Download Answers to Practice Problems

Answers to Practice Problems
1.
We employ Widmark’s equation as follows:
N=
N=
W [ C t +  t]
0.82(fl.oz./drink)
(180lbs.)(16ozs./lb.)(0.7L/Kg) [0.00139Kg/L + (0.00017Kg/L/hr)(7hr s)]
= 13.2 beers
0.82(0.48fl.oz./drink)
13.2 beers 25% = 13.2  3.3 = 9.9 to 16.5 beers
2.
We again employ Widmark’s equation, this time solving for Ct as
follows:
N=
Ct =
W [ C t + t]
N(0.82)(fl .oz./drink)
 Ct =
- t
0.82(fl.oz./drink)
W
(8 drinks)(0.82oz./fl.oz.)(0.4fl.oz./drink)
- (0.00015Kg/L/hr)(5) = 0.00144Kg/L = 0.144g/100ml
(125)(16)(
0.60)
0.144 g/100ml 20% = 0.144 0.029 = 0.115 to 0.173 g/100ml
3.
We employ Widmark’s equation but this time the number of drinks (N) is
the contribution of two types of drinks. We solve as follows:
N 1 (0.82(fl.oz./ drink1 )) + N 2 (0.82(fl.oz./ drink2 )) = W [ C t + t]
Put in the given data and solve for Ct:
9(0.82(0.48)) + 5(0.82(0.4)) = (160)(16)( 0.72)[ C t + (0.00018)(6)]
Ct = BAC = 0.173 g/100ml 25% = 0.173 0.043 = 0.130 to 0.216 g/100ml
4.
We begin by employing Widmark’s equation as follows:
N=
(190lbs.)( 16ozs./lb. )(0.7L/Kg) [0.000885Kg/L + (0.00014Kg /L/hr)(7hr s)]
= 10.1 beers
0.82(0.48fl.oz./drin k)
Assuming the ±25% uncertainty interval we obtain:
10.1 beers 25% = 10.1 2.5 = 7.6 to 12.6 beers
1
Solving for Widmark’s estimate of uncertainty we employ the following
equation for men:


0.68 C t W
1 = 0.015625 N + 0.050176  N 
0.82  fl .oz . / drink  

2
2
for men
Putting in our estimates we obtain:

0.680.000885 190 16  
1 = 0.015625 10.1 + 0.050176 10.1 
0.82 0.48 fl .oz . / drink  

2
2
1 
3.09  1.76 beers
Based on Widmark’s uncertainty estimate we would obtain:
10.1 beers ± 2(1.76) = 10.1 ± 3.52 = 6.5 to 13.6 beers
5.
We first begin by employing Widmark’s equation as follows:
N=
(170lbs.)(16ozs./lb. )(0.72L/Kg )[0.001485 Kg/L + (0.00018Kg/L/hr)(6hr s)]
= 12.7 beers
0.82(0.48fl.oz./drink)
Assuming the ±25% uncertainty interval we obtain:
12.7 beers 25% = 12.7  3.2 = 9.5 to 15.9 beers
We now estimate the number of drinks using the equation developed by
Watson, et.al. We must convert 5 feet 10 inches into 70 inches and
then into 177.8 cm given that 2.54cm = 1 inch. We must also convert
170lbs into 77.3Kg given that 1Kg = 2.2lbs. For a male the equation
for TBW is:
TBW  2.447  0.09516 38 yrs   0.1074 177.8 cm   0.3362 77.3 Kg   43.92 L
We now employ Widmark’s equation based on TBW as follows:
43.92 L
TBW
1.485 g / L  0.18 g / L / h6 h
( C t +  t)
140.8 g
0.8 L / L
0
.
8
N=


0.82(fl.oz. EtOH/drink)
0.82  fl .oz . EtOH / drink 
0.3936 oz / drink
4.97 oz
N 
 12.6 beers
 25%  9.4 to 15.8 beers
0.3936 oz / drink
2
6.
We begin by determining the systematic error associated with the test
and then correct for this amount. Taking the mean of the first three
simulator standards we find: mean = 0.0813 g/210L.
We then determine the systematic error according to:
 X - R
 0.0813 - 0.083 
SE = 
  100 = 
  100 = - 2.05%
0.083


 R 
Assuming there is no uncertainty in the estimate of the systematic
error, the individual’s mean breath alcohol results could then be
increased by 2.05% according to:
 X 
 0.1015
  0.1015 
X Corr  
  
 
  0.1036 g / 210 L
 1   0.0205    0.9795 
 1  bias 
We then employ the equation for a 99% confidence interval:
X Corr  t (1-/2)df = 
0.0058
SD i
= 0.1036  2.57
= 0.1036  0.011
n
2
This would yield a 99% confidence interval of: 0.0926 to 0.1146 g/210L
7.
We first determine the 99% confidence interval from:
X Corr  t (1 -/2)df =  uc = 0.0825  2.57
0.0032
= 0.0825  0.0058  0.0767 to 0.0883g / 210L
2
From this we see that the interval overlaps the critical 0.080 g/210L
limit. Next, we employ the basic equation for a confidence interval
which is:


 
P  X - Z (1 -/2)
   X + Z (1-/2)
=P

2
2 
We begin by noticing that we are interested only in the probability
that μ exceeds a lower limit and we do not care about the upper limit.


P  X - Z (1-/2)

2


=P

Next, we notice that we are interested in the probability that μ
exceeds 0.080 g/210L which is the same as the value for the lower limit
in the above probability equation. We set the two equal, introduce our
known information and solve for Z(1-α/2):
3
X - Z (1 -/2)

= 0.080  0.0825 - Z (1 -/2) 0.0032 = 0.080  Z (1 -/2) = 1.10
2
2
We then rearrange our probability statement and introduce our
determined value for Z(1-α/2):


X -

P X - Z 1-/2 u   = P 
 Z1 / 2  = P Z  Z 1-/2 = P Z  1.09 = 0.8621
c
 uc



Using the standard normal tables we see that P(Z < 1.09) = 0.8621.
There is a probability of 0.8621 that the individual’s true mean breath
alcohol concentration is greater than or equal to 0.08 g/210L.
8.
We first determine the standard deviation and variance from the
replicate set of n=12 breath alcohol measurements. This will be an
estimate of the total combined uncertainty because it includes both
components of sampling and analytical.
S = 0.00237 g/210L
S2 = 0.0000056
We also know the variance due to the analytical instrument based on the
historical results: σ2 = 0.000001. This was determined from replicate
measurements of the same control standard performed in the field over
several months. We can now determine the uncertainty due to the
sampling component by assuming independence (sampling and analytical)
and finding the sum of variances according to:
2
2
2
STotal
 S Sampling
 S Analytical

2
0.0000056  S Sampling
 0.000001
2
S Sampling
 0.0000046
Now we can determine the proportion that each component contributes to
the total according to:
2
S Sampling
2
STotal

