ES 25 Quantitative Thinking
Lab 3: Lets Practice and Review Graphs
Due: Tuesday, April 24,th before 12 noon (E-mail to your facilitator)
DOUBLE-check your answers, and make sure to always include UNITS. Label your graphs
clearly. Neatness and effort are positively correlated with high homework scores!
Bacteria Population Dynamics (pg 21 Workbook.. lets try it in Excel, and verify by hand later)
Given: p(t) = 5000 + 3000t -2000t2
1. Set up the data table in Excel, and use the equation given to calculate p(t).
Transpose, to have data in columns. (Copy, paste into Excel. Then select again,
cut, paste special, check “transpose”)
p(t) (individuals)
Avg. rate of change (slope
between points) (change in # of
t (time, in hours)
individuals/hr)
5000
0
6000
0.5
2000
6000
1
0
5000
1.5
-2000
3000
2
-4000
0
2.5
-6000
-4000
3
-8000
Note: the slope of 2000 is for the time period between 0 and 0.5 hours
2. Graph the data.
3. Estimate the equation of an appropriate trendline. Copy and paste your graph,
with trendline equation an associated R2 value, below.
Bacteria Population Dynamics
Population size (# of individuals)
8000
6000
4000
2
y = -2000x + 3000x + 5000
2
R =1
2000
0
0
0.5
1
1.5
2
-2000
-4000
-6000
Time (hours)
2.5
3
3.5
a. In a new column, calculate the average rate of change between 0 and 0.5
hours. Repeat for each half-hour thereafter (replace the data table above
with your completed data table from Excel).
b. Offer an explanation for changes that you observe in the slope (i.e. explain
the population dynamics).
The slope starts off positive, meaning that individuals are added to the population in that
time period (in the first 1/2 hour, the growth rate is 2000 individual per hour, and since
we are only looking at 1/2 of an hour, only 1000 individuals are added the population).
In the next time period (0.5 to 1 hour), the slope is zero, there is no growth (no
individuals are added, per unit time). From 1 hour onward, the slope is increasingly
negative, meaning that the death rate increases with time.
I think that maybe the population was growing initially, but then a predator discovered
the bacteria, and began consuming individuals. Then the predators may beckoned for
help eating the bacteria from their relatives, which would explain why the death rate is
increases. Certainly, you could probably come up with a better explanation that this, but
mine does make logical sense (which will give it full credit).
c. During what time period is the magnitude (absolute value) of the slope
largest? What does your answer mean, for this problem?
The (absolute value) of the slope is largest between 2.5 and 3 hours after the “start.”
During this time period, the death rate is 8000 individuals/hour (or, the growth rate is –
8000 indv/hr), which means that 8000 individuals are removed from the population each
hour. Since the time period is only 1/2 of an hour, the population should decrease by
4000 individuals between 2.5 and 3 hours after the start (I verified this using the
population numbers given, 0 to –4000 individuals. However, I can’t make sense of
“negative population,” so perhaps a death rate of 6000 indv/hr is the largest logical rate of
change, which occurs between time 2 and time period 2.5 hours).
d. What does a slope of 0 mean, for this problem?
It means that the average rate of change is zero (for the time period from 0.5 hours to 1
hour after the start). Thus, no individuals are added or subtracted, on average, from the
population.
e. Evaluate p(0.75), and interpret your answer in words.
p(.75) = 6125 individuals. This means that 45 minutes after the start (45 is .75 of 60
minutes), the population of bacteria is 6125 individuals.
f. Report and interpret the y-intercept of the graph in terms this problem.
p(t) = 5000 + 3000t -2000t2
p(0) = 5000 individuals; at the start of the experiment there are 5000 individuals (this is
the initial condition).
g. Report and interpret the x-intercept of the graph.
0 = 5000 + 3000t -2000t2
The above equation can be solved using the quadratic formula, to get t = 2.5 or t = -1.
Or, by looking at the graph we can find that the curve “intercepts” the x axis at 2.5 hours.
This means that after 2.5 hours, all of the bacteria were dead.
h. Why does the R2 value equal 1?
Since we actually used the polynomial “given” to generate our y-values, it is no surprise
that the equation “predicts” the y-values perfectly (in other words, all of the variance in y
is explained by our model).
i. Over what range of x values do you think the equation is valid? Explain.
The equations seems logical for times between 0 and 2.5 hours, after which there is a
negative amount of bacteria, which does not make intuitive sense to me.
