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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: 0.680.000885 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 / h6 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 0.11450.100 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 / dL2 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 29422 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 B1 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 B1 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 16332 5312 4.9 bc 10810 1080 OR ad 13339 4407 13.2 1113 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 0.10050.1025 0.0815 0.0842 bias 100 3.3% GC SimCorr 0.0842 1.230.0995 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 0.9920.65 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 0.0003582 0.008 2 0.001082 0.001162 0.001652 0.0006 2 0.00000062 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 1.645uc 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 1columns 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 SEˆ = 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 34 43. We solve the differential equation by integrating after the separation of variables: dB dB k B B0 k dt dt B B0 dB BB 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: B0 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 35