0.0000046
 0.82
0.0000056
2
S Analytical
2
STotal

0.000001
 0.18
0.0000056
The sampling component contributes 82% to the total uncertainty while
the analytical component contributes 18%.
9.
We first find the corrected BAC result from the measurement model as
follows:
Y Corr 
Y0 R
X
 f dil 
0.11450.100 1  0.1174g / 100ml
0.0975
Next, we compute the uncertainty associated with the original mean
4
results using the uncertainty function given:
S = 0.0108(0.1145) + 0.0008 = 0.0020 g/100ml
We now organize all of our information as follows:
Variable
Value
Uncertainty
n
Y0
R
0.1145 g/100ml
0.100 g/100ml
0.0020
0.0003
2
2
X
0.0975 g/100ml
0.0006
15
0.015
10
f dil
10.15 ml
Since the measurement model is multiplicative, we now estimate the
combined uncertainty using the method of root sum of squares (RSS) by
assuming independence for all variables as follows:
uY
Y Corr
where:

CVT2  CV R2  CV A2  CV f2
d il
CVT2  uncertainty due to the total method
CV R2  uncertainty due to the traceable reference value
CV A2  uncertainty due to the analytical instrument
CV f2
dil
 uncertainty due to the dilutor
We now solve for the combined uncertainty according to:
2
uY

0.1174
2
2
 0.0006 
 0.015 
 0.0020 
 0.0003 








15 
10 
2 
1 







 0.0975 
 10.15 
 0.1145 
 0.100 
















2
 uY  0.0015 g / 100ml
The 99% confidence interval would be found as follows:
Y Corr  k uc = 0.1174  3 0.0015 = 0.1174  0.0045  0.1129 to 0.1219g / 100ml
The uncertainty budget for this example is as follows:
5
Total Method
Reference
Analytical
Dilutor
93%
5%
1.5%
0.5%
10. We begin by writing down each of our five sources of uncertainty and
their estimates. They can be written down as follows:
Traceability
Mean 0.100g/100ml
SD 0.0003g/100ml
n
2
Analytical
0.1025g/100ml
0.0011g/100ml
8
Dilutor
Bias
Total Method
10.12µL 0.003g/100ml
0.05µL
10
0.0885g/100ml
0.0013g/100ml
2
We now incorporate these estimates into our equation for computing
combined uncertainty, assuming all components are independent and using
the method of combining CV’s squared as follows:
where:
CVT2  uncertainty due to traceability
CVA2  uncertainty due to the GC replicates
CVD2  uncertainty due to the dilutor
CVB2  uncertainty due to the bias
CV M2  uncertainty due to the total method
2
U
Y
0.0885

2
2
2
 0.0011 
 0.05 
 0.002 
 0.0003 
 0.0013 










8 
10 
3 
2 
2 





 0.1025 
 10.12 
 0.1025
 0.100 
 0.0885 




















2
 U  0.0014 g / 100ml
Y
The 95% confidence interval would be found from:
Y  k uc = 0.0885  1.96 0.0014 = 0.0885  0.0027  0.0858 to 0.0912g / 100ml
The sum of all terms under the radical is: 0.0002608
find the percent contribution from each component:
Traceability:
Analytical:
Dilutor:
Bias:
Method:
From this we
3%
6%
1%
49%
41%
6
11.
We begin by computing our total method uncertainty using the
uncertainty function determined from the proficiency test data:
 
S 2  2.42  0.000541C 2 mg / dL2  S 2  2.42  0.00054197 2
 S 2  7.5 mg / dL2  S  2.74 mg / dL  0.00274g / 100ml
We now write down each of our four sources of uncertainty and
their estimates as follows:
Traceability
Mean 0.100g/100ml
SD 0.0003g/100ml
n
2
Analytical
0.1046g/100ml
0.0010g/100ml
18
Bias
0.1046g/100ml
0.0012g/100ml
1
Total Method
0.0970g/100ml
0.0027g/100ml
2
We now incorporate these estimates into our equation for computing
combined uncertainty, assuming all components are independent and using
the method of combining CV’s squared (RSS) as follows:
SY
Y
where:

2
CVT2  CV A2  CVB2  CV M
CVT2  uncertainty due to traceability
CVA2  uncertainty due to the GC replicates
CVB2  uncertainty due to the bias
CV M2  uncertainty due to the total method
2
uY
0.0970

2
2
 0.0010 
 0.002 
 0.0003 
 0.00274 








18 
3 
2 
2 







 0.1046 
 0.1046
 0.100 
 0.0970 
















2
 uY  0.0023 g / 100ml
The 95% confidence interval would be found from:
Y  k uc = 0.0970  2 0.0023 = 0.0970  0.0046  0.0924to 0.1016 g / 100ml
12.
We first convert the maximum allowed standard deviation estimate to
35 2  1225ng / ml 2
its variance.
Next, we state our hypotheses as
follows for a one-tail test:
H0: σ2 ≤ 1225 ng/ml
7
HA: σ2 > 1225 ng/ml
Next, we compute our chi-square statistic since variances follow a chisquare distribution.
Test Statistic:

2

n  1 S 2


2

29422   41.76
35 2
Next, we determine our critical value for the chi-square distribution given
that α = 0.01.
Critical chi-square (found in Excel):
2
one  tail  crit
 2 , n 1
 CHIINV 0.01,29
 49.59
We now compare our computed chi-square value (41.76) to our critical chisquare value (49.59) for an level of α = 0.01
Conclusion: Since the computed value is less than our critical value we Accept
H0 p > 0.01 p-value: = CHIDIST(41.76,29) → 0.059
The chi-distribution appears as follows:
χ2 Distribution
 n2 1  41.76
  0.01
2  49.59
The 99% confidence interval for the experimental variance is as
follows:
The 99% confidence interval:
=CHIINV(0.005,29) 52.33
=CHIINV(0.995,29) 13.12
8

P


30  1 422     30  1 422  
52.33


13.12
 0.99
P 31.27    62.44  0.99
13. Assume the data below was generated in a proficiency test program.
The percent CV and bias have been determined according to equation 1.
Lab
1
2
3
4
5
6
7
8
Mean
0.143
0.148
0.144
0.147
0.142
0.149
0.144
0.147
St. Dev.
0.0012
0.0015
0.0009
0.0013
0.0015
0.0011
0.0014
0.0008
n
15
18
22
21
14
24
16
15
 SD 
%CV  
 100
 Y 
where:
% CV
0.837
1.016
0.624
0.887
1.056
0.739
0.976
0.543
Bias
0.982
1.012
0.988
1.004
0.974
1.02
0.984
1.01
Y
Bias  obs
R
Eq .1
The reference value has been determined from a weighted mean of the
measured results according to:
k
w Y
i
RW  i  1
where : w i 
k
w
i
ni
S i2
Eq . 2
i
i 1
Using the data in the table above to estimate the weighted mean we obtain:
Rw = 0.1459 g/100ml
For any individual laboratory, the combined uncertainty of their results
will be determined from the combined uncertainty of their mean observed

results Y and the bias.
The bias could be determined from PT results as
9
presented here or from the average of day-to-day control measurements.
The combined uncertainty is determined according to:


uY  Y 



2
SY 

2
 uB 
n 


Y 
 B 


where B 
u
Y
and B
RW
B







2
SY 
2

 u Rw 
n 


Y 
 Rw 


Eq . 3
The uncertainty in the weighted reference value is determined from:
1
uRw 
Eq .4
k
w
From our data above we obtain uRw = 0.000093
i
i 1
Let us now assume that we want the combined uncertainty for lab 4.
first determine the uncertainty in the bias according to Eq. 3:
uB  B






