一家能源公司正在研究日平均气温 $x$（摄氏度）与用电量 $y$（兆瓦时 MWh）之间的关系。随机抽取十天的记录如下表所示。

日平均气温（ $x^\circ$C ）	20	22	24	26	28	30	32	34	36	38
用电量（ $y$ MWh）	12.5	12.8	13.2	14.0	14.7	16.2	16.2	19.9	22.3	29.5

人们认为，用电量和日平均气温可以用以下公式之一进行建模： $y = a + b x^2, \qquad y = c + d x,$ 其中 $a$、$b$、$c$ 和 $d$ 是常数。 在某些地区，温度是用华氏度而不是摄氏度来衡量的。摄氏度温度 $C$ 相当于华氏度温度 $F$，其中 $F = \frac{9}{5}C + 32.$

1. Set up and label the axes appropriately

Draw a horizontal axis for $x$ (average daily temperature, in $^\circ$C) covering values from at least 20 to 38, and a vertical axis for $y$ (electricity consumption, in MWh) covering values from at least 12.5 to 30. Label the axes as '$x$ ($^\circ$C)' and '$y$ (MWh)'.

2. Plot the points from the given table

Plot the 10 coordinate points on the grid:
$(20, 12.5)$, $(22, 12.8)$, $(24, 13.2)$, $(26, 14.0)$, $(28, 14.7)$,
$(30, 16.2)$, $(32, 16.2)$, $(34, 19.9)$, $(36, 22.3)$, $(38, 29.5)$.

1. Use a graphing calculator to find the product moment correlation coefficient (PMCC) for the paired data.

By entering the $x$ and $y$ values into a graphing calculator, we obtain:
$r = 0.890557\dots$
$r \approx 0.891 \text{ (to 3 s.f.)}$

1. Calculate the product moment correlation coefficient for $x^2$ and $y$ using a graphing calculator.

Enter the paired data for $x^2$ and $y$ into the graphing calculator:
$X = x^2 = \{400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444\}$
$Y = y = \{12.5, 12.8, 13.2, 14.0, 14.7, 16.2, 16.2, 19.9, 22.3, 29.5\}$

From the calculator,
$r = 0.922289\dots$
$r \approx 0.922 \text{ (to 3 s.f.)}$

1. Explain which model is better based on the scatter diagram from part (a) and PMCC from part (b)

From part (b), we have the calculated PMCC values:

• For $x$ and $y$: $r \approx 0.891$
• For $x^2$ and $y$: $r \approx 0.922$

1. From part (a): The scatter diagram of $y$ against $x$ exhibits an upward curve, which suggests a non-linear relationship.
2. From part (b): The product moment correlation coefficient for $x^2$ and $y$ ($0.922$) is closer to $1$ than that for $x$ and $y$ ($0.891$), showing a stronger positive linear correlation.

Therefore, $y = a + bx^2$ is the better model.

2. Find the equation of the suitable regression line

Since $y = a + bx^2$ is the chosen model, we find the linear regression line of $y$ on $x^2$.
Using a graphing calculator, we input the squares of the temperature values ($x^2$) as the independent variable and the electricity consumption ($y$) as the dependent variable to find the least squares line $y = a + bx^2$.

From the calculator:
$a \approx 4.8291$
$b \approx 0.014074$

The equation of the regression line is:
$y = 4.83 + 0.0141x^2 \quad \text{(to 3 s.f.)}$

1. Estimate the electricity consumption when the average temperature is $37^\circ\text{C}$.

From part (c), the equation of the regression line of $y$ on $x^2$ is $y = 0.01407 x^2 + 4.830$.

When $x = 37$:
$y = 0.01407(37)^2 + 4.830$
$y = 0.01407(1369) + 4.830$
$y = 24.09183 \approx 24.1\text{ MWh}$

2. Comment on the reliability of the estimate.

Since $x = 37^\circ\text{C}$ lies within the given data range of $20^\circ\text{C}$ to $38^\circ\text{C}$, the process is interpolation. Therefore, the estimate is reliable.

1. Determine the regression equation for $y$ on $x^2$ (required from part c) and express the Celsius temperature $x$ in terms of the Fahrenheit temperature $F$.

To formulate the regression line $y = a + bx^2$ from part (c), we use the data points with $w = x^2$:
$\bar{w} = 874$, $\bar{y} = 17.13$
$b = \frac{S_{wy}}{S_{ww}} = \frac{15743}{1118568} \approx 0.014074$
$a = \bar{y} - b\bar{w} = 17.13 - (0.014074)(874) \approx 4.8291$
The regression equation is $y = 4.829 + 0.01407x^2$.

Since $x$ represents the temperature in Celsius ($C = x$), the conversion formula is $F = \frac{9}{5}x + 32$. Making $x$ the subject:
$F - 32 = \frac{9}{5}x$
$x = \frac{5}{9}(F - 32)$

2. Substitute the expression for $x$ into the regression equation to express $y$ in terms of $F$.

Substitute $x = \frac{5}{9}(F - 32)$ into the regression equation:
$y = 4.8291 + 0.014074 \left( \frac{5}{9}(F - 32) \right)^2$
$y = 4.8291 + 0.014074 \left( \frac{25}{81} \right) (F - 32)^2$
$y = 4.8291 + 0.0043439 (F - 32)^2$
Rounding to 3 significant figures, we get:
$y = 4.83 + 0.00434(F - 32)^2$