Regression Modeling - IB Mathematics Applications and Interpretation

1. To create a regression model for this prediction, how should the variables be assigned in a GDC?
   Answer: Independent variable (x): hours of training; Dependent variable (y): sprint time. In regression analysis, the independent variable (x) is the one that is used to make a prediction, and the dependent variable (y) is the one being predicted. The scientist wants to predict the sprint time (the outcome) based on the hours of training (the input). Therefore, hours of training should be the independent variable (x) and sprint time should be the dependent variable (y).
2. The researcher fits a linear regression model to this data. Using this model, what is the predicted resting heart rate for a person who exercises 4 hours per week?
   Answer: 65.9 bpm First, use a graphic display calculator (GDC) to find the linear regression equation for the given data points. The independent variable $x$ is hours of exercise, and the dependent variable $y$ is resting heart rate. The equation is approximately $y = -2.98x + 77.8$. To predict the heart rate for someone exercising 4 hours per week, substitute $x=4$ into the equation: $y = -2.98(4) + 77.8 = -11.92 + 77.8 = 65.88$. This rounds to 65.9 bpm.
3. What is the correct interpretation of the slope of this regression line?
   Answer: For each additional square metre of area, the price of the house is predicted to increase by $250. In a linear model $y = mx + c$, the slope $m$ represents the change in the dependent variable ($y$) for a one-unit change in the independent variable ($x$). Here, $P$ is in thousands of dollars and $A$ is in square metres. The slope is 0.25. This means for each 1 square metre increase in area, the price increases by 0.25 thousands of dollars, which is $0.25 \times 1000 = \$250$.
4. How will this new data point most likely affect the slope of the regression line and the value of the correlation coefficient, $r$?
   Answer: The slope will increase and $r$ will decrease. The new point is an outlier with a large x-value and a y-value that is much higher than the trend predicts. This point will act as a lever, pulling the regression line upwards on the right side, which increases the slope. However, because this point deviates significantly from the linear pattern of the other points (it lies 'well above' the line), it increases the overall scatter around the new line. This increased scatter will weaken the linear relationship, causing the correlation coefficient, $r$, to decrease (move closer to 0).
5. Which statement correctly interprets the parameter 1.09 in the model?
   Answer: The value of the cryptocurrency increases by 9% each month. In an exponential model $y = a \cdot b^x$, the base $b$ represents the growth factor. If $b = 1+r$, where $r$ is the growth rate, then the percentage growth is r \times 100\%. Here, \(b = 1.09, so $1+r = 1.09$, which means $r = 0.09$. This corresponds to a 9% increase each month.
6. Estimate the number of months it will take for the number of active users to first exceed 20,000.
   Answer: 15 months We need to solve for $t$ in the inequality $1500(1.2)^t > 20000$. First, divide by 1500: $(1.2)^t > 20000 / 1500 \Rightarrow (1.2)^t > 13.33...$. We can solve this using logarithms or a GDC solver. Using logarithms: $t \cdot \ln(1.2) > \ln(13.33...)$ \Rightarrow $t > \ln(13.33...) / \ln(1.2) \Rightarrow t > 2.59 / 0.182 \Rightarrow t > 14.2$. Since $t$ must be greater than 14.2 months, the number of users will first exceed 20,000 after 14.2 months, which is during the 15th month.
7. What is the correct interpretation of this $r^2$ value?
   Answer: 64% of the variation in monthly sales can be explained by the variation in the product's price. When you encounter questions about the coefficient of determination ($r^2$) in regression analysis, focus on what this statistic actually measures: the proportion of variance in the dependent variable that's explained by the independent variable. The coefficient of determination $r^2 = 0.64$ tells us that 64% of the variation in monthly sales can be explained by the variation in price. This means that when sales figures fluctuate from month to month, 64% of those fluctuations can be attributed to price changes, while the remaining 36% is due to other factors not included in the model. Let's examine why the other options are incorrect. Option A confuses $r^2$ with probability - the coefficient of determination doesn't give you the probability of a causal relationship occurring. Option B misinterprets $r^2$ as prediction accuracy. While a higher $r^2$ generally indicates better model fit, it doesn't mean the model will be "correct" 64% of the time. Option C makes two errors: it assumes $r = 0.64$ when actually $r = \pm\sqrt{0.64} = \pm 0.8$, and it assumes the relationship is positive when $r^2$ doesn't indicate direction. Remember that $r^2$ always represents "explained variation" - think of it as answering "What percentage of the ups and downs in my outcome variable can be attributed to changes in my predictor variable?" This interpretation applies whether you're looking at sales and price, test scores and study time, or any other regression relationship.
8. The company fits a quadratic regression model to the data. What is the predicted advertising spend that would maximize the weekly profit?
