1. Excellence Justify the choice of your model by commenting on at least 3 points. Your comments could include the following: a)Relate the solution to the problem. Explain clearly how well your model can be used to predict answers to a range of problems involving your variables. b)Explain how the theory of the situation relates to the model. c)Discuss any limitations of the model. d)Consider the nature of the underlying variables. e)Consider the possibility of using an alternative model. (highly unlikely you can do this one!)

2. Relate the solution to the problem. Make sure you state what the problem (experiment) is all about (use the Introduction to Part C to help you). Use your model (equation) to predict temperatures for the times that you measured the temperature at. Compare the values that the model gives you (theory) to the raw data you collected. [Do a table to summarise this information.] Does the model fit well all the way through the range of values? Are there values or a range of values where your model overestimates/underestimates the temperature?

3. Relate the solution to the problem. The model y = 48.1859e -0.0292x is used to predict the difference above room temperature of a cup of coffee as it cools. By comparing the raw data to predictions based on the model, we can decide whether the model is appropriate.

4. Relate the solution to the problem. Raw Data (from experiment) Theory (from exponential model) time (min) Diff above room temp (°C)time (min) Diff above room temp (°C)Difference 048.5048.2-0.3 443442.9-0.1 838838.10.1 1233.51233.90.4 16301630.20.2 2027.252026.9-0.4

5. Relate the solution to the problem. The predicted values from my model are very close to the actual data obtained in the experiment. This means that my model will provide good estimates of the temperature of the cup of coffee as it cools over the range of times up to 20 minutes. Initially my model slightly underestimates the temperature, then it slightly overestimates the temperature and at 20 minutes it starts to underestimate the temperature again.

6. Relate the solution to the problem. The first part of the problem is to predict the temperature of the coffee after 12 minutes. My model predicts that the difference above room temperature after 12 minutes will be 33.9 (°C) and in actuality the difference was 33.5 (°C). Since the difference is only 0.4 °C, my model is a very good fit for the data in this range. The second part of the problem is about a temperature difference of 30°C. This happened in my experiment at 16 minutes where the raw data and the model differ again by a very small amount (0.2 °C) showing that the model is a good fit in this range.

7. Explain how the theory of the situation relates to the model. The following answer is for an exponential model. Modify it accordingly for a power model. You will mostly memorise this!

8. Explain how the theory of the situation relates to the model. Linear regression is the most common tool for finding lines of best fit for data, but cannot be used directly for data which have a non-linear relationship. By transforming non-linear relationships into linear equations, we can then compare the graphs of the transformed data to see which one is a better model. The more appropriate model is the one for which the straight line is a better fit for the transformed data.

9. Explain how the theory of the situation relates to the model. For my exponential model, the transformation is: y = 48.1859e -0.0292x taking the natural log of both sides ln (y) = ln (48.1859e -0.0292x ) ln (y) = ln (48.1859) + ln (e -0.0292x ) ln (y) = ln (48.1859) + -0.0292x This is in the form of a linear equation ln (y) = ln (a) + mx where m is the gradient and ln (a) is the y-intercept

10. Explain how the theory of the situation relates to the model. The straight line fits the log-linear transformed graph better than the log-log graph so the exponential model is more appropriate for my data. When I plot time (x) against ln (y) I get a straight line graph and the parameters of the linear regression equation can be used to estimate the original parameters of the exponential equation with m = gradient = -0.0292 and ln (a) = y-intercept so a = 48.18599

11. Discuss any limitations of the model. Must relate to the MODEL, not the experimental process or the data. Suggest that you do two limitations to ensure you have one correct as it is easy for this section to overlap with other excellence answers. Consider things such as y-intercepts, asymptotes, problems with extrapolation, areas where the model does not fit the data well. Try to relate your comments to the actual context! (For example: What does your model predict if you let this cup of coffee sit for hours? Does this make sense?)

