ with equation [10.20]:

                                   .  f.7577-.5521^
                               M, =ln 	  =-.1641
                                1     ^   1-.7577  J

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         Then performing this step for each well group and using linear interpolation on
         u in Table 10-7, the approximate d statistics are:

             Welll:       MI =-.1641   Gl =-1.783
             Well 2:       w2 = -.3280   G2 =-1.932
             Well 3:       w3 = -.6425   G3 = -2.200
             Well 4:       u4=  .3502   G4 = -1.254
Step 3.   Compute the multiple group test statistic using equation [10.21]:

                    G = -)=[(-1.783)+(- 1.932M- 2.200)+ (-1.254)] = - 3.585
                       V4

Step 4.   Since a = 0.05, the lower a x WOth critical point from the standard normal
         distribution in Table 10-1 of Appendix D is 2.05 = -1.645. Clearly, G < z.o5 ; in
         fact G is equivalent to a Z-value probability of .0002. Thus, there is significant
         evidence of non-normality in at least one of these wells (and perhaps all of
         them). -4 "

Chapter 12, Example 12-1 Calculations for Screening with Probability Plots

In Figures 12-1   through  12-4 and  the accompanying text, normality correlation
coefficients have  been adjusted (using the method in UG Section 10.6) as follows:

       Figure 12-1 Raw Correlation Coefficient (N =20) - .502

       Figure 12-2 Log  "           "  (N =20)   - .973

       Figure 12-3 Raw  "    ", 1 outlier removed (N = 19)  - .854

       Figure 12-4 Log  "   ", 1 outlier removed (N = 19)  - .987

Chapter 13, Example 13-1 Tables of Iron Concentrations
p.13-3.  The median should be 50.06 ppm for Well  1; for Well 2, the mean is 55.74
ppm.

For comparative purposes, well sample standard deviations have been added at the
bottom of this table:  Well 1-12.40; Well 2-20.34; Well 3-59.35;  Well 4-25.95;
Well 5-92.16; and Well 6-51.20 ppm

p.13-4 (log iron concentration table) The median for Well 1 is 3.91 log(ppm).
Sample well log standard deviations are already found in the table on page 13-7.

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Chapter 13, Example 13-2 ANOVA Calculations for Log Iron Concentrations
(full revised example text in parentheses):

       " SOLUTION
Step 1.   With 6 wells and 4 observations per well, n; = 4 for all the wells.  The total
         sample size is n = 24 and p = 6. Compute the (overall) grand mean and the
         sample mean concentrations in each of the well groups using equations [17.1]
         and [17.2]. These values are listed (along with each group's standard deviation)
         in the above table.

Step 2.   Compute the sum  of squares due to well-to-well differences using equation
         [17.3]:

         SSwdls   =  [4(3.820)2 + 4(3.965)2 +  ...   + 4(5.000)2]- 24(4.354)2   =4.331

         This quantity has (6 - 1) = 5 degrees of freedom.

Step 3.   Compute the corrected total sum of squares using equation [17.4] with (n - 1)
         = 23 df:

                   SStotal   =  [(4.06)2+  ... +(5.08)2]-24(4.354)2   =8.935

Step 4.   Obtain the within-well  or error sum of squares by subtraction using equation
         [17.5]:

                                 SSerror =8.935-4.331 = 4.604

         This quantity has (n-p) = 24-6 =18 degrees of freedom.

Step 5.   Compute the well and error mean sum of squares using equations [17.6]  and
         [17.7]:

                                    MS wells= 4331/5= .866

                                   MSemr =4.604 718 = .256

Step 6.   Construct the F-statistic and the one-way ANOVA table, using Figure  13-3 as
         a guide:
Source of Variation
Between Wells
Error (within wells)
Total
Sums of Squares
4.331
4.604
8.935
Degrees of
Freedom
5
18
23
Mean Squares
0.866
0.256
F-Statistic
F= 0.866/0.256=3.38

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Step 7.   Compare the observed F-statistic of 3.38 against the critical point taken as the
         upper 95th percentage point from the F-distribution with 5 and 18 degrees of
         freedom. Using Table 17-1 of Appendix D, this gives a  value of ^.95,5,18 =
         2.77.  Since the F-statistic exceeds the critical point, the  null hypothesis of
         equal well means can be rejected, suggesting the presence of significant spatial
         variation.   -4  "

Chapter 13, Example 13-3 ANOVA Pooled Variance Used for Prediction Limit
Calculations for Log Iron Concentrations
With the slight modification in  ^MSerror = V.256 = .506 from Example 13.2, as shown
in Step 2 p.13-11, the resulting table with well-specific prediction limits is changed
to:

