Answers to Practice Problems
1. We employ Widmark’s equation as follows:
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:
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:
Put in the given data and solve for Ct:
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:
Assuming the ±25% uncertainty interval we obtain:
10.1 beers 25% = 10.1 2.5 = 7.6 to 12.6 beers
Solving for Widmark’s estimate of uncertainty we employ the following equation for men:
Putting in our estimates we obtain:
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:
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:
We now employ Widmark’s equation based on TBW as follows:
1
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:
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:
We then employ the equation for a 99% confidence interval:
This would yield a 99% confidence interval of: 0.0926 to 0.1146 g/210L
7. We first determine the 99% confidence interval from:
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:
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.
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):
1
We then rearrange our probability statement and introduce our
determined value for Z(1-α/2):
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:
Now we can determine the proportion that each component contributes to
the total according to:
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:
Next, we compute the uncertainty associated with the original mean
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
0.1145 g/100ml 0.0020 2
R 0.100 g/100ml 0.0003 2
0.0975 g/100ml 0.0006 15
10.15 ml 0.015 10
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:
where: uncertainty due to the total method
uncertainty due to the traceable reference value
uncertainty due to the analytical instrument
uncertainty due to the dilutor
We now solve for the combined uncertainty according to:
The 99% confidence interval would be found as follows:
The uncertainty budget for this example is as follows:
Total Method 93%
Reference 5%
Analytical 1.5%
Dilutor 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 Analytical Dilutor Bias Total Method
Mean 0.100g/100ml 0.1025g/100ml 10.12µL 0.003g/100ml 0.0885g/100ml
SD 0.0003g/100ml 0.0011g/100ml 0.05µL 0.0013g/100ml
n 2 8 10 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: uncertainty due to traceability
uncertainty due to the GC replicates
uncertainty due to the dilutor
uncertainty due to the bias
uncertainty due to the total method
The 95% confidence interval would be found from:
The sum of all terms under the radical is: 0.0002608 From this we
find the percent contribution from each component:
Traceability: 3%
Analytical: 6%
Dilutor: 1%
Bias: 49%
Method: 41%
11. We begin by computing our total method uncertainty using the
uncertainty function determined from the proficiency test data:
We now write down each of our four sources of uncertainty and
their estimates as follows:
Traceability Analytical Bias Total Method
Mean 0.100g/100ml 0.1046g/100ml 0.1046g/100ml 0.0970g/100ml
SD 0.0003g/100ml 0.0010g/100ml 0.0012g/100ml 0.0027g/100ml
n 2 18 1 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:
where: uncertainty due to traceability
uncertainty due to the GC replicates
uncertainty due to the bias
uncertainty due to the total method
The 95% confidence interval would be found from:
12. We first convert the maximum allowed standard deviation estimate to its variance. Next, we state our hypotheses as follows for a one-tail test:
H0: σ2 ≤ 1225 ng/ml
HA: σ2 > 1225 ng/ml
Next, we compute our chi-square statistic since variances follow a chi-square distribution.
Test Statistic:
Next, we determine our critical value for the chi-square distribution given that α = 0.01.
Critical chi-square (found in Excel):
We now compare our computed chi-square value (41.76) to our critical chi-square 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:
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
13. Assume the data below was generated in a proficiency test program. The percent CV and bias have been determined according to equation 1.
where:
The reference value has been determined from a weighted mean of the measured results according to:
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 and the bias. The bias could be determined from PT results as presented here or from the average of day-to-day control measurements.
The combined uncertainty is determined according to:
The uncertainty in the weighted reference value is determined from:
From our data above we obtain uRw = 0.000093
Let us now assume that we want the combined uncertainty for lab 4. We first determine the uncertainty in the bias according to Eq. 3:
Next we determine whether the bias (B) is significant by using the
t-distribution to test the null hypothesis that B = 1:
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:
We now express our final combined uncertainty from Eq. 3 as:
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:
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:
We now state our hypotheses: H0: δ = 0
HA: δ ≠ 0 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:
The computed t statistic is determined from:
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:
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. We estimate the necessary sample size from:
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:
Since this 95% confidence interval includes zero we conclude these
differences are acceptable.
For mean results at 0.250 g/210L we compute:
At this concentration our differences should not exceed 0.028 g/210L.
17. We take the information give and develop our contingency table as
follows:
We now compute our performance parameters as follows:
The power of the test is:
18. 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.
We proceed as follows:
Since this does not contain zero, the differences are significant – the duplicate test differences would have to be ≤ 0.0018 g/100ml to be non-significant
19. The odds ratio is the quantitative measure of association in case-control 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:
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:
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.
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:
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:
Knowing that we have a bias of +3.0%, we correct this mean result
according to:
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:
where: the variance determined from the equation given in the
problem which combines both analytical and biological
components
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:
We now put our variance estimates into the confidence interval
estimate:
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:
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:
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: µ1=µ2=µ3=µ4=µ5
HA: µ1≠µ2≠µ3≠µ4≠µ5
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:
From the table we see that the calculated F statistic
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:
where: MSbetween = the between groups value for mean squared (MS)
in the table
the within groups MS found in the table
n = the number of measurements performed at each
volume (n=10)
Putting the values from the table into the equation above we obtain:
The combined variance is the sum of these two components:
The percent contribution by each component is found from:
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.
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:
The Excel summary is as follows:
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 the mean difference is:
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: µ1=µ2=µ3=µ4
HA: 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:
From the table we see that the calculated F statistic
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:
where: MSbetween = the between groups value for mean squared (MS)
in the table
the within groups MS found in the table
n = the number of measurements performed at each
volume (n=10)
Putting the values from the table into the equation above we obtain:
The combined variance is the sum of these two components:
The percent contribution by each component is found from:
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