Ant Population Dynamics (pg 22 Workbook.. lets check our work in Excel)
Given: p(t) = 3000 – 2000/t2
4. Set up the data table in Excel, and use the equation given to calculate p(t).
p(t)
(individuals)
undefined
-5000
1000
2111.11111
2500
t (time, in days)
0
0.5
1
1.5
2
Avg. rate of
change (slope
between points)
(change in # of
individuals/day)
undefined
12000.00
2222.22
777.78
5. Graph the data. (note: do NOT use the data point p(0), t = 0 in your graph)
6. Estimate the equation of an appropriate trendline. Copy and paste your graph,
with trendline equation an associated R2 value, below.
Insect Population growth
3000
Population size (individuals)
2000
p(t) = 3000-2000/t2
R2 = 1
1000
0
0
0.5
1
1.5
2
2.5
-1000
-2000
-3000
-4000
-5000
-6000
Time (days)
a. Estimate the equation of the trendline. Why does the R2 NOT equal 1?
Because Excel can’t come up with the exact equation that we used to
generate the y-values, so the curve doesn’t explain all of the variance in y
anymore.
b. Lets remove the trendline, make excel connect the dots (by changing the
Graph Type to x-y scatter with the dots connected), and write in our own
trendline equation and R2 value (using view-toolbars-drawing- ‘text box’)
c. In a new column, calculate the average rate of change between 0 and 0.5
days (using the equation “given”. Repeat for each half-day thereafter
(replace the data table above with your completed data table from Excel).
Note: can’t get slope for 0 to 0.5 because the function is undefined (it approaches
negative infinity) at time = 0.
d. Is the absolute value of the slope larger between 1 and 1.5 days, or
between 1.5 and 2 days? What does your answer mean, for this problem?
The population is growing faster between 1 and 1.5 days (with an average growth rate of
2222 ind/day) than between 1.5 and 2 days (when the average growth rate is 778
indv/day). A growth rate of 2222 indv/day means that during this time period, 2222
individuals are added each day. Since the time period is only 1/2 of a day, only 1111
individuals are added (confirm with population values: 2111 indv. –1000 indv.). This is
clear on the graph because the slope (which represents the average rate of change) is
steeper between 1 and 1.5 days than between 1.5 and 2 days.
e. Evaluate p(0.75), using the given equation, and interpret your answer in
words.
p(.75) = 3000-2000/(0.75)2= -555.55
So, at 3/4 of an day (18 hours) past the start
of the experiment, the population was –555.55 individuals. Somehow the insect
population increased from this negative number to become positive after 0.816 days
(maybe insect eggs were incubating, and hatched??)
f. Report and interpret the y-intercept of the graph in terms this problem.
By definition, the curve intercepts the y-axis when x = 0. It answers the question, “when
time is zero (the initial condition), what is the population, p(t)?”
p(0) = 3000-2000/(0)2 which is UNDEFINED. From the graph, and from intuition, as t
gets smaller and smaller, 2000/t^2 gets larger and larger. Since p(t) = 3000 – an
increasingly large number, the function approaches negative infinity as t approaches zero.
g. Report and interpret the x-intercept of the graph.
To find out where the curve intercepts the x-axis, we need to realize that y is zero when x
is equal to the x-intercept. Replacing y, or “p(t)” in this case, with 0,
p(t) = 0 = 3000-2000/t2
we can solve for t to find t = sqrt (2/3)
So, the x-intercept is = 0.816 days. For this problem, after 0.816 days, the population
goes from being negative (hard to imagine) to being positive.
h. Over what range of x values do you think the equation is valid?
From 0.816 days onward
PCB’s:
As popularized in Rachel Carson’s Silent Spring, high levels of PCB (polychlorinated
biphenyl, an industrial pollutant) in the environment can damage pelican’s eggs by
thinning the shells. The data table below shows the relationship between the
concentration of PCB in the eggshells and the thickness of the eggshell1.