Log-mean
RMSE
df
t.99,18
99% PL
Well 1
3.820
0.5079
18
2.552
193.2
Adjusted
Well 2
3.965
0.5079
18
2.552
223.3
99% Prediction
Well 3
4.348
0.5079
18
2.552
327.5
Limits for Iron
Well 4
4.188
0.5079
18
2.552
279.1
(ppm)
Well 5
4.802
0.5079
18
2.552
515.8
Well 6
5.000
0.5079
18
2.552
628.7
Chapter 14, Section 14.2.2  Procedure for Estimating Sample Size for a
prediction limit with significant temporal variation, Step 9 (modifies equations
for total events):

" Step 9.  If there is no spatial variability but a significant temporal effect exists among a
         set of background wells, compute an appropriate interwell prediction or control
         chart limit as follows.  First replace the background sample standard deviation
         (s) with the following estimate built from the one-way ANOVA:
                                                                  [14.12]
                          W

         Then calculate the effective sample size for the prediction limit as:

                                              T2+(TK-l)-(W-l)^ [14.13]  "

Chapter 14. Example 14-2. Steps 5. 7 and 8:

Although the final calculation is correct, the value for (W-l) in equation [14.7]  is 3,
not 7:

" Step 5.  Compute the mean error sum of squares term using equation [14.7]:

       MSE  =  [(-1.150)2+(-.780)2+... + (l.338)2+(-.765)2]/(4-2)(3) =  1.87
                                       4

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The corrected degrees of freedom language in Step 7 is:

" Step 7.  Test  for a significant temporal effect,  computing the F-statistic in equation
         [14.11]:

                                    Fr= 7.55/1.87 = 4.04

The degrees of freedom associated with the numerator and denominator respectively are
         (TK-l) = 1 and TK(W-l) = 24. Just as with Levene's test run earlier, the 5%
         level critical point for the test is F_gsj^4 = 2.42. Since FT exceeds this value,
         there is evidence of a significant temporal effect in the manganese background
         data. "

In Step 8, the estimated adjusted standard deviation is:
                            -= J-[7.55+3 •( 1.87)] = 1.814/?/?m
Chapter 14. Example 14-7. Step 2:

Although the final calculation is correct, the value for Ne should be a decimal:

" Step 2.  To estimate the minimum time interval between sampling events enabling the
         collection  of physically independent samples of ground water, calculate the
         horizontal  component of the average linear groundwater velocity (Vh) using
         Darcy's equation [14.17]. With Kh = 15 ft/day, Ne = .15 (15%), and  i = 0.003
         ft/ft, the velocity becomes:

                   Vh=(l5ft/ dayxM3 ft /ft)/. 15 = .3 ft I day or 3.6 in I day  "

Chapter 16, Example 16-3, Shapiro-Wilk Calculations in Steps 1 and 2:

The G multiple group value in Step 1 is -6.671 using original benzene data.  The
corresponding value for the log transformed data is G = -.512 in Step 2. Other
results are correct.

Chapter 17. Example 17-6. Mann-Kendall Test Calculations. Steps 2 & 3:

In Step 2, the Mann-Kendall statistic S =194, not 196.  The standard deviation
calculation of 37.79  in Step 3  is correct.  The modified S changes the Z-
approximation to 5.11 in Step 3:
                             = (|l94|-l)/37.79 = 5.

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Chapter 18.  Example 18-2. Statistics in Table, and Step 2:

The joint well log mean for chrysene is 2.553, not 2.533 ppb. This changes the
future mean prediction limit calculation in Step 2:


             PL = 2.553+(2.998)(.706)J-+- = 3.85\og((ppb)
                                    V4  8

Chapter 19.  Example 19-2. Step 4:

The median value of the initial three data for well CW-2 is .36, not .41 ppb. Other
results are not affected.
Chapter 21.  Example 21-7. Step 4:

The correct value for (n-1) in the denominator should be 9, not 8 as applied with
equation [21.25]. This changes the UCL calculation results as follows:
" Step 4.  Since the comparison to the GWPS of 20 ppb is to be made at the a = 0.05
         significance level, the confidence limit is (l-oc)  = 95% confidence. Since the
         remediation effort aims to  demonstrate that the true mean TCE level has
         dropped below 20 ppb, a one-way UCL needs to  be determined using equation
         [21.25]. A logical point along the trend to examine is the last sampling event at
         to = 30. Using the estimated regression value at to = 30, and the fact that ^.90,2,8
         = 3.1131, the UCL on the mean TCE concentration at this point becomes:
               UCL95 =6.861+  2xl5.60x3.H31x
 J_  (30-15.3)2
 10  9x88.2333
 = 12.87 ppb
         Since  this  upper limit is less than the GWPS for TCE, conclude that the
         remediation goal has been achieved by to = 30. In fact, other times can also be
         tested  using the same equation. At the next to last sampling event (to = 26), the
         UCL is:
             UCL95 =13.272+  2xl5.60x3.H31x
J_  (26-15.3)2
10 + 9x88.2333
= 18.14 ppb "

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