Correlation and Regression Lines

Part (a): $r = 0.88$ represents a strong positive linear correlation ($0.88 \ge 0.8$).

Part (b): $r = -0.42$ represents a weak negative linear correlation ($|-0.42| < 0.5$).

Part (c): $r = -0.96$ represents a strong negative linear correlation ($|-0.96| \ge 0.8$).

Final Answer: (a) Strong positive; (b) Weak negative; (c) Strong negative.

Part (a): Substitute $x = 4$:

Part (b): Since $4\text{ km}$ lies within the observed domain $[1, 10]\text{ km}$, this is an interpolation and is therefore considered reliable.

Final Answer: (a) 3.4 visits; (b) Interpolation, reliable.

Step 1: The regression line has form $y = ax + b$ with $a = 2.5$.

Step 2: Use the property that the line passes through $(\bar{x}, \bar{y})$:

Final Answer: $y = 2.5x + 13$

Interpretation: For every additional 1 kilometer increase in distance from home to the gym, the predicted number of weekly visits decreases by an average of $0.45$.

Final Answer: An average decrease of 0.45 visits per additional kilometer.

Part (a): $f = 8 \implies h = 3.8(8) + 14.2 = 30.4 + 14.2 = 44.6\text{ cm}$.

Part (b): $f = 35 \implies h = 3.8(35) + 14.2 = 133 + 14.2 = 147.2\text{ cm}$.

Part (c): Reliability comparison:
• The estimate in (a) is an interpolation because $f = 8$ lies inside the experimental range $[2, 12]$. Given $r = 0.94$ (strong positive), this estimate is highly reliable.
• The estimate in (b) is an extrapolation because $f = 35$ lies far beyond the experimental range. In biological systems, excessive fertilizer may cause toxicity or burns rather than continuous linear growth. Thus, (b) is completely unreliable.

Final Answer: (a) $44.6\text{ cm}$; (b) $147.2\text{ cm}$; (c) (a) is reliable (interpolation with high $r$); (b) is unreliable (far extrapolation).

Part (a): Flawed causal claim:
Correlation does not imply causation. A high positive correlation between two variables only measures association and does not demonstrate that coffee consumption directly stimulates book reading.

Part (b): Lurking variable:
City population size: Larger cities have more residents, and therefore naturally have both more coffee shops and more library visitors/book loans. Total population is the confounding common cause.

Final Answer: (a) Correlation does not imply causation; (b) City population size.

Part (a): Find $\sum y$:
Mean $\bar{x} = \frac{\sum x}{n} = \frac{50}{10} = 5$.
Since $(\bar{x}, \bar{y})$ lies on the regression line:
$$\bar{y} = 1.6(5) - 4 = 8 - 4 = 4$$ $$\sum y = n\bar{y} = 10 \times 4 = 40$$

Part (b): Why rearranging is invalid:
The regression line of $y$ on $x$ was derived by minimizing the sum of squared vertical residuals $\sum(y_i - \hat{y}_i)^2$, treating $x$ as an exact known input. To predict $x$ from $y$, one must minimize the sum of squared horizontal residuals $\sum(x_i - \hat{x}_i)^2$, which produces a completely different line (the regression line of $x$ on $y$). Rearranging $y$ on $x$ does not yield the least-squares predictor for $x$.

Final Answer: (a) $\sum y = 40$; (b) Rearranging is invalid because regression of $y$ on $x$ minimizes vertical residuals, not horizontal residuals.

Part (a): Invariance under positive linear transformations:
Pearson's $r$ is scale- and translation-invariant under positive multipliers ($a > 0$ and $c > 0$):
Here $u = 1x + 10$ (shift by $+10$, scale by $+1$) and $v = 3y + 0$ (scale by $+3$). Both scale factors are positive.
Therefore, $r_{uv} = r_{xy} = 0.72$.

Part (b): Effect of negative scale factor:
If $v = -3y$, the scale factor for $y$ is negative ($-3 < 0$), which reflects the association across the vertical axis.
Therefore, the correlation reverses direction: $r_{uv} = -r_{xy} = -0.72$.

Final Answer: (a) $r_{uv} = 0.72$; (b) $r_{uv} = -0.72$.

