IB Maths AI HL · Paper 3

Only 11 of 415 doctors replied. Following up the non-responders, or repeating the study with a much larger sample, would make the findings more reliable.

The people who chose to reply may differ from those who didn't (self-selection), and 11 of 415 is a very low response rate. So the sample may not represent all doctors, which threatens validity.

An income of $123 456 789 is not realistic for a doctor, so it is probably an error. Another reason: it is extremely far from the other incomes and would dominate the correlation.

Add the ten incomes and divide by ten.

Enter the ten pairs into the calculator's regression.

She is investigating whether higher income goes with *greater* happiness, which is a statement about the direction of the correlation (positive).

Let $\rho$ be the population correlation coefficient.

Linear regression of happiness on income.

The gradient is the average increase in happiness score for each extra dollar of annual income: about 1.26 points for every extra $10 000.

Quadratic regression on the calculator.

The coefficient of determination for each model:

The quadratic model has the larger $R^2$, so it explains more of the variation in happiness (65.9% compared with 31.1%) and fits this data better.

This is not a valid way to choose. A quadratic model has an extra parameter, so it will always fit the data at least as well as the linear model. $R^2$ cannot tell you which model is better for predicting.

The population variance is unknown and the sample is small, so a one-sample $t$-test is used.

Let $\mu$ be the mean annual income of doctors in the city.

$\gamma MS$ is the growth rate of the shark population due to eating mackerel. Sharks reproduce or survive better in proportion to the number of encounters with mackerel.

$\delta S$ is the natural death rate of the sharks (when there is no mackerel to eat), which is proportional to the number of sharks.

At equilibrium $\frac{\mathrm{d}S}{\mathrm{d}t}=S(\gamma M-\delta )=0$. Since sharks are present, $S\ne 0$, so the bracket must be zero.

At equilibrium $\frac{\mathrm{d}M}{\mathrm{d}t}=M(\alpha -\beta S)=0$. Since mackerel are present, $M\ne 0$.

The average mackerel density is $\frac{\delta }{\gamma }$. Halving $\gamma$ doubles it, so the average mackerel population density doubles.

$\alpha$ does not appear in $\frac{\delta }{\gamma }$, so doubling $\alpha$ does not change the average mackerel density. It would double the average shark density $\frac{\alpha }{\beta }$.

With no sharks, $S=0$ and the mackerel grow at a rate proportional to their number.

The density triples in two years, so $M(2)=3M_0$.

Euler's method with $h=0.1$ and the given parameters.

Start from $M_0=5.7$, $S_0=2$ and take 10 steps (one year).

Plot the pairs $(M_n,S_n)$ for $n=0,\dots ,90$ with $M$ across and $S$ up. The path goes round in a loop that is not quite closed: it starts at $(5.7,2)$ and winds round the equilibrium point $\left(\frac{1.39}{0.244},\frac{0.549}{0.236}\right)\approx (5.7,2.33)$, making about one and a quarter turns in nine years.

From the table of values the smallest mackerel density is $M\approx 5.07$, that is 5070 per km³, which is above 5000. So the density does not fall below 5000 and sustainable fishing is supported.

The parameters are only estimates from observations. Euler's method with a step of 0.1 is only an approximation. The model also ignores other factors such as fishing, other predators, weather and migration.