OSH/5/Solution
Solution key
REMARK 1: There are various ways to set up a spreadsheet with the data given. Below, we make the choices to (1) number the years starting with $t=0$ for the smallest year in the data set, and (2) use natural logarithms. At the end of this page, we give parameters arising from different year conventions, and logarithm base 10. Other choices are possible, but we tried to cover the most obvious ones.
REMARK 2: The problem did not specify significant figures, so any reasonable significant figures used by students is OK. Note, however, that there should be enough significant figures to give reasonable projections. This is especially important if the student used the actual years as their $x$coordinatesit's quite easy in this case to have a rounding error that causes the predicted number of overdoses to be negative. Graders, be aware that students might copy down a truncated or rounded value from their spreadsheet, then use the spreadsheet's value for making their predictions.

 Using the year as the independent variable and the given rates as the dependent variable, the line of best fit is $y=ax+b$ with $a=0.31$ and $b=3.7$. (Remember, these might vary if you used a different naming convention for the years: see the table at the end for other possible answers.)
 Taking the natural log of the function $R(t)=Ce^{rt}$, we get $\ln(R(t)) = \ln(Ce^{rt})$. Log rules allow us to write this as $\ln(R(t)) = \ln(C) + rt$. The right hand side of this is a linear function of $t$ with slope $r$ and intercept $\ln(C)$.
 The line of best fit to the transformed data is $y=ax+b$ with $a=0.0233$ and $b=1.595$. (Again we are using $x=0$ for the year 1991.)
 The intercept of the line of best fit, as shown above, is $b=\ln(C)$. So $C=e^b \approx 4.9$ and $r=a=0.023$. These define for us $R(t)=Ce^{rt}$.
 Both slopes indicate an increasing trend but not a particularly rapid one. To see that the increase is not rapid, consider that a slope of $0.31$ means one additional person every three years which is a relative increase each year of less than 10%.
 As in Question 1, there are conventions to decide on that will change the equation of the lines of best fit, but they shouldn't change parts (c) or (d). We use the date convention $t=0$ is 2012, and we use the natural logarithm rather than log base 10.

First, we need to find the nonfentanylrelated death rates. We find these by subtracting the total death rate from the fentanylrelated death rate.
Year fentanyl rate overall rate nonfentanyl rate 2012 (t=0) 0.2 5.9 5.90.2=5.7 2013 1.1 7.3 6.2 2014 2.0 7.9 5.9 2015 3.2 11.1 7.9 2016 13.9 20.7 6.8 2017 25.6 31.6 6.0
The line of best fit for the data is $y=ax+b$, where $a=0.1514$ and $b=6.038$.
 The line of best fit for the transformed data is $y=ax+b$. Using our conventions, $a\approx 0.9240$ and $b\approx1.274$. If $\log_e(R(t))=at+b$, we have $R(t)=e^{at+b}=(e^b)e^{at}$. That is, $C=e^b\approx 0.2797$ and $r=a\approx 0.924$. Our rate of growth is approximately $r=0.9$; this is much larger than the value of $r$ found in 1(d). That is, according to our models, overdose deaths related to fentanyl are rising significantly faster than overdose deaths not related to fentanyl.

Note 1: Except for rounding issues, the choices of year convention (i.e. $t=0$ for 2012 versus $t=2012$, etc) and logarithm (natural versus base10) should not impact your answer for this part.
Note to graders: due to initially ambiguous wording, students may have used an exponential model or linear model where these solutions use the other. If you see a student do this, email Elyse to write up a solution and marking scheme.
The overdose rate not involving fentanyl is approximated by $y(t)=at+b$, where $a=0.1514$ and $b=6.038$. 2018: $y(6)\approx 6.95$ deaths per 100,000 people
 2020: $y(8)\approx 7.25$ deaths per 100,000 people
The overdose rate involving drugs contaminated with fentanyl is approximated by $R(t)=e^{at+b}$, with $a=0.9240$ and $b=1.274$. 2018: $R(6)\approx 71.5$ deaths per 100,000 people
 2020: $R(8)\approx 454$ deaths per 100,000 people
 2018: $R(6)+y(6)\approx 78$ deaths per 100,000 people
 2020: $R(8)+y(8)\approx 461$ deaths per 100,000 people
Sample interpretation: The data in Question 2 seems to fit our approximations better than the data in Question 1. So, we think the rate of overdose fatalities not involving fentanyl is rising very slowly, while the rate of fatalities involving fentanyl is rising exponentially. That is, fentanyl contamination has caused a fundamental shift in the overdose rate. 
A sample answer:
Due to increased public efforts to keep people safe from overdoses, we expect (and hope) that historical trends will not continue. Evolving public policy is likely to change the rates in ways not predicted by our model. So, we suspect our estimates are too high.