12. Discuss any limitations of the model. Although my model is very good for predicting values within the time range of the experiment which was 20 minutes, I need to be very careful about using my model for extrapolation as no data was collected for times more than 20 minutes. My model has an asymptote on the x-axis. This implies that the temperature will keep cooling if you let the time get longer and longer. However, in actuality, the temperature of the coffee would stop changing once it reached room temperature.

13. Consider the nature of the underlying variables. You should discuss at least two of the following: The nature of the phenomenon being modelled – Investigating changes which naturally happen over time are typically exponential. Situations where you make changes to the variable and measure the response (helicopter example) are typically power.

14. Consider the nature of the underlying variables. The presence of a y-intercept – Exponential curves have a y-intercept. i.e. they have a value when x=0 – Power curves will not have a y-intercept Proportional change – For exponential models, the response variable changes by the same proportion for a given change in the explanatory variable. Work out the proportional change in your response variable for each new time. How does this match with your model? – This is not true for power curves where the proportional change varies.

15. Consider the nature of the underlying variables. Example of determining proportional change... How to calculate the proportional change Results

16. Consider the nature of the underlying variables. The phenomenon of coffee cooling over time is a natural one so we would expect that an exponential model would fit the data best as the coffee should cool by the same proportion over equal time intervals. We can check this by investigating the proportional change in the difference above room temperature and compare this to the proportional change indicated in our model.

17. Consider the nature of the underlying variables. Proportional change in the data The temperature is changing by a factor of about 0.9 every 4 minutes which suggests that an exponential model may be appropriate. time (min)048121620 Difference above room temp (°C) 48.5433833.53027.25 Proportional change0.890.88 0.900.91

18. Consider the nature of the underlying variables. The model y = 48.1859e -0.0292x indicates a proportional change of e -0.0292x = 0.97. This value is quite different from the 0.9 expected from the data. This indicates that although an exponential model seems suitable as the difference above room temperature seems to be changing by a constant proportion, perhaps the equation of the model that we have obtained is not that appropriate for our data.

19. Consider the nature of the underlying variables. Exponential models have a y-intercept. In the coffee cooling experiment, we recorded a temperature of 48.5 °C when the time was 0 seconds. The presence of a y-intercept supports the fact that an exponential model is more appropriate for this data as a power model does not have a y-intercept.

20. Consider the possibility of using an alternative model. If the model tends to underestimate the data at one end and overestimate it at the other end, you may want to explore fitting two different curves. You can also look to see if the transformed graphs (straight line graphs) seem to have two sections with different gradients. Split your data into two sets and graph each set separately using List 1 and List 2 in order to get two different models (of the same type).

21. Consider the possibility of using an alternative model. You can find the equations of these two new models and obtain the graphs on your GC. You can do a table similar to the one for “Relate the solution to the problem” using the appropriate equation for each section of the table and comparing the values you obtain from your two new models to the raw data (exactly like you did in the Relate the solution to the problem). Are your predictions better?

22. Consider the possibility of using an alternative model. Alternatively, you can sketch these graphs along with the two new curved lines and discuss how the lines fit the data points more closely and therefore are more appropriate for making predictions in each of the ranges than a single model.

23. Consider the possibility of using an alternative model. If the data does not suggest the use of two models then you should say that the use of two models is not appropriate and give reference to the data, graph(s), and analysis to explain why. You can discuss the R 2 value.

24. Consider the possibility of using an alternative model. The original graph of the difference above room temperature, as well as the calculations done previously show that my model initially slightly underestimates the temperature of the coffee, then it slightly overestimates the temperature and at 20 minutes it starts to underestimate the temperature again.

25. Consider the possibility of using an alternative model. The predictions from my model are very close to the raw data and since there is not a permanent shift of the curve from below the data points at one end to above at the other end (as it shifts back to being below the point again at t=20), it is not suitable to fit two different curves to this data.

26. Consider the possibility of using an alternative model. My exponential model has an R 2 of 0.998 which is very close to 1 which also suggests that my model is a very good fit for the data. This is supported by the log-linear graph in which the points lie very close to the line of best fit. All of the above reasons mean that my model is very appropriate for this data.

27. Graph of raw data