2
SY 

 uR
n 
 w
Y 
 Rw


We
2



2
 0.0013 


2
0.147  21 
 0.000093


  0.0020 g / 100ml
0.1459  0.1470 
 0.1459 




Next we determine whether the bias (B) is significant by using the
t-distribution to test the null hypothesis that B = 1:
0.147
1
B1
0.0080
0
.
1459
t 


 4.0
uB
0.0020
0.0020
Eq . 5
The critical two tail t-statistic with α = 0.05 and 20 degrees of freedom
is 2.086. We therefore reject the null hypothesis that the bias is equal
to one and conclude that we have a significant bias effect. Rather than
correct for the bias we will add an additional term to the combined
uncertainty estimate. This additional term will be:
2
2
 B1
 1.008 1 
2
2
uB extra  

u

B


  0.0020  0.004
2
 2 


Eq .6
10
We now express our final combined uncertainty from Eq. 3 as:
uY  Y
 uB extra 


 B 
2
 0.004 
 0.147 

 1.008
2
 0.00058g / 100ml
The term for analytical uncertainty is not included here because it is
already part of uB computed earlier. If we assume the bias (a=0.0011)
follows a uniform distribution the combined uncertainty would be 0.00070
g/100ml. With the estimate of combined uncertainty determined above we can
report the uncertainty of future duplicate mean results having similar
concentrations with k=2 (95% CI) as:
Y n 2  2 0.00058
14. We begin by finding the difference between each of the paired results.
You can subtract in either order, just so you do all the same. The paired
differences are shown below:
BrAC
BAC
Difference
1
0.111
0.139
-0.028
2
0.076
0.111
-0.035
3
0.13
0.149
-0.019
4
0.079
0.111
-0.032
5
0.057
0.077
-0.02
6
0.092
0.133
-0.041
7
0.118
0.116
0.002
8
0.114
0.126
-0.012
9
0.156
0.16
-0.004
10
0.094
0.098
-0.004
H0: δ = 0
HA: δ ≠ 0
We now state our hypotheses:
11
0.089
0.106
-0.017
12
0.085
0.088
-0.003
13
0.077
0.076
0.001
14
0.104
0.099
0.005
15
0.109
0.11
-0.001
α = 0.05
We now draw our t-distribution and identify our regions for rejecting the
null hypothesis, show our critical t-values and show the computed t value:
t Distribution with df = 14
Accept H0

2
 3.60
 2.145

2
t
2.145
The computed t statistic is determined from:
11
t
d 
d
 0.0139


  3.6
S/ n
S / n 0.0149 / 15
From this we conclude that we reject the null hypothesis and p < 0.05.
The 95% confidence interval for the mean paired difference (δ) is found
from:
d  2.145
S
n

 0.0139  2.145
0.0149
 0.0139  0.0083
15
  0.022 to  0.0056
Since this 95% confidence interval does not bracket zero, it further
supports the results of the hypothesis test.
15. The t-test for paired data is a one sample test.
necessary sample size from:
2

S
n    Z 1 / 2  Z 1 
 

2
We estimate the
2

 0.005 
2
n 
 1.96  0.84  n  7.8  8
0
.
005


A sample size of 8 subjects providing paired blood and breath alcohol
samples would achieve 80% power to detect a difference (effect size) of
0.005.
16.
We begin by computing the mean of these results which is 0.0915
g/210L. Next, we need to estimate the standard deviation
associated with the difference as follows:
d  Y1  Y 2
Sd 
V d   V Y1   V Y 2 


V d   2 V Y i 
2 S Yi
d  1.96 S d
0.013  0.015

d  1.96 2 S Yi
0.013  2.77 0.0054

 0.002 to 0.028 g / 210 L
Since this 95% confidence interval includes zero we conclude these
differences are acceptable.
For mean results at 0.250 g/210L we compute:
d  1.96 S d

d  1.96 2 SYi

d  2.77 0.0102
d  0.028 g / 210 L
At this concentration our differences should not exceed 0.028 g/210L.
12
17.
We take the information give and develop our contingency table as
follows:
Contingency Table for PBT Diagnostics in Breath
Evidential Instrument
≥ 0.08 g/210L
< 0.08 g/210L
PBT Result
Positive
165
TP
38
FP
203
Negative
25
FN
172
TN
197
190
210
400
We now compute our performance parameters as follows:
Performance Parameters
TP
165

 86.8%
TP  FN
165  25
TN
172
Specificity :

 81.9%
TN  FP
172  38
TP
165
PPV :

 81.3%
TP  FP
165  38
TN
172
NPV :

 87.3%
TN  FN
172  25
TP  TN 165  172
Percent Efficiency :

 84.3%
Total
400
Sensitivity :
The power of the test is:
Power  P PBT  Evidential   
18.
TN
172

 81.9%  Sensitivity
TN  FP 172  38
We begin by finding the mean of the first results which is 0.1285
g/100ml and the mean of the second set is 0.1150 g/100ml. Next, we find
the variance of the differences between these two means and end with a
confidence interval for the difference.
13
We proceed as follows:
d  Y 1 Y 2
d  duplicate test difference
Y i  the mean of duplicate blood alcohol measurements
   

 

V d   V Y 1  V Y 2  2 Cov Y 1 ,Y 2

V d   2 V Y i  Cov Y 1 ,Y 2
 
V d   2 V Y i   SY SY
1
2


 

V d   2 V Y i  2Cov Y 1 ,Y 2



  
Cov Y 1 ,Y 2  Y ,Y V Y 1 V Y 2
1
2
 
 
V d   2 V Y i   V Y i
 
V d   2 V Y i 1   
 
diff  1.96 2 V Y i 1   
diff  1.96 2 1    SY
i
 0.0040 

0.0135  1.96 0.1 

2 

The 95% confidence int erval for the difference
Assume   0.95 and SY 
i
0.0135  0.0018
0.0040 g / 100m
2
0.0117 to 0.0153g / 100ml
Since this does not contain zero, the differences are significant – the
duplicate test differences would have to be ≤ 0.0018 g/100ml to be nonsignificant
19.
The odds ratio is the quantitative measure of association in casecontrol studies. They measure association only and not causation - as a
randomized controlled trial would do. The odds ratio is computed very
easily from tables as observed in this problem. We first identify each
cell within the table as follows:
14
Risk Factor
Outcome
Yes
No
a
b
c
d
Case
Control
We then take the values in each cell identified as a,b,c and d and
compute the odds ratios for each risk factor as follows:
Risk Factor :
OR 
0.02
0.08
ad 16332 5312


