Bivariate statistics

Unit 9 Bivariate statistics

Key unit competence: Extend understanding, analysis and interpretation of bivariate data to correlation coefficients and regression lines.

Key ideas

• Bivariate data are pairs $(x_i,y_i)$. A scatter diagram shows the pattern: positive, negative or no correlation; strong or weak.
• Covariance: $\mathrm{cov}(x,y)=\frac{1}{n}\sum (x_i-\bar{x})(y_i-\bar{y})=\frac{1}{n}\sum x_i{y}_i-\bar{x}\bar{y}$.
• Pearson correlation coefficient: $r=\frac{\mathrm{cov}(x,y)}{{\sigma }_x{\sigma }_y}$, with $-1\le r\le 1$. $r$ near 1: strong positive; near $-1$: strong negative; near 0: weak or no linear correlation.
• Regression line of $y$ on $x$: $y-\bar{y}=\frac{\mathrm{cov}(x,y)}{{\sigma }_x^2}(x-\bar{x})$. Of $x$ on $y$: $x-\bar{x}=\frac{\mathrm{cov}(x,y)}{{\sigma }_y^2}(y-\bar{y})$. Both lines pass through $(\bar{x},\bar{y})$; the product of their slopes $a\times c=r^2$.
• Spearman's rank correlation: $r_s=1-\frac{6\sum d^2}{n(n^2-1)}$ where $d$ is the difference in ranks.
• Use the regression line of $y$ on $x$ to predict $y$ from $x$ (interpolation is safer than extrapolation). Correlation does not prove causation.

Formulas to know

Common misconceptions

• "$r=0$ means there is no relationship." It means no linear relationship; there may be a curved one.
• "A strong correlation proves that $x$ causes $y$." No: a third factor may cause both.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. For the data $x$: 2, 4, 6, 8; $y$: 3, 7, 5, 9, find $\bar{x}$, $\bar{y}$, $\mathrm{cov}(x,y)$ and $r$.

   Show answer

   $\bar{x}=5$, $\bar{y}=6$. $\sum xy=6+28+30+72=136$; $\mathrm{cov}=34-30=4$. ${\sigma }_x^2=\frac{120}{4}-25=5$; ${\sigma }_y^2=\frac{164}{4}-36=5$. $r=\frac{4}{5}=0.8$.

   Common mistake: Using $n-1$ in some formulas and $n$ in others.

2. Give the value of $r$ for the points $(1,1),(2,2),(3,3)$ and justify.

   Show answer

   $r=1$: the points lie exactly on the line $y=x$ with positive slope.

   Common mistake: Computing a covariance and forgetting to divide by ${\sigma }_x{\sigma }_y$.

3. Find the regression line of $y$ on $x$ for the same data and predict $y$ when $x=5$.

   Show answer

   Slope $\frac{4}{5}=0.8$: $y-6=0.8(x-5)$, $y=0.8x+2$; at $x=5$, $y=6$.

   Common mistake: Using the slope $\frac{\mathrm{cov}}{{\sigma }_y^2}$.

4. Explain why extrapolating far beyond the data is risky, with an example.

   Show answer

   The linear pattern may not continue. A child's height grows with age up to about 18, but predicting the height at 40 with the same line gives a wrong answer.

   Common mistake: Assuming every relationship stays linear forever.

5. Ten students' ranks in Maths and Physics give $\sum d^2=36$. Find $r_s$ and interpret.

   Show answer

   $r_s=1-\frac{216}{990}\approx 0.78$: strong positive agreement.

   Common mistake: Using $n^2$ instead of $n(n^2-1)$.

6. Show that $r$ does not change if all $x$ values are increased by 10.

   Show answer

   Deviations $x-\bar{x}$ are unchanged, so $\mathrm{cov}$ and ${\sigma }_x$ are unchanged; $r$ is the same.

   Common mistake: Adding 10 to $r$.

7. Describe the correlation you expect between hours of exercise per week and resting heart rate.

   Show answer

   Negative: more exercise tends to go with a lower resting heart rate (other factors also matter).

   Common mistake: Claiming exercise is the only cause.

8. The two regression lines are $y=0.9x+1$ and $x=0.4y+2$. Find $r$ and the mean point.

   Show answer

   $r^2=0.9\times 0.4=0.36$, $r=0.6$ (positive slopes). Solve: $x=0.4(0.9x+1)+2$, $0.64x=2.4$, $x=3.75$, $y=4.375$.

   Common mistake: Taking $r=0.36$.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Using $y=0.8x-0.5$, estimate $y$ when $x=7$.	a. $(2,1)$
   2. The regression lines are $y=2x-3$ and $x=0.4y+1.6$. Find $(\bar{x},\bar{y})$.	b. $23$
   3. For 6 ranked pairs, $\sum d^2=14$. Find $r_s$.	c. $0.6$
   4. Same data ($r=0.6$, ${\sigma }_x=5$, ${\sigma }_y=10$, $\bar{x}=20$, $\bar{y}=50$). Estimate $x$ when $y=60$.	d. $5.1$

   Show answer

   1 → d ($5.1$) · 2 → a ($(2,1)$) · 3 → c ($0.6$) · 4 → b ($23$)

   • Step 1. $y=0.8\times 7-0.5$.
     Step 2. $5.6-0.5=5.1$.
   • Step 1. Both lines pass through $(\bar{x},\bar{y})$, so solve them together.
     Step 2. $x=0.4(2x-3)+1.6=0.8x+0.4\Rightarrow 0.2x=0.4\Rightarrow x=2$.
     Step 3. $y=2(2)-3=1$.
   • Step 1. $r_s=1-\frac{6\sum d^2}{n(n^2-1)}=1-\frac{84}{6\times 35}$.
     Step 2. $1-\frac{84}{210}=1-0.4=0.6$.
   • Step 1. Use $x$ on $y$: $b_{xy}=\frac{r{\sigma }_x}{{\sigma }_y}=\frac{0.6\times 5}{10}=0.3$.
     Step 2. $x=20+0.3(60-50)=23$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   unchanged, $r$  ·  a strong negative linear relationship  ·  $-r$  ·  $2r$
   1. Each $x$ is replaced by $2x+3$ and each $y$ by $y-5$. The new correlation coefficient is .
   2. Every $y$ value is multiplied by $-2$. The new correlation coefficient is .
   3. $r=-0.92$ means .
   Show answer

   1: unchanged, $r$ · 2: $-r$ · 3: a strong negative linear relationship

   • Step 1. Adding constants moves the data but not its shape.
     Step 2. Multiplying by a positive number does not change $r$. So $r$ is unchanged.
   • Step 1. Scaling by a positive number leaves $r$ unchanged.
     Step 2. A negative factor reverses the direction, so $r$ becomes $-r$.
   • Step 1. The sign gives the direction: negative.
     Step 2. $|r|$ close to 1 means strong. $r$ is not a slope.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. The regression line is $y=1.5x+2$. Find the residual for the point $(4,9)$.	a. $y=x+2$
   2. Data $x$: 2, 4, 6, 8 and $y$: 1, 3, 4, 6. Find $\mathrm{cov}(x,y)$.	b. $4$
   3. The two regression slopes are $b_{yx}=0.8$ and $b_{xy}=0.45$. Find $r$.	c. $1$
   4. With the same data and $\sum x^2=100$, find the regression line of $y$ on $x$.	d. $0.6$

   Show answer

   1 → c ($1$) · 2 → b ($4$) · 3 → d ($0.6$) · 4 → a ($y=x+2$)

   • Step 1. Predicted value: $1.5\times 4+2=8$.
     Step 2. Residual $=$ observed $-$ predicted $=9-8=1$.
   • Step 1. $\bar{x}=5$, $\bar{y}=3.5$, $\sum xy=2+12+24+48=86$.
     Step 2. $\mathrm{cov}=\frac{86}{4}-5\times 3.5=21.5-17.5=4$.
   • Step 1. $r^2=b_{yx}\times b_{xy}=0.36$.
     Step 2. Both slopes are positive, so $r=+0.6$.
   • Step 1. ${\sigma }_x^2=\frac{100}{5}-16=4$, so slope $=\frac{4}{4}=1$.
     Step 2. $c=\bar{y}-\bar{x}=6-4=2$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $0.4$  ·  $0.77$  ·  $1.2$  ·  $0.6$
   1. Which value cannot be a correlation coefficient? →
   2. For the same data $(1,2),(2,4),(3,5),(4,4),(5,5)$, $r\approx$ →
   3. Two judges rank 4 items. A: 1, 2, 3, 4; B: 2, 1, 4, 3. Find $r_s$. →
   Show answer

   1: $1.2$ · 2: $0.77$ · 3: $0.6$

   • Step 1. Always $-1\le r\le 1$.
     Step 2. $1.2$ is outside this interval.
   • Step 1. ${\sigma }_y^2=\frac{86}{5}-16=1.2$.
     Step 2. $r=\frac{1.2}{\sqrt{2\times 1.2}}=\frac{1.2}{1.549}\approx 0.77$.
   • Step 1. $d$: $-1,1,-1,1$, so $\sum d^2=4$.
     Step 2. $r_s=1-\frac{6\times 4}{4\times 15}=1-\frac{24}{60}=0.6$.

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