In an experiment involving two chemicals $x$ and $y$, a researcher recorded observations of values of $y$ for controlled values of $x$ as shown in the table

1. Draw the scatter diagram to illustrate the data

Set up a coordinate grid with $x$ on the horizontal axis and $y$ on the vertical axis. Scale the axes appropriately to encompass $4.4 \le x \le 12.3$ and $2.7 \le y \le 6.3$. Plot all 9 bivariate data points: $(4.4, 3.2), (5.1, 2.7), (6.3, 6.3), (7.5, 5.6), (8.1, 5.3), (9.4, 4.8), (10.6, 4.6), (11.7, 4.5),$ and $(12.3, 4.5)$.

2. Calculate the product moment correlation coefficient

Enter the 9 pairs of data into the statistics mode of a graphing calculator and compute the product moment correlation coefficient, $r$.
$r = 0.263346...$
$r \approx 0.263 \text{ (to 3 s.f.)}$

1. Calculate the product moment correlation coefficient $r$ for the 9 pairs of data (dependent value from part (i)).

$n = 9, \sum x = 75.4, \sum y = 41.5, \sum x^2 = 695.82, \sum y^2 = 201.37, \sum xy = 354.35$

$r = \frac{354.35 - \frac{(75.4)(41.5)}{9}}{\sqrt{\left(695.82 - \frac{75.4^2}{9}\right)\left(201.37 - \frac{41.5^2}{9}\right)}} \approx 0.263$

2. Explain the unsuitability of a linear model by referencing both the scatter diagram and the correlation coefficient.

The value of the product moment correlation coefficient, $r \approx 0.263$, is close to $0$, which indicates a very weak linear correlation between $x$ and $y$.
Furthermore, the points on the scatter diagram do not follow a straight line trend (they appear to rise initially and then fall).
Therefore, a linear model is not appropriate.

1. Filter the data to include only observations where $x > 6$ and calculate the PMCC for model (a).

Retain only the 7 pairs where $x > 6$:
$x$: $6.3, 7.5, 8.1, 9.4, 10.6, 11.7, 12.3$
$y$: $6.3, 5.6, 5.3, 4.8, 4.6, 4.5, 4.5$

For model (a) $y = ax^2 + b$, use a graphing calculator to find the correlation coefficient $r_a$ between $x^2$ and $y$:
$r_a = -0.912911\dots \approx -0.913$

2. Calculate the PMCC for model (b).

For model (b) $y = a\ln x + b$, use a graphing calculator to find the correlation coefficient $r_b$ between $\ln x$ and $y$:
$r_b = -0.966993\dots \approx -0.967$

3. Compare the correlation coefficients to determine the better model.

Since $|r_b| > |r_a|$ (i.e. $|-0.967|$ is closer to $1$ than $|-0.913|$), the linear correlation between $\ln x$ and $y$ is stronger than that between $x^2$ and $y$.
Therefore, model (b) $y = a \ln x + b$ is the better model.

1. Find the equation of the appropriate regression line

From the previous part, the better model is $y = a \ln x + b$.
Since $x$ is the controlled (independent) variable, we use the regression line of $y$ on $\ln x$.
Using the 7 pairs of data where $x > 6$:
$x$: $6.3, 12.3, 8.1, 7.5, 10.6, 11.7, 9.4$
$y$: $6.3, 4.5, 5.3, 5.6, 4.6, 4.5, 4.8$
From the graphing calculator, the equation is:
$y = 10.2087 - 2.3109 \ln x$

2. Substitute the given value to estimate $x$

Substitute $y = 6.1$ into the regression line equation:
$6.1 = 10.2087 - 2.3109 \ln x$
$2.3109 \ln x = 4.1087$
$\ln x \approx 1.77796$
$x = e^{1.77796} \approx 5.9178$
$x \approx 5.92 \text{ (to 3 s.f.)}$

3. Comment on the validity of the estimated value

The estimated value of $x = 5.92$ is outside the range of data used to construct the model ($6.3 \le x \le 12.3$), which makes it an extrapolation.
Furthermore, it was stated that observations where $x \le 6$ were discarded due to contaminated materials. Since $5.92 \le 6$, the value falls into this contaminated region.
Therefore, the estimate is not valid.