 4.9
bc 10810 1080
OR 
ad 13339 4407


 13.2
1113 333
bc
We now compute the 95% confidence intervals for these odds ratio estimates
as follows. We first compute it on the log scale and then convert back to
the OR scale.
ln OR  Z 1   / 2
1 1 1 1
  
a b c d
1.59  1.96 0.1748 
e 0.771 to e 2.41
ln 13.2  1.96
e 1.31 to e 3.85

ln 4.9  1.96
1
1
1
1



16 108 10 332
1.59  0.819  0.771 to 2.41
2.16 to 11.13
1
1
1
1

 
13 111 3 339



2.58  1.27  1.31 to 3.85
3.71 to 47.0
15
Since both Odds Ratios for the two risk factors are greater than one,
there is an increased risk in being a case when having an alcohol
concentration greater than 0.02 or greater than 0.08. At the 0.08
level the risk is even greater when comparing the OR of 13.2 to 4.9.
Moreover, both odds ratios have 95% confidence intervals that exceed
one. This indicates that the increased risk is significant at the 95%
level of confidence.
20. In this problem we do not have replication of the measurements so we
must use the two-way ANOVA without replication. There is only one
measurement in each cell of the table. We will not be able to obtain an
interaction term, only main effects for each factor. The two factors we
are evaluating are Subject and Tube Type. We put the data into Excel
exactly as seen in the table of data. In Excel we click on Data, then Data
Analysis and then Two-way ANOVA without replication. For the data input
range we highlight except the title of “Tube Type Used”. The results are
shown below:
ANOVA
Source of Variation
Rows
Columns
Error
SS
0.023615725
0.000807075
0.007037175
Total
0.031459975
df
9
3
27
MS
F
P-value F crit
0.002623969 10.068 1E-06 2.2501
0.000269025 1.0322 0.394 2.9604
0.000260636
39
The Rows represent the ten subject volunteers. The Columns represent the
four tube types. From the p-value in the table we see that the main effect
of subjects is highly significant (p = 0.000001) while the main effect for
the tube types is not significant (p = 0.394). There is much more
variation between subjects than within subjects across the four tube types.
21.
We begin by finding the mean result for the subject’s duplicate tests:
0.088 / 0.091 g / 210 L 
Y  0.0895 g / 210 L
Knowing that we have a bias of +3.0%, we correct this mean result
according to:
 Y 
 0.0895 
Y Corr  
  
  Y Corr  0.0869 g / 210 L
 1  0.03 
 1  bias 
We now want to find the 99% confidence interval for this corrected
mean. The standard error of this mean, however, is determined from two
sources: the combined analytical and biological component and the
reference standard. Therefore, our confidence interval estimate
appears as:
16
0.0869  2.57
where:
S 12
S2
 2
n1
n2
S 12  the variance determined from the equation given in the
problem which combines both analytical and biological
components
2
S 2  the variance representing the uncertainty in the
reference value
From the equation given in the problem we estimate the standard
deviation combining both analytical and biological components as:
S  0.0305 B  0.0026  0.0305 0.0869  0.0026  0.0053 g / 210 L
We now put our variance estimates into the confidence interval
estimate:
0.0053 2
0.0014 2
0.0869  2.57

2
10
 0.0869  0.0097
This results in the confidence interval of: 0.0772 to 0.0966 g/210L
We do not use a pooled estimate of the variance in this example because
the two variance components are largely different. The problem can
also be solved by estimating a combined uncertainty from the CV’s for
each contributing element. In this case we estimate the confidence
interval from:
0.0869  2.57 SY
2
where :
SY
0.0869

CVT2  CV R2
 0.0014 
 0.0053 




SY
10 
2 

 

0.0869
 0.082 
 0.0869 




2
 SY  0.0038 g / 210 L
Notice that we have solved directly for the standard deviation of the
mean by incorporating the appropriate sample sizes for each component.
We now find the confidence interval from:
0.0869  2.57 0.0038  0.0869  0.0098
This results in the confidence interval of: 0.0771 to 0.0967 g/210L
which are almost identical to those estimated above.
22. We first recognize that we are testing the following hypotheses where
we claim in the null hypothesis that all mean results are equal:
H0 :
HA :
µ1=µ2=µ3=µ4=µ5
µ1≠µ2≠µ3≠µ4≠µ5
17
We create a table in Microsoft Excel exactly like that shown in the
problem. Using the Data Analysis oneway ANOVA single factor, Microsoft
Excel provides the following table:
Anova: Single Factor
SUMMARY
Groups
400
450
500
550
600
Count Sum Average Variance
10 0.813
0.0813 1.12E-06
10 0.815
0.0815 1.17E-06
10 0.817
0.0817 2.01E-06
10 0.814
0.0814 1.82E-06
10 0.827
0.0827
9E-07
ANOVA
Source of Variation
Between Groups
Within Groups
SS
1E-05
6E-05
Total
8E-05
df
4
45
MS
F
P-value F crit
3.2E-06 2.292722 0.074 2.579
1.4E-06
49
From the table we see that the calculated F statistic

MSbetween 
 det er min ed from


MSwithin 

is 2.29 with a p-value of 0.074.
This means that
we would not reject the null hypothesis that all means were equal. The
p-value exceeds 0.05. The components of variance is determined from
the following equation:
2
MSbetween   within
2
 between 
n
where:
MSbetween = the between groups value for mean squared (MS)
in the table
2
 within

n
=
the within groups MS found in the table
the number of measurements performed at each
volume (n=10)
Putting the values from the table into the equation above we obtain:
MSbetween 
2
 between

n
2
 within

0.0000032 0.0000014
 0.00000018
10
The combined variance is the sum of these two components:
2
2
 between
  within
 0.00000018 0.0000014 0.00000158
18
The percent contribution by each component is found from:
2
% between

