Bivariate data

We’re told every $y$-value is the same. Call that fixed value $k$. Then for each data point $i$:

$y_i=k$

To use the covariance formula, we need $\bar{y}$. Because all $n$ values are the constant $k$:

$\begin{array}{rl}\bar{y}& =\frac{{\sum }_{i=1}^n{y}_i}{n}\\ \bar{y}& =\frac{k+k+\cdots +k}{n}\\ \bar{y}& =\frac{n\cdot k}{n}=k\end{array}$

So the mean $\bar{y}$ is the same constant $k$.

In the covariance sum we use $(y-\bar{y})$. For any data point, since $y_i=k$ and $\bar{y}=k$:

$y_i-\bar{y}=k-k=0$

Now substitute into the given covariance formula:

$\begin{array}{rl}{s}_{xy}& =\frac{\sum (x-\bar{x})(y-\bar{y})}{n}\\ s_{xy}& =\frac{\sum (x-\bar{x})\cdot 0}{n}\\ s_{xy}& =\frac{0}{n}\\ s_{xy}& =0\end{array}$

Because every product includes a factor of $0$, the entire numerator becomes $0$.

From the algebra, the covariance must be $0$, so the other options do not fit:

• Option B is incorrect because the result is exactly $0$, not $1$.
• Option C is incorrect because when the $y$-part is always $0$, the covariance cannot turn into $\bar{x}$.
• Option D is incorrect because we can determine the value without needing any specific $x$ data.

Since every term in the sum becomes zero when one variable is constant, the covariance is always $0$.

In statistics, covariance describes whether two variables tend to change together. When both increase/decrease together, covariance is positive; when one rises as the other falls, covariance is negative.

Here, every $y$-value is identical. A variable with no change has zero variance.

If all $y_i=C$ for some constant, then $\bar{y}=C$ as well, so for every data point: $(y-\bar{y})=(C-C)=0$

If $y$ does not vary, then $x$ and $y$ cannot “vary together.” Using $s_{xy}=\frac{\sum (x-\bar{x})(y-\bar{y})}{n}$

all the $(y-\bar{y})$ terms are $0$, so each product is $0$ and the sum in the numerator is also $0$, no matter what the $x$-values are.

• Option B: A value of $1$ is associated with correlation limits, not covariance.
• Option C: $\bar{x}$ is just an average of $x$ and doesn’t become the covariance.
• Option D: We do have enough information—constant $y$ forces every $(y-\bar{y})$ to be zero.

With one variable fixed, joint variation cannot happen, so the covariance is $0$.

To check the idea, create an example. Take 3 points with $x=\{2,4,6\}$ and constant $y=\{8,8,8\}$.

Find the mean of the $y$-values:

$\bar{y}=\frac{8+8+8}{3}=8$

Because all $y$-values match, the mean equals that same number.

Now compute deviations from the mean:

• Point 1: $8-8=0$
• Point 2: $8-8=0$
• Point 3: $8-8=0$

Every deviation is $0$.

Substitute into the formula:

$s_{xy}=\frac{\sum (x-\bar{x})(y-\bar{y})}{n}$

Since each $(y-\bar{y})$ is $0$, each product is $0$, so the total in the numerator is $0$:

$s_{xy}=\frac{0+0+0}{3}=0$

From the test dataset, the covariance came out as $0$.

• Option B: We got $0$, not $1$.
• Option C: Even though $\bar{x}$ here is $4$, the covariance is still $0$.
• Option D: The example clearly shows we can determine the covariance.

Whenever one variable has no spread (all values equal the mean), the covariance calculation always gives $0$.

In bivariate data, a line of best fit is a straight line that represents the overall relationship between two variables on a scatter diagram. When we talk about “balancing,” the aim is to make the line reflect the main pattern of the data instead of being pulled too much by a few unusual points.

Here’s why the other ideas are not what “balancing” means for a best-fit line:

• Option B: The trend line does not need to go through the origin $(0,0)$. It should be placed where the data points actually appear.
• Option C: Whether the totals of the $x$-values and $y$-values match is not what determines the position of a best-fit line, especially since $x$ and $y$ can measure different things.
• Option D: Scatter plots usually show natural variation, so a best-fit line is meant to show the general trend and typically will not pass through every point.

When drawing a line of best fit by eye, “balancing” means you place the line so that:

1. About an equal number of points are above and below the line.
2. The overall vertical gaps (residuals) above the line are roughly matched by the overall vertical gaps below the line.

This helps the line run through the middle of the cluster of points.

Since balancing is about keeping the data points fairly evenly split around the line to represent the central trend, Option A is the best description.

In Topic: Sampling Methods and Bivariate Data, we learn that a line of best fit is a straight line that models the overall trend in a scatter plot. When we sketch it “by eye,” the word “balancing” describes a visual rule. We can find the correct meaning by removing common incorrect ideas.

Some people think a line has to begin at $(0,0)$, but that isn’t true for a best-fit line. The line should follow the pattern of the plotted data. If the points mostly appear between $x=50$ and $x=100$, forcing the line through the origin would usually drag it away from the trend.

Another misconception is that the line should touch every point. Real data has randomness and measurement error, so one straight line almost never goes through all points. The goal is to show the average direction, not to join the dots.

Saying the sum of the $x$-values must equal the sum of the $y$-values is a distractor. Although a best-fit line passes through the mean point $(\bar{x},\bar{y})$, there is no rule that requires $\sum x=\sum y$.

After eliminating the incorrect choices, “balancing” refers to how the points are spread around the line. When drawing by eye, you place the line so there are roughly the same number of points above it as below it, keeping the line centered within the data cloud.

Start by noting the measurement units given:

• The independent variable $x$ is measured in metres (m).
• The dependent variable $y$ is measured in seconds (s).

To work out the unit of covariance ($s_{xy}$), look at its definition:

$s_{xy}=\frac{\sum (x_i-\bar{x})(y_i-\bar{y})}{n-1}$

This shows covariance is built from adding up products of how far $x$ values and $y$ values are from their means, then dividing by $n-1$.

Focus on the numerator term-by-term:

1. $(x_i-\bar{x})$ is a difference of distances, so it is measured in metres (m).
2. $(y_i-\bar{y})$ is a difference of times, so it is measured in seconds (s).
3. Multiplying them gives the unit

$\text{m}\times \text{s}=\text{m}\cdot \text{s}$

The denominator is $n-1$. Since $n$ is just the number of data pairs, it has no units.

Dividing by a unitless number does not affect units, so the covariance keeps the unit $\text{m}\cdot \text{s}$.

Using the unit result, we can reject the wrong choices:

• Option B is wrong because covariance is not unit-free (that’s a property of the correlation coefficient $r$).
• Option C is wrong because metres and seconds cannot be added; the formula uses multiplication.
• Option D is wrong because it describes a “per second” rate ($\text{m}/\text{s}$), but covariance comes from a product.

Covariance takes the product of the units of $x$ and $y$, so its unit is $\text{m}\cdot \text{s}$.

Another way is to use how covariance links to Pearson’s correlation coefficient $r$:

$r=\frac{s_{xy}}{s_x{s}_y}$

Rearranging gives:

$s_{xy}=r\cdot s_x\cdot s_y$

In $s_{xy}=r\cdot s_x\cdot s_y$, the value $r$ describes the strength/direction of a linear relationship and is dimensionless (it has no units and lies between $-1$ and $1$).

A standard deviation always uses the same unit as the variable it comes from.

So:

• Since $x$ is measured in metres (m), $s_x$ is also in m.
• Since $y$ is measured in seconds (s), $s_y$ is also in s.

Substitute units into the rearranged relationship:

$\text{Units of }{s}_{xy}=(\text{units of }r)\times (\text{units of }{s}_x)\times (\text{units of }{s}_y)$ $\text{Units of }{s}_{xy}=(1)\times (\text{m})\times (\text{s})$ $\text{Units of }{s}_{xy}=\text{m}\text{s}$

Compare with the answer choices:

• Option B is incorrect because only $r$ is unitless; $s_{xy}$ is not.
• Option C is incorrect because adding unlike units (metres and seconds) is not meaningful.
• Option D is incorrect because $\text{m}{\text{s}}^{-1}$ is a rate, not a covariance unit.

The unit you get is $\text{m}\text{s}$, which corresponds to the first option.

To decide the unit for covariance, begin with the units of the variables:

• The independent variable $x$ is in metres (m).
• The dependent variable $y$ is in seconds (s).

This fits with Bivariate Data and checking units carefully.

You can only add/subtract quantities with matching units. Since metres and seconds are different, writing $\text{m}+\text{s}$ does not make sense as a measurement.

So that option must be wrong.

The unit $\text{m}{\text{s}}^{-1}$ means “metres per second,” which is used for rates like speed or gradients.

Covariance is not found by dividing one variable by the other; it comes from multiplying their deviations, so this unit doesn’t fit.

A quantity is dimensionless only when units cancel. That happens for the correlation coefficient ($r$), but covariance itself is not scaled to cancel units.

So covariance cannot be unit-free.

Using the covariance formula:

$\text{cov}(x,y)=\frac{{\sum }_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{n}$

• $(x_i-\bar{x})$ carries m.
• $(y_i-\bar{y})$ carries s.
• Their product has units $\text{m}\cdot \text{s}$.
• The denominator $n$ (or $n-1$) is just a count, so it has no units.

So the covariance unit must be $\text{m}\text{s}$.

Whether $s_{xy}$ is positive or negative depends on adding up the products of each point’s deviations from the mean point $(\bar{x},\bar{y})$. The key relationship is:

$s_{xy}=\frac{1}{n}{\sum }_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})$

So we decide the sign by checking whether $(x_i-\bar{x})(y_i-\bar{y})$ is mostly positive or mostly negative for points in the given quadrants.

For points in the Top-Left Quadrant (TLQ):

• They are left of the mean in $x$, so $(x_i-\bar{x})<0$.
• They are above the mean in $y$, so $(y_i-\bar{y})>0$.

That makes the product $\text{negative}\times \text{positive}=\text{negative}$

So each TLQ point adds a negative term into the covariance sum.

For points in the Bottom-Right Quadrant (BRQ):

• They are right of the mean in $x$, so $(x_i-\bar{x})>0$.
• They are below the mean in $y$, so $(y_i-\bar{y})<0$.

So the product is $\text{positive}\times \text{negative}=\text{negative}$

Therefore BRQ points also contribute negative values to the total.

Because most of the points are in TLQ and BRQ, most terms in $\sum (x_i-\bar{x})(y_i-\bar{y})$ are negative, so the overall sum is expected to be negative.

• A Positive covariance would need more points in the Top-Right and Bottom-Left quadrants.
• A value of Zero would require the negative products to be balanced by enough positive products, which is not likely if the majority are negative.

Since the deviation products are mainly negative, $s_{xy}$ is negative overall. That corresponds to a downward (negative) linear pattern in the scatter plot.

Treat the mean point $(\bar{x},\bar{y})$ like the center that splits the scatter plot into quadrants.

• The Top-Left Quadrant (TLQ) has points with $x<\bar{x}$ and $y>\bar{y}$.
• The Bottom-Right Quadrant (BRQ) has points with $x>\bar{x}$ and $y<\bar{y}$.

If points mainly fall in these two quadrants, it suggests a negative relationship: when $x$ goes up, $y$ tends to go down.

Covariance tells us how two variables vary together. It is computed using:

$s_{xy}=\frac{1}{n}{\sum }_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})$

So the sign of $s_{xy}$ depends on whether the products $(x_i-\bar{x})(y_i-\bar{y})$ are mostly positive or mostly negative.

Now determine the sign of $(x_i-\bar{x})(y_i-\bar{y})$ in each relevant quadrant:

1. TLQ: $(x_i-\bar{x})$ is negative while $(y_i-\bar{y})$ is positive, so the product is negative.
2. BRQ: $(x_i-\bar{x})$ is positive while $(y_i-\bar{y})$ is negative, so the product is negative.

With most points in TLQ and BRQ, most products are negative, so the total sum is negative and the covariance is negative.

Using the pattern of the scatter plot:

• Option A is wrong because a positive covariance would match points mostly in the Top-Right and Bottom-Left quadrants.
• Option D is wrong because a clear diagonal pattern through TLQ and BRQ would not usually cancel perfectly to give zero.
• Option B is wrong because covariance can still be computed for any finite set of paired data.

Since the data cluster in TLQ and BRQ indicates a negative association, the covariance must be negative.