A group of climbers collected information about the height above sea level, $h$ metres, and the air temperature, $t ext{ }^ ext{C}$, at the same time at 8 different points on the same mountain - Edexcel - A-Level Maths Statistics - Question 6 - 2018 - Paper 1

The product moment correlation coefficient is -0.985, which is very close to -1. This indicates a strong negative correlation between height and temperature, suggesting that as height increases, the temperature decreases. Therefore, it does support the use of a regression equation to predict air temperature at different heights.

The equation of the regression line is given by:

$t = a + bh$

To find $a$ and $b$, we use the formulas:

1. The slope $b$ is calculated using:

$b = -\frac{S_{ht}}{S_h}$

Given $S_{ht} = -0.985 \cdot \sqrt{S_h \cdot S_t}$: $b = -\frac{-0.985 \cdot \sqrt{-17501.25 \cdot 227.875}}{-17501.25}$

Calculating gives: $b \approx 0.0177$

2. The intercept $a$ is calculated as follows:

$a = \bar{t} - b\bar{h}$

Calculating $\bar{t} = \frac{61}{8} = 7.625$ and $\bar{h} = \frac{6370}{8} = 796.25$:

$a = 7.625 - (0.0177 \cdot 796.25) \approx 1.390$

Thus, the regression equation is: $t = 1.390 + 0.0177h$

$a = 1.390$ can be interpreted as the estimate of the air temperature at sea level (i.e., when height $h = 0$). Therefore, at sea level, the expected air temperature is approximately 1.39 °C.

To estimate the drop in temperature over a 150 metre climb, we apply the equation of the regression line:

For a height increase of 150 metres, we can substitute:

$t = 1.390 + 0.0177(150)$

Calculation: $t \approx 1.390 + 2.655 = 4.045 ext{ °C}$

To determine the drop in temperature, we can find: $\Delta t = \text{Initial Temperature} - \text{Final Temperature} ext{ (using height increase of 150m)}$

Given the values: $\Delta t \approx 4.045 - 1.390 = 2.655 ext{ °C}$

Thus, the estimated drop in temperature over this climb is approximately 2.655 °C.