# OSH/5/Solution

< OSH‎ | 5

### 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 $x$-coordinates--it'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.

1. 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.)
2. 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 $\ln(C)$.
3. 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.)
4. 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 $R(t)=Ce^{rt}$.
5. 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%.
1. 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.
1. First, we need to find the non-fentanyl-related death rates. We find these by subtracting the total death rate from the fentanyl-related death rate.
Year fentanyl rate overall rate non-fentanyl rate
2012 (t=0) 0.2 5.9 5.9-0.2=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 $b=6.038$.

2. The line of best fit for the transformed data is $y=ax+b$. Using our conventions, $a\approx 0.9240$ and $b\approx-1.274$. 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.
3. 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 base-10) 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
Note the small value of $a$: these deaths, in our model, are rising slowly.
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
So, all together, we predict a combined fatal overdose rate of:
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.
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.
2. If you or someone you know struggles with substance abuse, you can find resources here, here, and here.
3. 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 reasonable--for 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 non-fentanyl 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 in-between

Subtotal: 3 points

Presentation and communication: 3 points

Total: 29 pts.

### Tables of Alternate Parameter Values

1. 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
2. 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

3. 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 $b$.

1. 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 $b=778.96$.

2. 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 base-10 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
• $C=10^b$.
Once again the logarithm used should not affect the final answer, except for potentially different rounding errors.
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 $r=a=0.96$.

3. If the student used the absolute year but omitted 2017, the predicted non-fentanyl-related overdose rates for 2018 are 8.06 and 8.84, respectively. The predicted fentanyl-related overdose rates for 2018 are 82.68 and 558.42, respectively.