0.00000018
 11%
0.00000158
2
% within

0.00000018
 89%
0.0000158
From this we see that most of the variation (89%) is from the within
groups. That is, the variation within each run of n=10 contributes
more than the variation between the means for each of the five
groups.
Difference (PBT-BrAC g/210L)
23. After determining the mean and differences for the n=450 data pairs, we
plot the difference as the Y variable and the mean as the X variable. We
then generate the linear regression line in Excel. The plot will look like
as follows:
Plot of Difference Against Means
y = -0.0326x + 0.0117
n = 450
mean difference = 0.0072g/210L
SD = 0.021g/210L
0.15
0.1
0.05
0
-0.05
-0.1
0
0.1
0.2
0.3
Mean Result (g/210L)
0.4
The Excel summary is as follows:
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.07160655
R Square
0.0051275
Adjusted R Square
0.0029068
Standard Error
0.02134439
Observations
450
ANOVA
df
Regression
Residual
Total
Intercept
Slope
1
448
449
SS
MS
F
Significance F
0.001051922 0.001 2.308958
0.129335613
0.204101109 5E-04
0.205153031
Coefficients Standard Error t Stat P-value
0.011689
0.003098717 3.772 0.000184
-0.0326312
0.021474625 -1.52 0.129336
Difference
Lower 95% Upper 95%
0.005599169 0.017779
-0.074834752 0.009572
-0.034
0.049
19
From the Excel summary we see that the 95% confidence interval for the
slope brackets zero and therefore implies there is no proportional error.
The 95% confidence interval for the intercept, however, does not bracket
zero and thus implies a fixed bias. Also, the 95% confidence interval for

0.021
 0.0053 to 0.0091 g / 210 L
the mean difference is: d  1.96 d  0.0072  1.96
n
450
This does not bracket zero and further suggests a fixed bias.
24.
This problem is nearly identical to that of problem 13. We are testing
the same hypotheses as follows:
H0:
HA:
µ1=µ2=µ3=µ4
µ1≠µ2≠µ3≠µ4
We enter the data into the Excel, Data Analysis Toolpak exactly as it
is shown in the table. We allow the first row to be the column labels.
We obtain the following table:
Anova: Single Factor
SUMMARY
Groups
bottle 1
bottle 2
bottle 3
bottle 4
Count Sum
Average Variance
10 0.978
0.0978 6.222E-07
10 0.978
0.0978 1.511E-06
10 0.993
0.0993
9E-07
10 0.983
0.0983
9E-07
ANOVA
Source of Variation
Between Groups
Within Groups
SS
2E-05
4E-05
Total
5E-05
df
3
36
MS
F
P-value F crit
5E-06 5.0847458 0.0049 2.866
9.83E-07
39
From the table we see that the calculated F statistic

MSbetween 
 det er min ed from


MSwithin 

is 5.08 with a p-value of 0.0049.
This means that
we would reject the null hypothesis that all means were equal. The
p-value is very small. We conclude there is a significant difference
between two or more of the mean values. The components of variance is
determined from the following equation:
20
2
 between

where:
2
MSbetween   within
n
MSbetween = the between groups value for mean squared (MS)
in the table
2
 within

n
=
the within groups MS found in the table
the number of measurements performed at each
volume (n=10)
Putting the values from the table into the equation above we obtain:
2
 between

MSbetween  MSwithin
0.000005 0.00000098

 0.000000402
n
10
The combined variance is the sum of these two components:
2
2
 between
  within
 0.000000402 0.00000098 0.00000138
The percent contribution by each component is found from:
2
% between

0.000000402
 29%
0.00000138
2
% within

0.00000098
 71%
0.0000138
From this we see that most of the variation (71%) is from the within
groups. That is, the variation within each run of n=10 contributes
more than the variation between the means for each of the four
groups. Next we want to do a post-hoc analysis where we try to
identify which of the bottles might be significantly different from the
others. To do this we employ a different statistical program (STATA V
12.0, STATA Corp., College Station, TX). This program will do all
pairwise comparisons (there are six) using an adjusted t-test which
accounts for the multiple testing of the same data. The table below
shows the results. From the table we see that the ANOVA results are
the same as that above determined from Excel. We also see a test
(Bartlett’s) for equality of variance. This is an assumption of oneway
ANOVA and from the results we see that the assumption of equal
variance is valid since the p-value is large 0.623. We also see the
table for the pairwise t-test comparisons using the Bonferroni
correction. The first value in each cell is the row mean – the column
mean value. The second value in each cell are p-values. Small pvalues (<0.05) would mean that these two bottles were significantly
different. We see that bottles 1 and 3 are significantly different and
2 and 3 are also significantly different.
21
Analysis of Variance
SS
df
MS
Source
Between groups
Within groups
Total
.000015
.0000354
3
36
5.0000e-06
9.8333e-07
.0000504
39
1.2923e-06
Bartlett's test for equal variances:
chi2(3) =
F
Prob > F
5.08
1.7643
0.0049
Prob>chi2 = 0.623
Comparison of alc by bottle
(Bonferroni)
Row MeanCol Mean
1
2
2
0
1.000
3
.0015
0.010
.0015
0.010
4
.0005
1.000
.0005
1.000
3
-.001
0.182
.
25. This problem presents six sources of combined uncertainty that should
be reported as part of the final breath test instrument calibration
results. These six sources are: (1) the CRM traceability, (2) the
Toxicology Lab bias, (3) the Toxicology Lab uncertainty in measuring
the CRM, (4) the Toxicology Lab uncertainty in measuring the simulator
solution, (5) the partition coefficient uncertainty and (6) the breath
test instrument uncertainty. Each component was also determined based
on differing number of measurements. We incorporate these different
values of n into the square root sign and arrive directly at a standard
deviation (uncertainty) for the mean. The value for each of these
expressed as a CV are:
0.003
CVCRM 
0.0012
0.1025
CVToxBias 
3
0.0995
0.0007
CVToxCRM 
5
0.0995
0.0006
CVToxSol 
15
0.1005
0.0010
CV PartCoef 
0.0124
1.23
CV Inst 
10
0.0815
Notice that we have incorporated the sample sizes only for those
components that we had information for. Notice also that we have
assumed a uniform distribution for the Toxicology Lab bias and
estimated its uncertainty by dividing by the square root of three.
The vapor alcohol reference value in the simulator is found by dividing
0.1005 by 1.23 and obtaining 0.0817 g/210L. This, however, is biased
because the GC results are not corrected for in its bias. We account
for this uncorrected bias below. We now simply combine these sources
of uncertainty as follows:
22
uC
X

2
2
2
2
2
2
CVCRM
 CVToxBias
 CVToxCRM
 CVToxSol
 CV PartCoef
 CV Inst
2
uC

0.0815
2
2
 0.003 
 0.0007 
 0.0006 
 0.0010 






2 
2 
 0.0012
 0.0124
3 
5 
15 
10 




 0.1025   0.0995   0.0995    0.1005    1.23    0.0815 




















uC
 0.0239
0.0815

uC  0.0019 g / 210L
2
the s tan dard error of the mean
 X  R
GC sim R 0.10050.1025
 0.0815 0.0842
bias  
 100   3.3% GC SimCorr 

