一位研究人员正在调查学生在考试前一周的每日平均睡眠时间与他们的考试成绩之间的关系。下表给出了 9 名学生的每日平均睡眠时间 $t$ 小时以及他们的考试成绩 $x$（满分 100 分）。

$t$	5.5	5.8	6.1	6.5	7.2	7.4	8.0	8.2	9.1
$x$	39	43	52	60	66	68	77	78	80

1. State whether the PMCC will differ and explain why based on the properties of PMCC.

The PMCC will not differ. Converting hours to minutes involves multiplying the $t$ values by a positive constant ($60$), which is a linear transformation. The product moment correlation coefficient is independent of units of measurement and remains unchanged under such linear transformations.

1. Tabulate the transformed values $\ln(100 - x)$ and find the product moment correlation coefficient $r$ using a graphing calculator

By keying the values of $t$ and $\ln(100 - x)$ into a graphing calculator:
$r = -0.95904 \dots$
$r \approx -0.959$ (to 3 s.f.)

2. State the conclusion about the relationship with justification

Since $r \approx -0.959$, which is close to $-1$, there is a strong negative linear relationship between $t$ and $\ln(100 - x)$.

1. Calculate the values of $\ln(100 - x)$ for each corresponding $t$.

2. Plot the points on a Cartesian plane and draw the line of best fit by eye.

• Draw a pair of axes, labelling the horizontal axis as $t$ and the vertical axis as $\ln(100 - x)$.
• Plot the 9 points calculated in the table.
• Draw a straight line of best fit passing as closely as possible through the plotted points. The line should have roughly an equal number of points scattered above and below it, and should ideally pass through or near the mean point $(\bar{t}, \overline{\ln(100-x)}) \approx (7.09, 3.55)$.

1. Enter the given data into a graphing calculator and obtain the linear regression equation.

Let $Y = \ln(100 - x)$.
Inputting the values of $t$ into list $L_1$ and computing the corresponding $Y$ values into list $L_2$ on the graphing calculator:
From the calculator, $Y = a + bt$ where:
$a \approx 5.860$
$b \approx -0.3263$

Equation: $\ln(100 - x) = 5.86 - 0.326t$ (to 3 s.f.)

1. Retrieve the regression equation from part (d) and substitute $t = 10$ to find the estimated test marks.

From part (d), calculating the regression of $\ln(100 - x)$ on $t$ using the provided data yields the equation:
$\ln(100 - x) = 5.9662 - 0.34118t$

Substitute $t = 10$ into the equation:
$\ln(100 - x) = 5.9662 - 0.34118(10)$
$\ln(100 - x) = 2.5544$
$100 - x = e^{2.5544}$
$100 - x \approx 12.863$
$x = 100 - 12.863 = 87.137$

The estimated test mark is $87.1$ (to 3 s.f.).

2. Determine the reliability of the estimate by checking if it involves interpolation or extrapolation.

The given data range for average daily duration of sleep is $5.5 \le t \le 9.1$.
Since $t = 10$ is outside this data range, predicting the test mark involves extrapolation.
Hence, we would not expect this estimate to be reliable.