First, we need to find the nonfentanylrelated death rates. We find these by subtracting the total death rate from the fentanylrelated death rate.
 If you or someone you know struggles with substance abuse, you can find resources here, here, and here.
 Students should get 1 pt if they only listed the problems and type of error(s) made on each one. For a more complete analysis, including evidence of future plans to avoid such errors, students should get 3 pts. Something in between is worth 2 pts.
Marking
For full credit, students should use enough significant digits that their predictions are reasonablefor example, a predicted overdose rate should not be negative. Students should: [not entirely updated for 2017 yet]
 Problem 1
 1a. (3pt) Find the line of best fit for the data.
 1b. (3 pt) Derive the linear relationship between $\ln(R(t))$ and $t$. One point for using the product log rule. One point for using the exponent log rule. One point for the rest being coherent and correct.
 1c. (3 pts) Find a line of best fit for the transformed data. Two points for correctly finding the log of best fit (one off for minor errors, two off if it's totally wrong). One point for clear presentation of the data, a well organized table, etc.
 1d. (2 pts) Report the $r$ and $C$ values. One point for each correct value.
 1e. (2 pts) Communicate clearly.
Subtotal: 13 points
 Problem 2
 2a. (2pt) 1 point: finding the line of best fit. 1 point: comparing it to 1a.
 2b. (2pt) 1 point: finding the line of best fit for the transformed data. 1 point: comparing it to 1d.
 2c. (5pt) 1 point each for fentanyl and nonfentanyl predictions, 1 point for adding them together; 2 points for a reasonable interpretation. No credit for predictions that make no sense, even if they match previous work, e.g. rates going down or being negative.
 2d. (1pt) Communicate clearly.
Subtotal: 10 points
 Problem 3: nothing to submit, no points
 Problem 4: (3 pts) 1 point for classifying all errors; 3 points for having a clear plan to address them; 2 points for something inbetween
Subtotal: 3 points
Presentation and communication: 3 points
Total: 29 pts.
Tables of Alternate Parameter Values


Parameter Absolute year (1991, 1992...) Relative year (0,1,2,...) Relative year (1,2,...) a 0.3100 0.3100 0.3100 b 613.5 (610 is ok too based on significant figures) 3.722 3.412  The line of best fit to the transformed data is $y=ax+b$ with
Parameter Absolute year (1991, 1992...) Relative year (0,1,2,...) Relative year (1,2,...) $\log=\log_e$ $\log=\log_{10}$ $\log=\log_e$ $\log=\log_{10}$ $\log=\log_e$ $\log=\log_{10}$ a 0.0233 0.0101 0.0233 0.0101 0.0233 0.0101 b $$44.85 $$19.48 ($$20 OK) 1.595 0.6928 1.572 0.6827