 0.0842
  

1.230.0995
0.0842
K GC R


 R 
From these results we can set up an uncertainty budget and list the
percent contribution from each factor:
Factor
Uncertainty
Type
Percent
CRM
Tox Lab Bias
Tox Lab CRM
Tox Lab Sol
Part Coef
Inst
0.0012
0.0017
0.0003
0.00016
0.0124
0.0003
B
A
A
A
B
A
24%
53%
1%
1%
18%
3%
26. We first need to determine the uncertainty (standard deviation)
estimates for each of the three input variables. They are as follows:
Mass of ethanol:
Purity:
Density:
Mass of solvent:
Estimate
3.00 g
0.992
0.65 g/ml
1.8 Kg
Uncertainty
0.008 g
uP 
0.002
 0.00116
3
0.0006 g/ml
0.009 Kg
We first solve for the concentration C according to:
m Etoh P D
3 g 0.9920.65 g / ml   0.00107 g / ml

m Solvent
1800 g
We now combine these four components using the method of general
error propagation and assuming independence of all components:
C 
23
2
2
uC 
 C 
 C 
 C 
 C 
2
2
S P2  
S D2  

 S m Etoh  
 S m Solvent


 P 
 D 
 m Etoh 
 m Solvent 
uC 
 m Etoh PD  2
 PD 
2  m Etoh D 
2  m Etoh P 
2
0
.
008

0
.
00116

0
.
0006


 9






2
 m Solvent 
 m Solvent 
 m Solution 
 m Solvent 
uC 
0.0003582 0.008 2  0.001082 0.001162  0.001652 0.0006 2   0.00000062 9 2
uC 
0.00000000004  0.0000063 g / ml  0.00063 g / 100ml
2
2
2
2
2
2
Solving for the uncertainty by the method of summing the CV squared
yields:
2
uC

C
CV m2 Etoh  CV P2  CV D2  CV m2 Solvent 

uC

0.00107
 0.002 
2
2
2


 0.008 
 0.0006 
 9 
3 

 3    0.992    0.65    1800 










uC
 0.00586
0.00107
uC  0.0000063 g / L  0.00063 g / 100ml

CV  0.59%
These results are the same because the preparation function for the
ethanol concentration is multiplicative.
From these results we can set up an uncertainty budget and list the
percent contribution from each factor:
Factor
Mass of ethanol
Purity
Density
Mass of solvent
Uncertainty
Type
Percent
0.008
0.0012
0.0006
9.0
A
B
B
A
21%
4%
2%
73%
24
27.
Here we are interested in first estimating the combined uncertainty,
uc , from the equation: 0.080 2 k uc  0.086 considering both the Type I
And Type II error at 5% each. First we must look up the value of k
from the standard normal table. If we want a 5% probability for both
errors the value of k will be 1.645, having area (probability) to the
left (of the null distribution) of 0.95.
obtain:
Solving then for
uc
we
0.080 2 1.645uc  0.086  3.29 uc  0.006  uc  0.0018g / 100ml
The reason for the value of 2 in the equation is because we want to
guard against both the Type I and Type II errors (the false-positive
and false-negative errors). Including 2 in the equation will set both
error probabilities equal. The value of k will also be 1.645 which
provides a probability of 5% for both error types.
28.
We first compute the confidence intervals for each gender as follows:
For women:
p  Z 1 / 2
p 1  p
64
0.14 0.86 

 0.14  1.96
 0.14  0.032  0.108 to 0.172
n
458
458
For men:
p 1  p
298
0.16 0.84 

 0.16  1.96
 0.16  0.017  0.143 to 0.177
n
1860
1860
From this we see a slight overlap between the two confidence intervals
p  Z 1 / 2
Next, we find the confidence interval for the difference between the
two proportions according to:
p 1  p 2  Z 1  / 2
  0.02  0.036
p1 1  p1  p 2 1  p 2 
0.14 0.86  0.16 0.84 

 0.14  0.16  1.96

n1
n2
458
1860

 0.056 to 0.016
From this we see that the 95% confidence interval for the difference
includes zero. Thus we conclude there is not significant difference.
We now set up the two-way contingency table and enter the observed and
expected values as follows:
25
Provide Test
Refused Test
394
64
386.5
71.5
1562
298
1569.5
290.5
Women
Men
1956
Our test hypotheses are:
362
Proportion
Refusing
458
14.0%
1860
16.0%
2,318
H 0 : gender and refusal rates are independent
H A : gender and refusal rates are not independent
From the table values we compute the  2 statistic as follows:
k
 =
2
i =1
2
( Oi - E i )
Ei

394  386.5 2  64  71.5 2  1562  1569.5 2  298  290.5 2  1.17
386.5
71.5
1569.5
290.5
For a two-way table the degrees of freedom are: df  rows  1columns  1  1
The critical  2 statistic for df=1 and α=0.05 is found from the table
to be:  02.95 ,df 1  3.841 . Since our
than this table value, we do not
is no association between gender
supports the confidence interval
proportions.
computed  2 statistic is much less
reject the null hypothesis that there
and refusal rate. This further
for differences between the two
29. This problem illustrates the more complete analysis for determining the
uncertainty in forensic breath alcohol results. This will incorporate
the uncertainty from the three major sources: the subject, the breath
test instrument and the gas chromatograph used to measure the simulator
solution standard. We will begin by identifying the following
variables: Y = subject BrAC results, X = simulator results,
R = reference value determined from gas chromatograph, Z = subject BrAC
results after correcting for any systematic error and S = standard
deviation estimates. We now find the mean and standard deviation of
the subject’s results: mean = 0.095 g/210L and from the equation for
the standard deviation we obtain:
S = 0.0305(0.095) + 0.0026 = 0.0055 g/210L
The standard deviation for the mean is found from:
S
0.0055
SY  Y 
 0.0039 g / 210 L
n
2
Next, given the mean reference standard result of 0.082 g/210L and the
26
standard deviation of 0.0010 g/210L with n=30 measurements, we find the
standard deviation of the mean according to:
SR 
SR
n

0.0010
30
 0.00018 g / 210 L
Next, we find the mean and standard deviation resulting from the
instrument’s measurements of the first five simulator standards:
mean = 0.0844 g/210L and S = 0.0015 g/210L from which we find that the
S
0.0015
 0.00067 g / 210 L
standard deviation of the mean is: S X  X 
n
5
S
Z  t (1-/2)df =  Z
The confidence interval we will employ is:
n
where Z is the mean BrAC result corrected for any systematic error.
SZ
Since
is the standard deviation of the mean we will simply ren
express this as: S Z . Our corrected confidence interval expression
will be: Z  t (1-/2)df = S Z .
Based on the information given in the
 Y  YR
Z 

X
 1  bias 
The bias is found to be:
problem we find the corrected BrAC according to:
where:
the bias = the systematic error.
 X  R   0.0844  0.082 
bias  

   0.029
0.082

 R  
We now find our corrected mean to
be:
 Y

Z 
  0.095 0.9718   0.0923 g / 210 L
 1  0.029 
Next, we express more fully the equation for the corrected mean BrAC:



 YR
Y

Z 
X R
X

1


R 
Based on this equation, we see that Z
is a function of three
measured variables: the mean of the subject’s results, the mean of the
instrument measurement of the simulator standards and the mean of the
gas chromatograph measurements of the simulator standard. All three
have uncertainty that must be included in the uncertainty estimate for
Z , S Z . To determine this we use equation 5 from the attached page
27
and assume that all three sources of uncertainty are independent:
2
2
2
Z  2 Z  2  Z  2
SZ  
 SY  
 SR  
 SX
 Y 
 R 
 X 
Introducing our data into this equation we obtain:
 0.082   0.0055   0.095   0.0010   0.095 0.082    0.00152 
SZ  
  
 
 
 
0.0844 2   5 
2   0.0844   30  
 0.0844  
2
This results in:
2
2
2
2
2
S Z  0.00001487  0.00386 g / 210 L
Another approach to finding the standard deviation estimate for Z
is
to employ coefficient of variations since the model is multiplicative.
Some find this approach easier. We write the general uncertainty
equation in the form:
CVZ 
SZ
Z
 CVY2  CVR2  CV X2
We now incorporate our estimates into this equation for the CV’s:
2
 0.0010 
 0.0055 




SZ
30 
2 




0.0922
 0.082 
 0.095 




2
 0.0015 


5 


 0.0844 


2
 0.04174
S Z  0.0922 0.04174   0.00385 g / 210 L
Now we compute our corrected confidence interval:
Z  t (1-/2)df = S Z
 0.0922  2.57 ( 0.0039 )  0.0822 to 0.1022
Our confidence interval now includes uncertainty from both random and
systematic sources and is therefore a more complete analysis.
30.
In this problem we want to determine the time necessary for the
individual to eliminate sufficient alcohol so that the mean of
duplicate measurements can be sufficiently distinguished. We begin by
solving for the critical difference necessary for two means to have
(given the measurement variability) in order to be able to conclude
that they will be measured as different:
28
 0.0053 
 = 0.010 g/210L
2 

 cr = 2.77 SD x = 2.77 
We now put this critical difference value into an equation that will
solve for the time necessary to wait: 0.010 = 0.015t  t = 0.66 hrs.
31. We are interested here in determining the 99% confidence interval for a
population parameter δ, the true difference between the mean of the
instrument results and the mean of the reference standard
determinations. The relationship is as follows:
 =  X -  R  ˆ = ˆ X - ˆ R  ˆ = X - R
Our 99% confidence interval for δ which assumes equal variance for both
X and R is:
(n - 1) S 2X + (m - 1) S 2R  1 1 
ˆ  t 1-/2 SEˆ = ˆ  t 1-/2
n + m
n+ m - 2


Putting our data into this equation yields:
- 0.004  2.712
(9)(0.0012 )2 + (29)(0.001 )2  1
1
+  = - 0.004  0.0010

38
 10 30 
The 99% confidence interval for δ would be: -0.005 to -0.003 g/210L.
In percentage of systematic error this corresponds to:
  0.005 
 0.103  100   4.85%


  0.003 
 0.103  100   2.9%


-4.85% to -2.9%
32.
Much of the biomedical literature reports concentrations of substances
in many different units (i.e., as mM or milliMoles in this case).
Since mM is actually an expression of mM/L, we must first convert the
mM to grams. We do this by observing that 1 Mole of acetone weighs
58 grams. Therefore we obtain:
8.91 mM  0.00891 M

X
0.00891 M

58 g
1M

X  0.517 g / L
Assuming the value of Kbl/br = 300 we obtain:
300 
0.517 g / L
Y

Y  0.00172 g / L  1720 g / L in the breath
29
If 642 g/L are necessary to yield 0.01 g/210L ethanol equivalent
then:
1720 g / L
Z

642 g / L
0.01 g / 210 L
33.

Z  0.027 g / 210 L
We begin by identifying the following variable which has the Poisson
distribution:
X = the number of tests performed on the instrument per day
Since X can be viewed as the sum of several other Poisson random
variables (e.g., the number of tests per minute or per hour) and
λ = 40 is fairly large (e.g., > 20) we will assume that X has a normal
distribution by the Central Limit Theorem. Since we are using a
continuous distribution (the normal) to approximate a discrete random
variable (X) we will use a continuity correction of ½. We
incorporate our data and solve as follows:
34.
1


 X - E(X) (55 - 2 ) - E(X) 
54.5 - 40 

P [X  55] = P 
>
 = P Z >
 = P[Z > 2.29] = 0.011
V(X)
V(X)
40






We obtain the probability of 0.011 from the standard normal tables.
This is a very small probability which suggests that it is unlikely
that 55 or more tests will be performed on the instrument in one day.
Here we have two random variables of interest: XF = the alcohol
elimination rate for women and XM = the alcohol elimination rate for
men. Since we do not know what their exact distributions are we do
know that by the Central Limit Theorem their means will be
approximately normally distributed for sample sizes greater than
approximately 30. We write each of their distributions showing the
expected values (means) and variances as follows:
F )
2
X F ~ (  F ,  F )  X F ~ N(  F ,
n
2
M )
2
X M ~ (  M ,  M )  X M ~ N(  M ,
n
2
We now identify the hypotheses that we want to test:
H0: βF = βM
H1: βF  βM
We use the two sample t-test since we do not know σF2 or σM2 exactly
and have to use our sample variances SF2 and SM2 as estimates. We will
assumed equal variances. We use equation 8:
XF - XM
t=
(n1  1) S F2 + ( n 2  1) S M2  1
1
  
n1 + n 2 - 2
 n1 n 2 
30
Since the study presented the standard errors for each mean we need to
determine the standard deviations as follows:
SE X F =
SD
 SD = 0.55( 59 ) = 4.22
n
SE X M =
SD
_ SD = 0.40( 75 ) = 3.46
n
We then introduce our information into the equation for t as follows:
t=
20.77 - 17.24
(59 - 1)(17.8) + (75 - 1)(12.0)  1
1
  
59 + 75 - 2
 59 75 
= 5.34
Our degrees of freedom for this problems are n+m-2 = 59+75-2 = 132.
From the t-table we find t0.995,132 = 2.576. Our calculated t value is
5.34 which is even larger than 2.576. The probability of obtaining a
value as large as 5.34 when the null hypothesis is true is even
smaller than 0.005. Therefore, we conclude that there is a significant
difference between the elimination rates of men and women in this
study.
35.
We use equation 9. We note that α=0.05 for the two-tailed test and so
we look up the value for Z0.975 in the standard normal table. This
value is 1.96. We use this two tailed value because we did not
specify that our critical difference of 0.005 g/100ml was only in one
direction. Next we look up the Z0.8 in the standard normal table which
is 0.85. We then put these values along with the information given
into equation 9 as follows:
2
 0.007 g / 100 ml 
2
n 
 1.96 + 0.85  = 15.5  16
0.005
g
/
100
ml