Concentration, c, in parts per million (ppm) 87 147 204 289 356 452
Thickness, h, in millimeters (mm)
.44 .39 .28 .23 .22 .14
1. Copy and paste the data table into Excel. Transpose, to have data in columns (or
you can type it in by hand). (Transpose: Copy, paste into Excel. Then select
again, cut, click on a new cell, edit, paste special, check “transpose”)
2. Decide which column should be y, and which should be x (include units).
y = Thickness, h, in millimeters (mm)
x = Concentration, c, in parts per million (ppm)
3. Predict whether the relationship will be positive or negative: negative
and whether it will be linear or non-linear: linear
Explain your logic in making these predictions: I expect that as PCB concentrations
increase, the thinkness of pelican’s eggs should decrease. I think it should be linear,
because I expect that each additional increase in concentration (ppm) will cause the same
amount of thinning of eggshells (change in thickness). If I thought the rate of change in
thickness, with respect to concentration, was not constant, I would predict a non-linear
relationship.
4. Graph the data.
5. Estimate the equation of an appropriate trendline. Copy and paste your graph,
with trendline equation an associated R2 value, below.
1
Risebrough, R. W., “Effects of environmental pollutants upon animals other than man.”
Proceedings of the 6th Berkeley Symposium on Mathematics and Statistics, VI, p. 443463, (Berkeley: University of California Press, 1972)
PCBs
0.5
Eggshell thickness (mm)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
y = -0.0008x + 0.488
2
R = 0.9362
0.1
0.05
0
0
100
200
300
400
500
Concentration of PCB (ppm)
a. Report and interpret the slope of the graph in terms of this problem.
The slope = -0.0008 mm/ppm. This means that we expect the eggshell thickness to decrease by
0.0008 mm for every 1 ppm increase in concentration of PCB. Notice that this rate of change is
the same for all concentrations of PCB’s (as opposed to the previous two problems, where the
average rate of change varied depending on what range of x’s we were looking at)
b. Report and interpret the y-intercept of the graph in terms of this problem.
When the concentration of PCBs is 0, y(0) = 0.488 mm. This means that without any PCBs, the
average pelican egg is 0.488 mm thick. We could think of this as the “baseline” thickness for
pelican eggshells.
c. Report and interpret the x-intercept of the graph in terms of this problem.
The x-intercept will tell us “at what concentration the eggshell thickness will be equal to zero.”
We can solve by setting y equal to zero, and solving for x.
x = 0.488/.0008 ppm = 610 ppm.
d. Report the R2 value for your trendline:0.9362
i. Interpret the R2 value, in terms of eggshells and PCB’s.
93 % of the variance (a measure of spread) in eggshell thickness is explained by the
model (the relationship between thickness and concentration of PCB’s)
ii. What could account for the “unexplained variance” in y?
Genetic differences in pelicans which leads to naturally occurring differences in eggshell
thickness.
e. Calculate f(204) using your trendline equation. Interpret your answer, in
terms of eggshells and PCB’s.
y= -0.0008x + 0.488
remember y = f(x), y is a function of x,
y(204) = -0.0008(204) + 0.488 = 0.325 mm
f. Calculate the residual associated with f(204). Interpret your answer, in
terms of eggshells and PCB’s.
Our predicted value of eggshell thickness for a PCB concentration of 204 ppm was 0.325
mm. Looking at our data table, the actual thickness associated with this PCB
concentration was 0.28 mm.
Thus, the residual (yactual – ypredicted) is 0.28 – 0.325 = -0.045 mm. We overpredicted the
eggshell thickness by .045 mm.
g. Do you think that the trendline will continue to make accurate predictions
for y as x gets larger (in other words, do you feel confident extrapolating
for this data set)? Why/why not?
It can’t go on forever, because the eggshell thickness can not become negative. Thus, the
function only makes sense for concentrations between 0 and 610 ppm.
Explore Global Warming
In Lab 2, we looked at trends in global temperature anomaly through time, which
scientists believe is a result of increases in CO2 (primarily from burning fossil fuels). Lets
look at the relationship between the temperature anomaly and CO2.
Open the Lab3.xls spreadsheet, ExploreGW
*note, since CO2 varies dramatically between summer and winter, I took the average of the two values to
compare to the yearly temperature anomaly data (this may not be the best approach, but it was convenient).
I matched the CO2 and temperature data by years (I only had CO2 data for 1959 through 2002).
