Advanced Algebra Linear Regression Calculator Worksheet 2 5 - Fill ... - Free Printable

It looks like you've uploaded a worksheet titled "Advanced Algebra: Linear Regression Calculator Worksheet 2.5" that contains several problems related to linear regression, which is a statistical method used to model the relationship between two variables by fitting a linear equation to observed data.

Since I can't view images directly, I’ll solve each problem based on the text you provided. Let’s go through each one step-by-step and explain the solution process.

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Problem 1: Anthropologist uses a bone model to estimate height

Given:
- A model relates femur length to height.
- Data table of femur lengths (in cm) and corresponding heights (in cm).

We are to:
a) Find and graph a linear equation that models the data.
b) Use the equation to predict the height of a person with a femur of 38 cm.

#### Step 1: Identify the variables
Let:
- $ x $ = femur length (cm)
- $ y $ = height (cm)

We'll use linear regression to find the line of best fit:
$$ y = mx + b $$

We need to calculate:
- Slope $ m $
- Y-intercept $ b $

We can use the formulas:
$$
m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}
$$
$$
b = \frac{\sum y - m \sum x}{n}
$$

But since this is likely meant for calculator use, we’ll assume we’re using a calculator or software to compute the regression line.

Let’s first compute the necessary sums from the data:

| Femur (x) | Height (y) |
|-----------|------------|
| 50.1 | 165.5 |
| 48.3 | 173.6 |
| 52.2 | 180.0 |
| 46.0 | 168.3 |
| 49.7 | 176.7 |
| 47.7 | 165.0 |
| 50.0 | 175.8 |

Now calculate:

- $ n = 7 $
- $ \sum x = 50.1 + 48.3 + 52.2 + 46.0 + 49.7 + 47.7 + 50.0 = 344.0 $
- $ \sum y = 165.5 + 173.6 + 180.0 + 168.3 + 176.7 + 165.0 + 175.8 = 1205.9 $
- $ \sum xy = 50.1×165.5 + 48.3×173.6 + ... $
- Compute each:
- 50.1 × 165.5 = 8281.55
- 48.3 × 173.6 = 8389.88
- 52.2 × 180.0 = 9400.0
- 46.0 × 168.3 = 7741.8
- 49.7 × 176.7 = 8777.99
- 47.7 × 165.0 = 7870.5
- 50.0 × 175.8 = 8790.0
- Sum: $ \sum xy = 8281.55 + 8389.88 + 9400.0 + 7741.8 + 8777.99 + 7870.5 + 8790.0 = 59,251.72 $
- $ \sum x^2 = 50.1² + 48.3² + 52.2² + 46.0² + 49.7² + 47.7² + 50.0² $
- = 2510.01 + 2332.89 + 2724.84 + 2116.0 + 2470.09 + 2275.29 + 2500.0 = 17,930.12

Now plug into formula:

$$
m = \frac{7(59251.72) - (344)(1205.9)}{7(17930.12) - (344)^2}
$$

Numerator:
$ 7×59251.72 = 414,762.04 $
$ 344×1205.9 = 414,457.6 $
→ $ 414,762.04 - 414,457.6 = 304.44 $

Denominator:
$ 7×17930.12 = 125,510.84 $
$ 344² = 118,336 $
→ $ 125,510.84 - 118,336 = 7,174.84 $

So,
$$
m = \frac{304.44}{7174.84} \approx 0.0424
$$

Now $ b $:
$$
b = \frac{1205.9 - 0.0424×344}{7} = \frac{1205.9 - 14.576}{7} = \frac{1191.324}{7} \approx 170.19
$$

So the equation is approximately:
$$
\boxed{y = 0.0424x + 170.19}
$$

#### Part a) Answer:
The linear model is:
$$
\boxed{y = 0.0424x + 170.19}
$$

You would graph this line along with the data points.

