Pearson's correlation coefficient $r$ for HSC Maths Standard 2

The negative sign means the relationship is inverse: as $x$ increases, $y$ tends to decrease.

The magnitude $|r| = 0.86$ is close to $1$, indicating a strong linear relationship.

Overall, $r = -0.86$ indicates a strong, negative, linear association between the two variables.

Markers reward identification of sign (direction), magnitude (strength) and the linear qualifier.

Dataset A: strong positive linear relationship. As $x$ increases, $y$ increases, with points closely clustered around the line.

Dataset B: very weak or no linear relationship. The points show essentially no straight-line pattern.

A low value of $r$ measures only the linear association. A scatterplot may show a strong non-linear pattern (for example, parabolic, exponential or U-shaped), in which case Pearson's $r$ will be near zero despite a clear relationship. Always look at the scatterplot before relying on $r$.

Markers reward describing both datasets correctly with sign, strength and linear qualifier, and the caveat that low $r$ does not preclude non-linear patterns.

Read the sign for direction (a)

The value $r = 0.83$ is positive, so the relationship is positive: as one variable increases, the other tends to increase.

Read the magnitude for strength (b)

The size is $|r| = 0.83$, which falls in the $0.8$ to $1.0$ band, so the relationship is very strong.

Write the full interpretation (c)

Combine all three parts: $r = 0.83$ indicates a very strong, positive, linear relationship between the two variables.

Check

The sentence names strength (very strong), direction (positive) and includes the word "linear", which is exactly what markers reward.

Read the direction from the slope (a)

The cloud rises to the right (drink sales go up as temperature goes up), so the trend is positive and $r > 0$.

Read the strength from the spread (b)

The points lie in a fairly tight band close to a straight line, so the linear fit is strong and $|r|$ is close to $1$, not close to $0$.

Choose a consistent value (c)

A tight rising band matches a large positive value, for example $r = 0.9$. (Any value such as $0.85$ to $0.95$ is reasonable.)

Check

Positive sign matches the rising cloud and a magnitude near $0.9$ matches the tight band, so $r = 0.9$ fits the description.

Enter the data into statistics mode (a)

Clear any old data, switch to $2$-variable statistics mode, and enter the five pairs into the $x$ and $y$ lists in order: $x = 1, 2, 3, 4, 5$ and $y = 12, 19, 23, 31, 35$.

Read the coefficient (a)

From the regression-results menu the calculator returns $r = 0.99$ (the unrounded value is $0.9947$).

Describe the relationship (b)

The sign is positive and $|r| = 0.99$ is in the $0.8$ to $1.0$ band, so this is a very strong, positive, linear relationship: more water is associated with greater height.

Check

Both $x$ and $y$ increase together and the points lie almost exactly on a line, so a value very close to $+1$ is expected, agreeing with $r = 0.99$.

Compare magnitudes for strength (a)

Strength depends only on the size $|r|$. Here $|-0.78| = 0.78$ and $|0.41| = 0.41$, and $0.78 > 0.41$, so study P shows the stronger linear relationship.

Compare signs for direction (b)

The sign gives the direction. Study P has a negative value, so study P shows the negative relationship.

Explain the separation (c)

Sign and magnitude answer different questions: the sign is only the direction, while the magnitude is the strength. A value of $-0.78$ sits closer to $-1$ than $0.41$ sits to $1$, so its points hug the line more tightly and the relationship is stronger, even though it slopes downward.

Check

Study P is both stronger (larger magnitude) and negative, which is consistent: strength and direction are read separately.

Enter the pairs into statistics mode (a)

Clear old data, then enter $x = 1, 2, 3, 4, 5, 6$ and $y = 45, 52, 49, 63, 58, 70$ as six $(x, y)$ pairs.

Read the coefficient (a)

The calculator gives $r = 0.90$ (the unrounded value is $0.8999$, rounded to two decimal places).

Interpret in context (b)

The sign is positive and $|r| = 0.90$ lies in the $0.8$ to $1.0$ band, so there is a very strong, positive, linear relationship: more revision hours are associated with higher test marks.

Check

The marks generally climb as revision hours rise, with only small dips, so a large positive value near $0.9$ is sensible.

Enter the data (a)

In $2$-variable statistics mode enter $x = 1, 2, 3, 4, 5, 6$ and $y = 9, 7.5, 8, 6, 6.5, 5$.

Read the coefficient (a)

The calculator returns $r = -0.92$ (unrounded $-0.9221$).

Describe direction and strength (b)

The sign is negative, so the direction is negative; $|r| = 0.92$ is in the $0.8$ to $1.0$ band, so the relationship is very strong. Overall this is a very strong, negative, linear relationship.

Explain the sign (c)

The negative sign means the variables move in opposite directions: as screen time increases, nightly sleep tends to decrease.

Check

Sleep hours fall as screen time rises, so a strong negative value near $-0.9$ is expected, matching $r = -0.92$.

Enter the eight pairs (a)

In statistics mode enter $x = 1, 2, 3, 4, 5, 6, 7, 8$ and $y = 3, 5, 4, 7, 6, 9, 7, 10$.