   Answer: €7,700 Using a GDC, a quadratic regression model is fitted to the data points, giving an equation of the form $P = as^2 + bs + c$. The approximate equation is $P = -1.90s^2 + 29.3s - 28.5$. The maximum profit occurs at the vertex of the parabola. The s-coordinate of the vertex is given by $s = -b / (2a)$. $s = -29.3 / (2 \times -1.90) = -29.3 / -3.8 \approx 7.71$. Since $s$ is in thousands of euros, the advertising spend is approximately €7,710. The closest answer is €7,700.
9. The results are: Linear Model $r^2=0.78$, Quadratic Model $r^2=0.91$, and Exponential Model $r^2=0.85$. Based on these values, which model provides the best fit for the data?
   Answer: The quadratic model, because it has the highest $r^2$ value. The coefficient of determination, $r^2$, measures the proportion of the variance in the dependent variable that is predictable from the independent variable(s). A higher $r^2$ value indicates a better fit of the model to the data. Since the quadratic model has the highest $r^2$ value (0.91), it provides the best statistical fit among the three options.
10. According to this model, what is the optimal number of hours of sleep to maximize the test score?
    Answer: 8 hours When you encounter a quadratic function modeling a real-world situation and need to find the maximum value, you're looking for the vertex of the parabola. Since the coefficient of $h^2$ is negative (-1.5), this parabola opens downward, confirming it has a maximum point. For a quadratic function in the form $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex (which gives the maximum or minimum) is found using $x = -\frac{b}{2a}$. In this function $S(h) = -1.5h^2 + 24h + 12$, we have $a = -1.5$ and $b = 24$. Therefore: $h = -\frac{24}{2(-1.5)} = -\frac{24}{-3} = 8$ The optimal number of hours of sleep is 8 hours, making D correct. Looking at the wrong answers: A) 6 hours might tempt students who estimate or use trial-and-error rather than finding the exact vertex. C) 12 hours is the constant term from the function, which has no relationship to the optimal input value. B) 108 hours is completely unrealistic (there are only 24 hours in a day) and likely represents a calculation error, possibly confusing the maximum output value with the optimal input. Remember that for optimization problems involving quadratic functions, always find the vertex using the formula $x = -\frac{b}{2a}$. Don't rely on plugging in answer choices or visual estimation—the algebra gives you the precise answer every time.
11. Using an exponential regression model of the form $N(t) = a \cdot b^t$, predict the number of transistors in the year 2022 ($t=12$).
    Answer: 125,500 million First, enter the data into a GDC to find the exponential regression model. The model is approximately $N(t) = 1195 \cdot (1.46)^t$. To predict the number of transistors in the year 2022, we need to evaluate the model at $t=12$. $N(12) = 1195 \cdot (1.46)^{12} \approx 1195 \cdot 105.02 \approx 125500$. So, the predicted number is 125,500 million.
12. Calculate the residual for this data point.
    Answer: -150 The residual is the difference between the actual value and the predicted value (Residual = Actual - Predicted). First, calculate the predicted sales revenue for an advertising spend of $x=500$: $y_{predicted} = 3.5(500) + 1200 = 1750 + 1200 = 2950$. The actual sales revenue was $y_{actual} = 2800$. Therefore, the residual is $2800 - 2950 = -150$.
13. According to this model, in which year is the population predicted to reach 34,000?
    Answer: 2030 We need to find the value of $t$ when $P = 34000$. We set up the equation: $34000 = 450t + 25000$. Subtract 25000 from both sides: $9000 = 450t$. Divide by 450: $t = 9000 / 450 = 20$. This means 20 years after 2010. The year is $2010 + 20 = 2030$.
14. Using an exponential regression model, at the end of which year will the car's value first be less than $15,000?
    Answer: Year 6 Using the data points including the initial value (0, 40000), the exponential regression model is approximately $V(t) = 40000(0.85)^t$. We need to find when $V(t) < 15000$. Calculating: At $t = 5$: $V(5) = 40000(0.85)^5 ≈ 17,804$. At $t = 6$: $V(6) = 40000(0.85)^6 ≈ 15,134$. At $t = 7$: $V(7) = 40000(0.85)^7 ≈ 12,864$. The value first drops below $15,000 at the end of year 6.
15. The biologist uses the model to predict the plant's height after 40 days. Which statement best describes this prediction?
    Answer: The prediction is an extrapolation and is likely to be unreliable as the plant's growth may slow down. The model was created using data for days 3 to 20. Predicting for day 40 is using an input value far outside the original data range. This is called extrapolation. Extrapolation is often unreliable because the trend observed in the data range may not continue. In this context, a plant's growth typically slows down and stops, so a linear model would likely overestimate the height significantly after a long period.
16. Which type of regression model would be most appropriate to represent the medication's concentration over time?
    Answer: A quadratic model, as it can represent a quantity that increases to a maximum and then decreases. The description of the medication's concentration—increasing to a peak and then decreasing—describes the shape of a parabola that opens downwards. A quadratic function is the most suitable model for this type of behavior. A linear model would not show a peak, and an exponential model would only show continuous growth or decay, not both.

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