Data was collected to investigate the behaviour of tides in Sydney Harbour.

There are usually two high tides and two low tides each day.

The variables in this study were:

• Day: the day number in the sample
• LLT: the height of the lowest low tide for that day (in metres)
• HHT: the height of the highest high tide for that day (in metres)

Table 1 displays the data collected for a sample of 14 consecutive days in February 2021.

Table 1

\begin{array}{|c|c|c|}
\hline
 \rule{0pt}{2.5ex}\ \ \ \textit{Day}\ \ \ \rule[-1ex]{0pt}{0pt}& \textit{LLT (m)} & \textit{HHT (m)}\\
\hline \rule{0pt}{2.5ex}1 \rule[-1ex]{0pt}{0pt}& 0.43 & 1.65 \\
\hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 0.49 & 1.55 \\
\hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 0.55 & 1.44 \\
\hline \rule{0pt}{2.5ex}4 \rule[-1ex]{0pt}{0pt}& 0.61 & 1.42 \\
\hline \rule{0pt}{2.5ex}5 \rule[-1ex]{0pt}{0pt}& 0.68 & 1.42 \\
\hline \rule{0pt}{2.5ex}6 \rule[-1ex]{0pt}{0pt}& 0.73 & 1.42 \\
\hline \rule{0pt}{2.5ex}7 \rule[-1ex]{0pt}{0pt}& 0.72 & 1.42 \\
\hline \rule{0pt}{2.5ex}8 \rule[-1ex]{0pt}{0pt}& 0.65 & 1.47 \\
\hline \rule{0pt}{2.5ex}9 \rule[-1ex]{0pt}{0pt}& 0.57 & 1.55 \\
\hline \rule{0pt}{2.5ex}10 \rule[-1ex]{0pt}{0pt}& 0.48 & 1.64 \\
\hline \rule{0pt}{2.5ex}11 \rule[-1ex]{0pt}{0pt}& 0.39 & 1.74 \\
\hline \rule{0pt}{2.5ex}12 \rule[-1ex]{0pt}{0pt}& 0.30 & 1.83 \\
\hline \rule{0pt}{2.5ex}13 \rule[-1ex]{0pt}{0pt}& 0.25 & 1.90 \\
\hline \rule{0pt}{2.5ex}14 \rule[-1ex]{0pt}{0pt}& 0.22 & 1.92 \\
\hline
\end{array}

1. For the \(H H T\) values in Table 1:
   1. Calculate the mean, in metres.
   2. Round your answer to one decimal place.   (1 mark)
      

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   2. Calculate the standard deviation, in metres.
   3. Round your answer to three decimal places.    (1 mark)
      

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2. Use the \(HHT\) data from Table 1 to construct a boxplot on the grid below.    (2 marks)
   

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3. The five-number summary of the \(L L T\) data is shown in Table 2 below.
4. Table 2

\begin{array}{|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\textbf{Minimum} \rule[-1ex]{0pt}{0pt}& \ \ \textbf{Q1} \ \ & \textbf{Median} & \ \  \textbf{Q3} \ \ & \textbf{Maximum} \\
\hline \rule{0pt}{2.5ex}0.22 \rule[-1ex]{0pt}{0pt}& 0.39 & 0.52 & 0.65 & 0.73 \\
\hline
\end{array}

1. Show that the minimum \(L L T\) value of 0.22 m is not an outlier.    (2 marks)
   

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4. A least squares line can be used to model the association between \(L L T\) and \(H H T\). In this model, \(H H T\) is the response variable.
5. Use the data from Table 1 to determine the equation of this least squares line.
6. Round the values of the intercept and slope to four significant figures.
7. Write your answers in the boxes provided.    (2 marks)
   

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