6. Decide which column should be y, and which should be x (include units).
y = average temperature anomaly (deg C)
x =CO2 (ppm)
7. Predict whether the relationship will be positive or negative: positive
and whether it will be linear or non-linear: non-linear.l
Explain your logic in making these predictions:
Based on my ES classes, I believe that increasing levels of CO2 will lead to
greater average temperature anomalies. I think it will be non-linear, because
many complex, dynamic systems are governed by feedbacks and thresholds,
which lead to non-linear responses by the dependent variable. Since I think it will
be non-linear, I graphed it as follows:
8. Graph the data.
Does CO2 Cause Global Warming? Data for years
(1959-2002)
Average Temperature Anomoly (deg
C)
0.700
Non-linear model
y = 1E-05x3 - 0.0108x2 + 3.6104x - 402.47
0.600
R2 = 0.4972
0.500
0.400
0.300
Linear model
y = 0.0063x - 2.035
0.200
R2 = 0.3164
0.100
0.000
310
-0.100
320
330
340
350
360
370
380
-0.200
-0.300
CO2 (ppm)
9. Estimate the equation of an appropriate trendline. Copy and paste your graph,
with trendline equation an associated R2 value, below.
a. What does each data point on the graph represent?
Each point represents a year (between 1959 and 2002, but note, they are not plotted in
chronological order, they are plotted in coordinates of (temp. anomaly, C02). Thus, the pair of
values represents the “average” temperature and CO2 values for a year.
b. Report and interpret the slope of the graph in terms of this problem.
Since I used a non-linear plot (which I feel reflects the underlying pattern in the data more
accurately), The slope is changing (not constant). A positive slope indicates and increase in
temperature anomaly when CO2 increases, whereas a negative slope corresponds to a decrease
in temp. anomaly for a increase in CO2.
For a linear plot, you would have gotten the equation: y = 0.0063x - 2.035. The slope of .0063
degC/ppm means that for a one unit increase in CO2, we would expect a 0.0063 deg C increase
in the average temperature anomaly. If you got an equation of y = 0.125x – 0.815, convince
yourself that neither the slope nor the intercept makes sense (try intercept at CO2= 0 ppm, and
for CO2=315 ppm)
c. Report and interpret the y-intercept of the graph in terms of this probable
For non-linear plot, y = 1E-05x3 - 0.0108x2 + 3.6104x - 402.47 with R^2 = 0.4972
When x (CO2) = 0, y (temperature anomaly) = -402.47deg C. Thus, when there is no CO2 in
the atmosphere, the planet is a very very cold ice-ball. I would want to check and see what the
minimum historic CO2 levels have been (and associated temperatures) before deciding how far
out to extrapolate.
Linear plot:
y= 0.0063x - 2.035
y intercept is –2.035 deg C, which means that when the concentration of CO2 is zero ppm, the
temperature anomaly is –2.035 de C.
d. Report and interpret the x-intercept of the graph in terms of this problem.
My non-linear plot looks like it crosses the x-axis around 317 ppm. This is the concentration at
which the temperature anomaly changes from negative to positive.
My linear plot looks like it crosses the x-axis around 325 ppm. This is the concentration at
which the temperature anomaly changes from negative to positive. I could also solve for this
value by setting the y = 0 and solving for x.
non-linear answer:
e. Report the R2 value for your trendline: 0.4972
i. Interpret the R2 value, in terms of this problem.
49.7% of the variance (a measure of spread) in the global average temperature anomaly
was explained by my model (3rd degree polynomial with CO2 as the independent
variable).
ii. What could account for the “unexplained variance” in y?
Many factors are not accounted for in my model, such as deforestation, creation of urban
heat islands, etc. My model might account for factors such as ocean uptake of heat,
since that factor might be correlated with CO2 presence in the atmosphere.
f. The Intergovernmental Panel on Climate Change reports that the carbon
dioxide concentration is rising by 1.5 ppm per year.
i. Based on the data provided, do you agree with this value? Show
your calculation clearly.
This is a kind of tricky. We are asked for the rate of change of CO2, relative to time.
Thus, we need to graph CO2 vs. time, and estimate the average rate of change (slope).
CO2 increase from 1959 to 2002
380
y = 0.8534x + 321.37
370
360
CO2 (ppm)
350
340
330
320
310
300
290
01
99
20
97
19
95
19
93
19
91
19
89
19
87
19
85
19
83
19
81
19
79
19
77
19
75
19
73
19
71
19
69
19
67
19
65
19
63
19
61
19
19
19
59
280
Y ear (for linear equation, 1959 is year 0)
According to my linear model, CO2 is rising by 0.85 ppm each year (slope is average
rate of change). I don’t know why my graph has two high values and only one low value,
seems weird. I investigated (using the link to the original data on the spreadsheet), and
found that when yearly averages were taken, there was a miscalculation. The graph
should look like:
CO2 increase from 1959 to 2002
y = 1.3653x + 312.41
380.00
370.00
360.00
CO2 (ppm)
350.00
340.00
330.00
320.00
310.00
300.00
290.00
.0
19
59
19
61
19 .0
63
.
