The shear strength of soil is the ability of the soil to resist tensional stresses and sustain loads. The water content of soil is the ratio of the weight of water to the weight of the solids in a mass of soil. A soil investigation laboratory collected soil from a reclaimed site to study the effect of water content on shear strength. The information gathered about the shear strength $P$, measured in KiloPascal (KPa), at different water content $w$, measured in percentage is presented in the table.

Water Content, $w$ (%)	10.0	12.5	15.0	17.5	20.0	22.5	25.0	27.5
Shear Strength, $P$ (KPa)	2500	2100	1820	1630	1500	1410	1320	1270

1. Draw the scatter diagram by plotting the given data points

Plot the points $(10.0, 2500)$, $(12.5, 2100)$, $(15.0, 1820)$, $(17.5, 1630)$, $(20.0, 1500)$, $(22.5, 1410)$, $(25.0, 1320)$, $(27.5, 1270)$ on a set of axes with $w$ on the horizontal axis and $P$ on the vertical axis.

2. Explain whether the relationship is well modelled by a linear equation

Observe from the scatter diagram that as $w$ increases, $P$ decreases at a decreasing rate. The points exhibit a non-linear, curvilinear pattern rather than lying close to a straight line. Therefore, it is not likely to be well modelled by $P = aw + b$.

1. Calculate the product moment correlation coefficient (PMCC) for the linear model $P = aw + b$.

Using a graphing calculator to find the PMCC $r$ for the data pairs $(w, P)$:
$r_1 = -0.94866 \text{ (to 5 s.f.)}$

2. Calculate the PMCC for the non-linear model $P = a\sqrt{w} + b$.

Using a graphing calculator to find the PMCC $r$ for the data pairs $(\sqrt{w}, P)$:
$r_2 = -0.96828 \text{ (to 5 s.f.)}$

3. Determine and explain which is the better model by comparing the correlation coefficients.

Since $|r_2| > |r_1|$ (i.e. $-0.96828$ is closer to $-1$ than $-0.94866$), there is a stronger linear correlation between $P$ and $\sqrt{w}$ than between $P$ and $w$.
Therefore, $P = a\sqrt{w} + b$ is the better model.

4. Find the values of $a$ and $b$ to state the equation for the chosen model.

From the graphing calculator, the linear regression of $P$ on $\sqrt{w}$ gives the coefficients:
$a = -569.52 \text{ (to 5 s.f.)}$
$b = 4129.4 \text{ (to 5 s.f.)}$
Rounding to 3 significant figures, we get $a = -570$ and $b = 4130$.
The equation is $P = -570\sqrt{w} + 4130$.

1. Estimate the shear strength using the regression equation from (ii)

From (ii), the appropriate equation is $P = a\sqrt{w} + b$ (since its product moment correlation coefficient $|-0.969| > |-0.952|$), with $a \approx -569.59$ and $b \approx 4129.7$.

Substitute $w = 40$ into the equation:
$P = -569.59\sqrt{40} + 4129.7$
$P \approx 527.31 \approx 527 \text{ KPa}$

2. Explain the reliability of the estimate based on the data range

The value $w = 40$ is outside the range of given water content data ($10.0 \le w \le 27.5$).
This means the prediction requires extrapolation, so the estimate is not reliable.

1. Identify that the missing data point corresponds to the sample mean of the given data.

A property of the least squares regression line is that it always passes through the sample mean point of the dataset, $(\bar{x}, \bar{y})$. Adding the mean point to the dataset will not change the means or the covariance, meaning the regression equation remains entirely unaffected. Since no specific independent variable value ($w$) is provided for the missing point, it must be the mean point, meaning its shear strength is $\bar{P}$.

2. Calculate the mean shear strength, $\bar{P}$.

$\bar{P} = \frac{2500 + 2100 + 1820 + 1630 + 1500 + 1410 + 1320 + 1270}{8}$
$\bar{P} = \frac{13550}{8} = 1693.75$
Rounding to the nearest integer, the shear strength is $1694$ KPa.

1. State the original regression equation from (ii) and convert it to MegaPascal.

From (ii), comparing the PMCC of $P$ on $w$ ($r \approx -0.949$) and $P$ on $\sqrt{w}$ ($r \approx -0.968$), the better model is $P = a\sqrt{w} + b$.
Calculating the regression coefficients gives the equation $P = -569.63\sqrt{w} + 4129.9$.

Let $P_{\text{MPa}}$ be the shear strength measured in MegaPascal. Since $1 \text{ MPa} = 1000 \text{ KPa}$, we divide the $P$ values by 1000:
$P_{\text{MPa}} = \frac{P}{1000}$
$P_{\text{MPa}} = \frac{-569.63\sqrt{w} + 4129.9}{1000}$
$P_{\text{MPa}} = -0.570\sqrt{w} + 4.13 \quad \text{(to 3 s.f.)}$

2. Explain the effect of the unit conversion on the product moment correlation coefficient.

The product moment correlation coefficient (PMCC) would not differ.

Converting the units by dividing the shear strength by 1000 is a linear transformation (scaling by a positive constant). The PMCC is invariant under linear transformations with a positive scale factor because it purely measures the strength of the linear correlation between the two variables, completely independent of the units used.