Read the coefficient (a)

The calculator gives $r = 0.89$ (unrounded $0.8919$).

Give the strength band (b)

The magnitude $|r| = 0.89$ lies in the $0.8$ to $1.0$ band, so the linear relationship is very strong and positive.

Explain why it is not exactly $1$ (c)

A coefficient of exactly $1$ needs every point to sit on one straight line. Here the points rise overall but zig-zag slightly (for example $y$ dips from $5$ down to $4$, and from $9$ down to $7$), so the fit is very strong but not perfect, giving $r = 0.89$ rather than $1$.

Check

The upward trend matches the positive sign, and the small wobbles explain why the value is high but below $1$.

Enter the data (a)

In statistics mode enter $x = 1, 2, 3, 4, 5, 6, 7$ and $y = 6, 9, 5, 8, 5, 9, 6$.

Read the coefficient (a)

The calculator returns $r = 0.00$ (unrounded value $0.0000$).

Interpret the result (b)

With $|r| = 0.00$ the value sits in the $0.0$ to $0.2$ band, so there is essentially no linear relationship between shoe size and spelling mark.

Answer the effect question (c)

A value near $0$ tells us there is no straight-line link in this sample; it does not "prove" anything about cause. We would not expect shoe size to affect spelling, and the near-zero $r$ is consistent with two unrelated variables.

Check

The marks bounce up and down with no upward or downward drift as shoe size grows, so a coefficient close to $0$ is exactly what we expect.

Enter the data into statistics mode (a)

Clear old data, then enter $x = 2, 3, 4, 5, 6, 7, 8$ and $y = 30, 24, 27, 18, 16, 12, 9$ as seven pairs.

Read the coefficient (a)

The calculator gives $r = -0.97$ (unrounded $-0.9658$).

Interpret fully (b)

The sign is negative, so the direction is negative; $|r| = 0.97$ is in the $0.8$ to $1.0$ band, so the relationship is very strong. In context: there is a very strong, negative, linear relationship, so older cars of this model tend to be cheaper.

Comment on the causal claim (c)

A strong $r$ shows association, not proof of cause. While age plausibly contributes here, $r$ alone cannot establish causation: other factors (kilometres driven, condition, demand) move with age and also affect price. State the link as an association and use cautious language.

Check

Price falls steadily as age rises, so a value very close to $-1$ is expected, agreeing with $r = -0.97$.

Enter the eight pairs (a)

In statistics mode enter $x = 5, 8, 12, 15, 18, 22, 26, 30$ and $y = 42, 55, 51, 68, 72, 79, 88, 95$.

Read the coefficient (a)

The calculator gives $r = 0.98$ (unrounded $0.9798$).

Describe the relationship (b)

The sign is positive and $|r| = 0.98$ is in the $0.8$ to $1.0$ band, so there is a very strong, positive, linear relationship: higher attendance is associated with higher exam marks.

State the check before trusting it (c)

Always look at the scatterplot first. A high $r$ measures only linear association, so a curved pattern or a single outlier could distort the picture. Confirm the cloud genuinely follows a straight-line trend before relying on $r = 0.98$.

Check

Marks climb steadily as attendance rises with only small wobbles, so a value very close to $+1$ is sensible.

Find $r$ for data set A (a)

Enter $x = 1, 2, 3, 4, 5, 6$ and $y = 11, 13, 18, 21, 27, 30$ in statistics mode; the calculator gives $r_A = 0.99$ (unrounded $0.9929$).

Find $r$ for data set B (a)

Clear the lists, then enter $x = 1, 2, 3, 4, 5, 6$ and $y = 30, 22, 24, 18, 15, 9$; the calculator gives $r_B = -0.96$ (unrounded $-0.9613$).

Compare magnitudes (b)

Strength is set by the size only: $|r_A| = 0.99$ and $|r_B| = 0.96$. Since $0.99 > 0.96$, data set A has the stronger linear relationship, even though B is negative.

Describe each set (c)

Data set A: a very strong, positive, linear relationship (more advertising is associated with more sales). Data set B: a very strong, negative, linear relationship (more downtime is associated with fewer units produced).

Check

A rises together so its sign is positive, B falls so its sign is negative, and both clouds hug their lines tightly, so two large magnitudes are expected with A slightly larger.

Enter the data (a)

In statistics mode enter $x = 1, 2, 3, 4, 5, 6, 7$ and $y = 20, 11, 5, 2, 5, 11, 20$.

Read the coefficient (a)

The calculator returns $r = 0.00$ (unrounded $0.0000$).

Explain why the conclusion is wrong (b)

A value of $r = 0$ means there is no linear relationship, not no relationship at all. Plotting the points shows a clear U-shape (a strong non-linear pattern): $y$ falls to a minimum at $x = 4$ and then rises symmetrically. The fall on the left cancels the rise on the right, forcing $r$ to $0$ even though the pattern is obvious.

State the lesson (c)

Pearson's $r$ measures linear association only, so always look at the scatterplot before trusting it. A small $r$ rules out a straight-line trend but not a curved one.

Check

The $y$ values are symmetric about $x = 4$, so by symmetry the linear trend must cancel to give $r = 0$, confirming the calculator value.