19 0
65
.
19 0
67
.
19 0
69
.
19 0
71
.
19 0
73
.
19 0
75
19 .0
77
.
19 0
79
19 .0
81
.
19 0
83
.
19 0
85
.
19 0
87
.
19 0
89
.
19 0
91
19 .0
93
.
19 0
95
.
19 0
97
.
19 0
99
.
20 0
01
.0
280.00
Y ear (for linear equation, 1959 is year 0)
According to my new linear model, CO2 is rising by 1.36 ppm each year (slope is
average rate of change). This is close to the IPCC value. Interestingly, some students
figured out the slope (rate of change) between each pair of years, and then averaged those
twenty or so measurements, to arrive at an average slope of 1.3 ppm/yr.
Therefore, the data is flawed and I can not trust my original predictions. I changed my
“Source Data” to the properly calculated CO2 averages (with corresponding years and
temperatures) to revise my models as follows:
Average Temperature Anomoly (deg
C)
0.700
Does CO2 Cause Global Warming? Revised Data
for years (1959-2002)
Non-linear Equation
y = 0.0001x2 - 0.0917x + 14.26
0.600
2
R = 0.7655
0.500
0.400
0.300
0.200
Linear equation
y = 0.0095x - 3.1495
0.100
R2 = 0.7282
0.000
310.00
-0.100
320.00
330.00
340.00
350.00
360.00
370.00
-0.200
-0.300
CO2 (ppm)
This model has a much better R^2 value (it fits the data better), and does not have those
strange vertical grouping of data on the left hand side. I should re-interpret my intercepts
with this new model.
ii. If the amount of CO2 increase per year does, in fact, continue at 1.5
ppm/year, what will be the concentration of CO2 in the year 2100?
(notice that the CO2 concentration in 2002 was 372.55 ppm)
If the average rate of change is constant, at 1.5 ppm/year (note: this corresponds to a
constant rate of increase, linear model with slope = 1.5 ppm/year),
We will have added (1.5 ppm/year)*98 years = 147 ppm to the 2002 value of 372.55
ppm. Thus, the new concentration of CO2 will be 519.55 ppm.
iii. According to your model, what will be the temperature anomaly in the
year 2100, based on the concentration of CO2 calculated above? (you
can do a quick internet search for temperature predictions of more
sophisticated models to see if your estimate is in the ballpark)
According to my revised linear model, the new temperature anomaly should be:
380.00
According to my revised linear model, the new temperature anomaly should be:
y(519.55) = 0.0095(519.55) - 3.1495
= 1.78 deg. C (I have heard the temperature will rise by about 2-2.5 deg. C this
century, so my “back-of-the-envelope” calculation doesn’t seem too bad.
According you the old linear model (this is the one you may have used), the new temp.
anomaly should be:
y= 0.0063x - 2.035
= 1.24 deg C .
You could also calculate the predicted temperature anomaly using nonlinear equation:
y(x) = .001x2-0.0917x+14.26
y(519.55) = 0.001(519.55)2- 0.0917(519.55) + 14.26 = 88.84 deg C (check out this curve on
the graph to see why it gets so hot).
iv. Do you feel confident extrapolating for this data set? Why/why not?
I am not extremely confident, because I know that temperature change is governed by other
cycles besides just CO2. There could be negative feedbacks that act to regulate the effect of
CO2 on climate. Also, some students mentioned that they feel our society might undergo a
major paradigm shift which will change our CO2 and climate future. Sounds like an
interesting idea…
g. Offer a reasonable explanation for the vertical group of data points on the
left-hand side of the graph.
I found the actual error- it was a data entry problem (as is often the case… think about
how this relates to the outliers in the student reported data… are they really consuming
that much more water?). You could have offered any reasonable explanation, such as any
effect that would magnify the heating effect of CO2 (perhaps El Nino/La Nina events?).
**Disclaimer: temperature change is a very complicated process, with
feedbacks, thresholds, and non-linearities. This model is meant to provide a
“back-of-the-envelope” look at the relationship between C02 and temperature.