Part (a): Means:
$$\sum a = 1.0 + 2.5 + 4.0 + 6.0 + 8.5 + 10.0 = 32.0 \implies \bar{a} = \frac{32.0}{6} = 5.333\text{ km}$$ $$\sum T = 8.0 - 1.5 - 11.0 - 24.0 - 40.5 - 50.0 = -119.0 \implies \bar{T} = \frac{-119.0}{6} \approx -19.833^\circ\text{C}$$

Part (b): Regression equation:
Gradient $m = -6.45$. Line passes through $(\bar{a}, \bar{T})$: $$\bar{T} = m\bar{a} + c \implies -19.833 = -6.45\left(\frac{32}{6}\right) + c$$ $$-19.833 = -34.40 + c \implies c = 34.40 - 19.833 = 14.567 \approx 14.6$$ $$T = -6.45a + 14.6$$

Part (c): Prediction at $a = 5.0\text{ km}$:
$$T = -6.45(5.0) + 14.57 = -32.25 + 14.57 = -17.68 \approx -17.7^\circ\text{C}$$ Valid because $5.0\text{ km}$ is within the observed interval $[1.0, 10.0]$ (interpolation) where atmospheric temperature decreases linearly with altitude (lapse rate).

Part (d): Predicting altitude from $T = 0^\circ\text{C}$:
If using the rearranged formula as an approximation: $$0 = -6.45a + 14.57 \implies a = \frac{14.57}{6.45} \approx 2.26\text{ km}$$ Limitation: This uses the regression line of $T$ on $a$ in reverse; strictly, the regression line of $a$ on $T$ should be used to predict altitude from a given temperature.

Final Answer: (a) $\bar{a} = 5.33\text{ km}, \bar{T} = -19.8^\circ\text{C}$; (b) $T = -6.45a + 14.6$; (c) $-17.7^\circ\text{C}$ (valid interpolation); (d) $\approx 2.26\text{ km}$ (limited by reverse regression).

Part (a): Logarithmic linearization:
Take $\ln$ of both sides: $$\ln P = \ln(A b^t) = \ln A + \ln(b^t) = \ln A + t\ln b$$ Let $Y = \ln P$, $m = \ln b$, and $c = \ln A$: $$Y = mt + c$$

Part (b): Find $A$ and $b$:
Comparing with $\ln P = 0.42t + 3.50$:
$c = \ln A = 3.50 \implies A = e^{3.50} \approx 33.115 \approx 33.1$.
$m = \ln b = 0.42 \implies b = e^{0.42} \approx 1.52196 \approx 1.52$.

Part (c): Estimate at $t = 5$:
Method 1: $\ln P = 0.42(5) + 3.50 = 2.10 + 3.50 = 5.60 \implies P = e^{5.60} \approx 270.4 \approx 270$.
Method 2: $P = 33.115 \times (1.522)^5 \approx 270$.

Final Answer: (a) $Y = \ln P, m = \ln b, c = \ln A$; (b) $A \approx 33.1, b \approx 1.52$; (c) $P \approx 270$ microbes.

(a) Correlation description:
Since $r = -0.89$, there is a strong negative linear correlation between engine capacity and fuel efficiency. [A1 A1]

(b) Prediction:
Substitute $C = 1.8$:
$F = -4.2(1.8) + 24.8 = -7.56 + 24.8 = 17.24 \approx 17.2\text{ km/L}$. [M1 A1]

(c) Validity critique:
• $C = 6.0$ is an extrapolation far beyond the observed range $[1.2, 2.5]$. [R1]
• A negative fuel efficiency ($F < 0$) is physically impossible in the real world. [R1]

(a) Mean point coordinates:
$\bar{x} = \frac{72}{12} = 6$. [A1]
$\bar{y} = \frac{144}{12} = 12$. [A1]
Mean point is $(6, 12)$.

(b) Regression equation:
Line passes through $(6, 12)$ with gradient $a = 1.35$:
$12 = 1.35(6) + b \implies 12 = 8.1 + b \implies b = 3.9$. [M1]
Equation: $y = 1.35x + 3.9$. [A1]

(c) Gradient interpretation:
For each additional $\$100$ spent on advertising, weekly sales revenue is predicted to increase by an average of $\$1350$ ($1.35$ thousand dollars). [A1]

(d) Causation vs correlation:
Correlation measures linear association, not causation. [R1]
External factors such as seasonal market demand, promotional campaigns, or competitor actions could simultaneously drive higher advertising spend and higher sales. [R1]