So, we would need 16 individuals for our study.
36.
The average value of a smooth continuous function
determined by integrating the given function over
involved and then dividing by the time interval.
over the time interval of 5 seconds is determined
t
0
t
- kt
- 2t
B0 (1 - e ) + Ct dt = 0 0.15(1 - e ) + 0.003t dt = 0.15t +
of time is
the time interval
The average value
by:
0.15 - 2t 0.003 t 2
e +
2
2
5
0
= 0.7125
Note that the middle term involving e-2t goes to zero when t=5 so it
can be disregarded. Now divide this integral by the time interval of
5 seconds:
0.7125
= 0.1425 g/210L
5
When the same procedure is done for the exhalation time of 10 seconds
31
we obtain an average value of 0.1575 g/210L. Comparing these average
values to the actual BrAC values from equation 1 in the problem and
determining the percent differences we obtain:
 0.143 - 0.165 

  100 = - 13.6%
0.165

t =5
 0.158 - 0.18 
 100 = - 12.5%


0.18

t =10
_
So, no, it is not a constant percentage.
in the figure below:
BrAC
This problem is illustrated
Breath Alcohol Exhalation
0.2
0.15
0.1
Average Value
0.05
0
0
2
4
6
8
10
12
Exhalation Time (sec)
37.
First we must determine the percent by volume of alcohol associated
with each of the five measurements. We first use equation 10 to
determine the partition coefficient for ethanol in water heated to
340C in the simulator. Placing T=34 into equation 10 we obtain:
K w/a = 23017.268 e
-0.0643(34)
= 2586
We use this result to next determine the concentration of the alcohol
in the beer using the following equation:
K w/a =
C w  2586 =
Cw
 C w = 0.351g/100ml
0.285 g/210L
Ca
We then need to determine what volume 0.351 g/100ml of ethanol
occupies by using the density equation as follows:
D=
g
0.351 g/100ml
 0.789 =
 X = 0.445ml/100ml
ml
X ml/100ml
Our result is 0.445 ml/100ml which is equivalent to 0.445% by volume.
Doing this for the remaining four measurements we obtain the following
percent by volume estimates: 0.439, 0.434, 0.440 and 0.428. Their
mean is: 0.4372 with a standard deviation of 0.0065. We use
the following t-test since we do not know the exact population
32
standard deviation (σ) and must estimate it with the sample standard
deviation (S):
t (1-/2)df= 4 =
X -
0.4372 - 0.50
=
= - 21.7
S/ n
0.0065/ 5
From the table of critical t values we observe that for 4 degrees of
freedom (n-1) a t value of -4.604 would put an area in the lower tail
of 0.005. Our t result of -21.7 is much lower than the table value.
This means the area to the left of -21.7 is much smaller than 0.005.
This is interpreted to mean that the probability of obtaining the
results we did if the null hypothesis were true is much less than
0.005. We therefore reject the null hypothesis and conclude that
there is evidence in support of the alternate hypothesis, that the
population mean alcohol concentration for this “alcohol free” beer is
less than 0.5% by volume.
38.
Using the truncated two digit and three digit results we obtain the
following:
Two digit:
mean = 0.070 g/210L
sd = 0
CV = 0%
Three digit:
mean = 0.0767 g/210L sd = 0.0019 g/210L
CV = 2.45%
The significant differences are due to using the different number of
digits. One should clearly use the three digit data in this example.
39.
We would compute the confidence interval in the same manner as we did
in problem #5. We assume the one result is a random sample from a
population having the same standard deviation as we would estimate if
they had provided two samples. The one sample was still
received from the same instrument method and general protocol. We now
use the following equation for the 99% confidence interval:
0.0061
X  t (1-/2)df = SDi = 0.116  2.57
= 0.116  0.016
n
1
This would yield a 99% confidence interval of: 0.100 to 0.132 g/210L
40.
First computing the breath alcohol from equation 1 we obtain:
V 
0.08 


) = 0.108396 g/210L
X a =  1.3 ln( 1  ) = X a =  1.3 ln( 1 
5 
5 


Next, computing the result from the approximation we obtain:
f(x) =  1.3(  x 
x2 x3
0.08 2 0.08 3

) =  1.3(  0.08 

) = 0.108382 g/210L
2
3
2
3
Computing the results with a v=1.1 we obtain: 0.322999 g/210L from the
direct method and 0.322075 from the approximation method. At all
concentrations the approximation method used by the instrument will be
33
less than that found by direct calculation using equation 1.
Since the approximation is an expansion about v=0, the approximation
result will deviate further from the direct calculation as results
increase above zero. The deviation, however, is always to become
lower than the direct calculation. Moreover, this illustrates that
using three terms in the expanded series is more than enough for three
digit final results.
41.
We have three different estimates (three labs) of this person’s BAC
where each has different n and different variance (Si2) estimates. We
will want to use a weighted mean in which the weights will be
determined according to n and the variance for each lab.
n
 wi Y i
Yw=
i= 1
n
 wi
where : wi
=
ni
S i2
i= 1
where:
wi = the weight for the ith lab
2
5
4






 0.000002 (0.0850) +  0.0000008 (0.0836) +  0.0000029 (0.0858)
= 0.0839 g/100ml
Yw=
2
5
4

 
 

 0.000002  +  0.0000008  +  0.0000029 
42.
If the thermometer is reading 0.30 C too low, then the simulator
headspace actually contained more alcohol than was thought.
Therefore, a higher concentration than realized was introduced into
the instrument during calibration. This would result in the
instrument measuring the alcohol concentration systematically low.
Given that a one degree centigrade change results in a 6.5% change in
headspace alcohol concentration, we can determine the percent change
resulting from a 0.30 degree centigrade change according to:
0.065
X

 X  0.0195  1.95%
0
1 C 0 .3 0 C
Based on this result, the instrument is reading all results 1.95%
systematically low. Therefore, the subject’s mean BrAC should be
adjusted up by 1.95%. This is done by:
0.125 / 0.132  X  0.1285
 0.1285 ( 1.0195 )  0.131 g / 210 L
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43.
We solve the differential equation by integrating after the separation
of variables:
dB
dB
  k B  B0  
  k dt 
dt
B  B0
dB
 BB

  k dt  ln B  B0   kt  C
0
Now exponentiating both sides we obtain:
B  B0  e  kt  C  B  B0  C 0 e  kt  B  C 0 e  kt  B0
Given the initial conditions of:
B0  C max  C max  C 0 e  k 0   B0  C max  C 0  B0  C 0  C max  B0
From the non-linear regression we obtained the three parameter
estimates which we now put into our model:
B  0.85e 0.5 t  0.150
We solve for t when B = 0.165 g/210L:
0.165  0.85e 0.5 t  0.150  0.015  0.85e 0.5 t  0.0176  e 0.5 t
Taking the natural log of both sides we obtain:
ln 0.0176  ln e 0.5 t   4.04   0.5t
 t  8.1 min utes
R. G. Gullberg
[email protected]
9/1/2016
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