Bivariate Data Analysis Topic Test

(a) Sessions attended goes on the x-axis. It is the independent (explanatory) variable, the number of sessions is the predictor, and the pain score is expected to respond to it.

(b) The point $(6, 2)$ represents a patient who attended $6$ rehabilitation sessions and reported a pain score of $2$ out of $10$.

(c) Points falling from upper-left to lower-right = negative direction; tightly grouped around a line = strong. So there is a strong negative linear correlation between sessions attended and pain score: as sessions increase, pain scores tend to decrease.

(a) The sign of $r$ is negative, so the direction is negative: as exercise increases, resting heart rate tends to decrease.

(b) $|r| = 0.82 \ge 0.8$, so the correlation is strong. Overall: a strong negative linear correlation between daily exercise and resting heart rate.

(c) Correlation is not causation. A strong $r$ shows only an association. A confounding variable, for example overall fitness or age, could influence both. Establishing causation would require a controlled experiment.

(a) $m = \dfrac{55 - 35}{6 - 2} = \dfrac{20}{4} = 5$.

(b) Using $(2, 35)$: $35 = 5(2) + b \Rightarrow b = 35 - 10 = 25$. Equation: $y = 5x + 25$.

(c) $\$8000$ means $x = 8$: $y = 5(8) + 25 = 40 + 25 = 65$, so predicted sales are $\$65\,000$.

(a) $a = 38$ is the y-intercept: an employee with $0$ years of experience is predicted to earn $\$38\,000$ per year (a plausible starting salary, since $x = 0$ is realistic here).

(b) $b = 2.8$ is the gradient: for each additional year of experience, the predicted annual salary increases by $2.8$ thousand dollars, that is $\$2800$.

(c) $y = 38 + 2.8(10) = 38 + 28 = 66$, so the predicted salary is $\$66\,000$.

(a) $y = 4 + 0.3(40) = 4 + 12 = 16$ days. Since $40$ is within the data range $25$ to $65$, this is interpolation, so the prediction is reliable.

(b) $y = 4 + 0.3(80) = 4 + 24 = 28$ days. Since $80 > 65$ is outside the data range, this is extrapolation, so the prediction is less reliable.

(c) The regression line is fitted only to data within the observed range. Outside that range the linear pattern may not continue, it could level off, reverse, or be affected by other factors, so extrapolated predictions are less trustworthy.

(a) $r = 0.86$ indicates a strong positive linear correlation: towns with more ice-cream vans tend to have higher sunburn rates.

(b) Correlation does not prove causation. A strong $r$ shows only an association. Ice-cream vans do not physically cause sunburn, so a third factor is more likely to explain the pattern.

(c) The confounding variable is hot, sunny weather. Sunny days bring out more ice-cream vans and lead to more time in the sun (hence more sunburn). Both variables are driven by weather, producing a spurious correlation.

(a) $\bar{x} = \dfrac{6 + 10 + 12 + 14 + 18}{5} = \dfrac{60}{5} = 12$. $\bar{y} = \dfrac{92 + 80 + 74 + 66 + 48}{5} = \dfrac{360}{5} = 72$. Mean point $(12, 72)$.

(b) Gradient $b = -3.65$: for each $1$°C increase in maximum temperature, predicted hot-chocolate sales decrease by $3.65$ units. Because the gradient is negative, Pearson's $r$ must also be negative (gradient and $r$ always share the same sign). Check the line against part (a): at $x = 12$ it gives $115.8 - 3.65(12) = 115.8 - 43.8 = 72 = \bar{y}$, so it passes through the mean point $(12, 72)$ as every least-squares line must.

(c) $y = 115.8 - 3.65(11) = 115.8 - 40.15 = 75.65$, so about $76$ sales. Since $11$°C is within the data range $6$ to $18$°C, this is interpolation, so the prediction is reliable.