The line of best fit for the transformed data is $y=at+b$, for the values of $a$ and $b$ found above.
That is, $\log(R(t))=at+b$.
 If the natural logarithm was used to transform the data, then $R(t)=e^{at+b}=Ce^{rt}$ for
 ${r=a}$ and
 ${C=e^b}$.
 If the logarithm base 10 was used to transform the data, then $R(t)=10^{at+b}
=10^b\cdot10^{at}=10^b\cdot e^{(a\ln 10)t}=Ce^{rt}$ for
 ${r=a\ln 10}$ and
 ${C=10^b}$.
Parameter Absolute year (1991, 1992...) Relative year (0,1,2,...) Relative year (1,2,...) $\log=\log_e$ $\log=\log_{10}$ $\log=\log_e$ $\log=\log_{10}$ $\log=\log_e$ $\log=\log_{10}$ r a=0.0233 a$\ln$10=0.0101$\ln$ 10 $\approx$0.02326 a=0.0233 a$\ln$10=0.0101$\ln$ 10 $\approx$0.02326 a=0.0233 a$\ln$10=0.0101$\ln$ 10 $\approx$0.02326 C $e^b=e^{44.85}\approx 3.326 \times 10^{20}$ $10^b=10^{19.48}\approx 3.311 \times 10^{20}$ $e^b=e^{1.595}\approx 4.928$ $10^b=10^{0.6928}\approx 4.929$ e^b=$e^{1.572}\approx 4.816$ $10^b=10^{0.6827}\approx 4.816$ Note: choosing $\log=\log_{10}$ or $\log=\log_e$ ideally has no impact on our values of $r$ and $C$ above; the differences in the table are due to rounding when we found $a$ and $b$.
 If the natural logarithm was used to transform the data, then $R(t)=e^{at+b}=Ce^{rt}$ for



The line of best fit for the data is $y=ax+b$, where $a=0.1514$ and $b$ is approximately:
Parameter Absolute year (1991, 1992...) Relative year (0,1,2,...) Relative year (1,2,...) b $298.6$ $6.038$ $5.887$ If the student used the absolute year but omitted 2017, $a=0.39$ and $b=778.96$.

The line of best fit for the transformed data is $y=ax+b$, for the parameters shown below.
 If we use natural logarithms, then $\log_e(R(t))=at+b$, so $R(t)=e^{at+b}=(e^b)e^at$. In this case,
 $r=a$ and
 $C=e^b$.
 If we use base10 logarithms, then $\log_{10}(R(t))=at+b$ gives us $R(t)=10^{at+b}=10^be^{ta\ln(10)}$. In this case,
 $r=a\ln 10$ and
 $C=10^b$.
Parameter Absolute year (2012, 2103...) Relative year (0,1,2,...) Relative year (1,2,...) $\log=\log_e$ $\log=\log_{10}$ $\log=\log_e$ $\log=\log_{10}$ $\log=\log_e$ $\log=\log_{10}$ a 0.9240 0.4013 0.9240 0.4013 0.9240 0.4013 b 1860.4 807.94 1.2739 0.5532 2.1979 .95453 $r$ $0.9240$ $0.4013\ln(10) \approx 0.9240$ $0.9240$ $0.4013\ln(10) \approx 0.9240$ $0.9240$ $0.4013\ln(10) \approx 0.9240$ $C$ $e^{1860.4}\approx 1.0928 \times 10^{808}$ $10^{807.94}\approx 1.1482 \times 10^{808}$ $e^{1.2739}\approx 0.2797$ $10^{0.5532}\approx0.2798$ $e^{2.1979}\approx 0.1110$ $10^{0.95453}\approx 0.1110$ If the student used the absolute year but omitted 2017, the line of best fit for the transformed data is $ax+b$ with $a=0.96$ and $b=1,922.87$. That corresponds to an exponential model $y=Ce^{rt}$ with $C=e^b=e^{1923}$ and $r=a=0.96$.
 If we use natural logarithms, then $\log_e(R(t))=at+b$, so $R(t)=e^{at+b}=(e^b)e^at$. In this case,
 If the student used the absolute year but omitted 2017, the predicted nonfentanylrelated overdose rates for 2018 are 8.06 and 8.84, respectively. The predicted fentanylrelated overdose rates for 2018 are 82.68 and 558.42, respectively.

The line of best fit for the data is $y=ax+b$, where $a=0.1514$ and $b$ is approximately: