Scatter plots worksheet for students to plot data points and find lines of best fit.

Let's solve the scatter plot problems step by step.

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Problem 1: First Data Set

Data:
| x | y |
|---|---|
| 2 | 1 |
| 5 | 5 |
| 7 | 7 |
| 3 | 3 |
| 1 | 5 |
| 6 | 2 |

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#### Step 1: Plot the Points

We will plot each (x, y) pair on the coordinate plane:

- (2, 1)
- (5, 5)
- (7, 7)
- (3, 3)
- (1, 5)
- (6, 2)

These points are scattered. Let’s analyze their pattern.

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#### Step 2: Look for a Trend

Let’s look at the relationship between x and y:

- When x increases, y sometimes increases, sometimes decreases.
- For example:
- (1,5), (2,1): x goes up, y drops
- (2,1), (3,3): x up, y up
- (3,3), (5,5): x up, y up
- (5,5), (6,2): x up, y down
- (6,2), (7,7): x up, y up

There is some variability, but overall, there might be a weak positive trend. However, notice that (1,5) and (2,1) are outliers in terms of the trend.

Let’s compute the line of best fit using linear regression.

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#### Step 3: Find Line of Best Fit (Linear Regression)

The line of best fit has the form:

$$
y = mx + b
$$

Where:
- $ m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} $
- $ b = \frac{\sum y - m\sum x}{n} $

Let’s compute:

| x | y | x*y | x² |
|---|---|-----|----|
| 2 | 1 | 2 | 4 |
| 5 | 5 | 25 | 25 |
| 7 | 7 | 49 | 49 |
| 3 | 3 | 9 | 9 |
| 1 | 5 | 5 | 1 |
| 6 | 2 | 12 | 36 |

Now sum up:

- $ \sum x = 2+5+7+3+1+6 = 24 $
- $ \sum y = 1+5+7+3+5+2 = 23 $
- $ \sum xy = 2+25+49+9+5+12 = 102 $
- $ \sum x^2 = 4+25+49+9+1+36 = 124 $
- $ n = 6 $

Now compute slope $ m $:

$$
m = \frac{6(102) - (24)(23)}{6(124) - (24)^2} = \frac{612 - 552}{744 - 576} = \frac{60}{168} = \frac{5}{14} \approx 0.357
$$

Now compute intercept $ b $:

$$
b = \frac{23 - (5/14)(24)}{6} = \frac{23 - 120/14}{6} = \frac{23 - 8.571}{6} = \frac{14.429}{6} \approx 2.405
$$

So, the line of best fit is approximately:

$$
y = 0.357x + 2.405
$$

You can round it to:

$$
\boxed{y \approx 0.36x + 2.4}
$$

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Problem 2: Second Data Set

Data:
| x | y |
|---|---|
| 4 | 6 |
| 6 | 2 |
| 1 | 1 |
| 3 | 5 |
| 0 | 3 |
| 2 | 4 |

---

#### Step 1: Plot the Points

Points:
- (4,6)
- (6,2)
- (1,1)
- (3,5)
- (0,3)
- (2,4)

Plot these on the grid.

---

#### Step 2: Look for a Trend

Looking at the data:

- As x increases from 0 to 6, y varies: 3 → 4 → 5 → 6 → 2 → 1
- There seems to be a negative correlation: as x increases, y tends to decrease.
- But not perfectly.

Let’s do linear regression again.

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#### Step 3: Linear Regression

| x | y | x*y | x² |
|---|---|-----|----|
| 4 | 6 | 24 | 16 |
| 6 | 2 | 12 | 36 |
| 1 | 1 | 1 | 1 |
| 3 | 5 | 15 | 9 |
| 0 | 3 | 0 | 0 |
| 2 | 4 | 8 | 4 |

Sums:

- $ \sum x = 4+6+1+3+0+2 = 16 $
- $ \sum y = 6+2+1+5+3+4 = 21 $
- $ \sum xy = 24+12+1+15+0+8 = 60 $
- $ \sum x^2 = 16+36+1+9+0+4 = 66 $
- $ n = 6 $

Compute slope $ m $:

$$
m = \frac{6(60) - (16)(21)}{6(66) - (16)^2} = \frac{360 - 336}{396 - 256} = \frac{24}{140} = \frac{6}{35} \approx 0.1714
$$

Wait — this is positive, but visually, it looks like a negative trend?

Let’s check the values:

- From (0,3) to (6,2): x up, y down → negative
- From (1,1) to (4,6): x up, y up → positive
- So mixed.

But let’s double-check calculations.

Wait: $ \sum xy = 24+12+1+15+0+8 = 60 $? Yes.

$ \sum x = 16 $, $ \sum y = 21 $

Numerator: $ 6*60 = 360 $, $ 16*21 = 336 $, so $ 360 - 336 = 24 $

Denominator: $ 6*66 = 396 $, $ 16^2 = 256 $, $ 396 - 256 = 140 $

So $ m = 24 / 140 = 0.1714 $

Positive? That suggests a slight positive trend.

But let’s plot mentally:

- (0,3)
- (1,1)
- (2,4)
- (3,5)
- (4,6)
- (6,2)

From (0,3) to (1,1): down
(1,1) to (2,4): up
(2,4) to (3,5): up
(3,5) to (4,6): up
(4,6) to (6,2): down

So maybe no strong trend.

But let’s compute the correlation coefficient or just accept that the line is weak.

But wait — maybe I made a mistake.

Let me recompute $ \sum xy $:

- (4,6): 4×6 = 24
- (6,2): 6×2 = 12
- (1,1): 1×1 = 1
- (3,5): 3×5 = 15
- (0,3): 0×3 = 0
- (2,4): 2×4 = 8

Total: 24+12=36; +1=37; +15=52; +0=52; +8=60 → correct.

So $ m = 24 / 140 ≈ 0.1714 $

Then $ b = \frac{21 - (0.1714)(16)}{6} = \frac{21 - 2.742}{6} = \frac{18.258}{6} ≈ 3.043 $

So line: $ y ≈ 0.17x + 3.04 $

But this seems odd because (6,2) is below the line: 0.17×6 + 3.04 = 1.02 + 3.04 = 4.06, but actual y=2 → way below.

Wait — maybe the data is not linear.

Alternatively, perhaps we should use visual estimation.

Let’s try plotting and estimating.

Visual Estimation Approach (Recommended for Students)

For both scatter plots, students are expected to:

1. Plot all points.
2. Draw a line that passes through the "center" of the data, minimizing distances to points.

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First Scatter Plot (Top)

Points:
- (1,5), (2,1), (3,3), (5,5), (6,2), (7,7)

Plot them:
- (1,5) — high y
- (2,1) — low y
- (3,3)
- (5,5)
- (6,2)
- (7,7)

Try to draw a line that balances the points above and below.

Notice:
- (3,3), (5,5), (7,7) are on the line y = x
- (1,5) is above
- (2,1) is below
- (6,2) is below

So maybe the line is close to y = x, but slightly lower.

Try y = x - 1?

- At x=1: y=0 → too low
- At x=2: y=1 → matches (2,1)
- At x=3: y=2 → but point is (3,3) → too low
- At x=5: y=4 → point is (5,5) → too low

So maybe y = 0.8x + 1?

Try:
- x=1: 0.8+1=1.8 → vs (1,5) → too low
- x=2: 1.6+1=2.6 → vs (2,1) → too high

Not good.

Better to go back to regression result:

We had $ y ≈ 0.36x + 2.4 $

Let’s test:
- x=1: 0.36 + 2.4 = 2.76 → actual y=5 → too low
- x=2: 0.72 + 2.4 = 3.12 → actual y=1 → too high
- x=3: 1.08 + 2.4 = 3.48 → actual y=3 → close
- x=5: 1.8 + 2.4 = 4.2 → actual y=5 → close
- x=6: 2.16 + 2.4 = 4.56 → actual y=2 → too high
- x=7: 2.52 + 2.4 = 4.92 → actual y=7 → too low

So not very good fit.

Perhaps there's no strong linear trend.

But let’s try a different approach.

Maybe the line should pass near (3,3), (5,5), (7,7) — which are on y=x.

And (1,5) is an outlier, (2,1) and (6,2) are below.

So perhaps y = x is a reasonable visual line of best fit.

Check:
- (1,5): predicted 1 → actual 5 → far off
- (2,1): predicted 2 → actual 1 → off
- (3,3): perfect
- (5,5): perfect
- (6,2): predicted 6 → actual 2 → bad
- (7,7): perfect

Only 3 points on the line.

But (1,5) and (6,2) suggest maybe a curve?

But since it’s a scatter plot with line of best fit, we assume linear.

Given the variation, a reasonable line of best fit could be:

$$
\boxed{y = x}
$$

Or slightly adjusted.

But based on regression, we got $ y = 0.36x + 2.4 $, which doesn’t make sense.

Wait — let’s recalculate regression carefully.

Earlier for first set:

- $ \sum x = 24 $
- $ \sum y = 23 $
- $ \sum xy = 2+25+49+9+5+12 = 102 $
- $ \sum x^2 = 4+25+49+9+1+36 = 124 $
- $ n = 6 $

$ m = \frac{6*102 - 24*23}{6*124 - 24^2} = \frac{612 - 552}{744 - 576} = \frac{60}{168} = 0.357 $

$ b = \frac{23 - 0.357*24}{6} = \frac{23 - 8.568}{6} = \frac{14.432}{6} = 2.405 $

So $ y = 0.357x + 2.405 $

Now test:
- x=1: 0.357 + 2.405 = 2.762 → actual y=5 → error
- x=2: 0.714 + 2.405 = 3.119 → actual y=1 → big error
- x=3: 1.071 + 2.405 = 3.476 → actual y=3 → ok
- x=5: 1.785 + 2.405 = 4.19 → actual y=5 → ok
- x=6: 2.142 + 2.405 = 4.547 → actual y=2 → error
- x=7: 2.499 + 2.405 = 4.904 → actual y=7 → error

So it's not a good fit.

But this is mathematically correct.

However, in practice, for students, they are expected to draw a line by eye.

So for first plot, a better visual line might be one that goes through (3,3), (5,5), (7,7), ignoring (1,5), (2,1), (6,2).

So y = x is a reasonable choice.

Similarly, for second plot:

Points: (0,3), (1,1), (2,4), (3,5), (4,6), (6,2)

Let’s try to estimate.

- (0,3)
- (1,1)
- (2,4)
- (3,5)
- (4,6)
- (6,2)

If we ignore (6,2), the others go from (0,3) to (4,6): slope = (6-3)/(4-0) = 3/4 = 0.75

So y = 0.75x + 3

Test:
- x=0: y=3 → matches
- x=1: 0.75+3=3.75 → actual y=1 → too high
- x=2: 1.5+3=4.5 → actual y=4 → close
- x=3: 2.25+3=5.25 → actual y=5 → close
- x=4: 3+3=6 → matches
- x=6: 4.5+3=7.5 → actual y=2 → way off

So (6,2) is an outlier.

Maybe the line should be drawn through (0,3), (2,4), (3,5), (4,6) — which are roughly on a line.

(0,3), (2,4): slope = (4-3)/(2-0) = 0.5

(2,4), (3,5): slope = 1

(3,5), (4,6): slope = 1

So increasing slope.

But if we force a line, maybe y = 0.75x + 3 is acceptable.

Or y = x + 3? No, (0,3) is on it, but (1,1) is below.

Best guess: y = 0.5x + 3

Test:
- x=0: 3 → match
- x=1: 3.5 → actual 1 → too high
- x=2: 4 → match
- x=3: 4.5 → actual 5 → close
- x=4: 5 → actual 6 → close
- x=6: 6 → actual 2 → off

Still not great.

But (6,2) is likely an outlier.

So visual line of best fit for second plot might be:

$$
\boxed{y = 0.75x + 3}
$$

or simply estimated by drawing a line through (0,3), (2,4), (4,6)

That gives slope = (6-3)/(4-0) = 3/4 = 0.75, y-intercept = 3.

So $ y = 0.75x + 3 $

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✔ Final Answer (Student-Friendly)

Since this is a worksheet for kids, they are expected to plot the points and draw a line by eye.

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#### First Scatter Plot (Top)

- Plot all six points.
- The points (3,3), (5,5), (7,7) lie on the line $ y = x $
- Other points are scattered around.
- A reasonable line of best fit is:

$$
\boxed{y = x}
$$

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#### Second Scatter Plot (Bottom)

- Plot all six points.
- Points (0,3), (2,4), (4,6) seem to follow a trend.
- Slope = (6-3)/(4-0) = 3/4 = 0.75
- Y-intercept = 3
- So line: $ y = 0.75x + 3 $

Even though (1,1) and (6,2) don't fit, they may be outliers.

So line of best fit:

$$
\boxed{y = 0.75x + 3}
$$

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📌 Instructions for Student:

1. Plot each (x,y) point on the grid.
2. Draw a straight line that goes through the middle of the points, balancing the number of points above and below.
3. For the first plot, the line $ y = x $ is a good choice.
4. For the second plot, the line $ y = 0.75x + 3 $ fits most points well.

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✔ Summary

| Scatter Plot | Line of Best Fit |
|--------------|------------------|
| Top | $ y = x $ |
| Bottom | $ y = 0.75x + 3 $ |

These are approximate lines of best fit based on visual estimation, suitable for educational purposes.

Let me know if you'd like a printable version with the points plotted!