#### Part b) Predict height for femur = 38 cm:
$$
y = 0.0424(38) + 170.19 = 1.6092 + 170.19 = \boxed{171.799} \text{ cm}
$$

Answer: Approximately 171.8 cm

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Problem 2: Store manager tracks soft drinks sold vs temperature

Data:

| Temp (°F) x | Sales (cans) y |
|-------------|----------------|
| 55 | 240 |
| 60 | 280 |
| 65 | 420 |
| 70 | 480 |
| 75 | 510 |
| 80 | 590 |

We want:
a) Find and graph a linear equation.
b) Predict sales at 90°F.

Again, use linear regression.

Calculate:
- $ n = 6 $
- $ \sum x = 55+60+65+70+75+80 = 405 $
- $ \sum y = 240+280+420+480+510+590 = 2520 $
- $ \sum xy = 55×240 + 60×280 + ... $
- = 13,200 + 16,800 + 27,300 + 33,600 + 38,250 + 47,200 = 176,350
- $ \sum x^2 = 55² + 60² + 65² + 70² + 75² + 80² = 3025 + 3600 + 4225 + 4900 + 5625 + 6400 = 27,775 $

Now:
$$
m = \frac{6(176,350) - (405)(2520)}{6(27,775) - (405)^2}
$$

Numerator:
$ 6×176,350 = 1,058,100 $
$ 405×2520 = 1,021,200 $
→ $ 1,058,100 - 1,021,200 = 36,900 $

Denominator:
$ 6×27,775 = 166,650 $
$ 405² = 164,025 $
→ $ 166,650 - 164,025 = 2,625 $

$$
m = \frac{36,900}{2,625} = 14.04
$$

Now $ b $:
$$
b = \frac{2520 - 14.04×405}{6} = \frac{2520 - 5691.2}{6} = \frac{-3171.2}{6} = -528.53
$$

So the equation is:
$$
\boxed{y = 14.04x - 528.53}
$$

#### Part a) Answer:
$$
\boxed{y = 14.04x - 528.53}
$$

#### Part b) Predict sales at 90°F:
$$
y = 14.04(90) - 528.53 = 1263.6 - 528.53 = \boxed{735.07} \text{ cans}
$$

Answer: Approximately 735 cans

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Problem 3: Estimate age of oak tree from diameter

Data:

| Diameter (in) x | Age (years) y |
|------------------|---------------|
| 4.0 | 14 |
| 4.5 | 18 |
| 5.0 | 22 |
| 5.5 | 27 |
| 6.0 | 33 |
| 6.5 | 38 |
| 7.0 | 42 |
| 7.5 | 47 |
| 8.0 | 52 |

We want:
a) Find and graph a linear equation.
b) Estimate age if diameter is 10 in.

Use linear regression again.

Compute:
- $ n = 9 $
- $ \sum x = 4.0+4.5+...+8.0 = 54.0 $
- $ \sum y = 14+18+22+27+33+38+42+47+52 = 333 $
- $ \sum xy = 4.0×14 + 4.5×18 + ... $
- = 56 + 81 + 110 + 148.5 + 198 + 247 + 294 + 352.5 + 416 = 2005.5
- $ \sum x^2 = 16 + 20.25 + 25 + 30.25 + 36 + 42.25 + 49 + 56.25 + 64 = 338.75 $

Now:
$$
m = \frac{9(2005.5) - (54)(333)}{9(338.75) - (54)^2}
$$

Numerator:
$ 9×2005.5 = 18,049.5 $
$ 54×333 = 18,000 $
→ $ 18,049.5 - 18,000 = 49.5 $

Denominator:
$ 9×338.75 = 3,048.75 $
$ 54² = 2,916 $
→ $ 3,048.75 - 2,916 = 132.75 $

$$
m = \frac{49.5}{132.75} \approx 0.3727
$$

Now $ b $:
$$
b = \frac{333 - 0.3727×54}{9} = \frac{333 - 20.1258}{9} = \frac{312.8742}{9} \approx 34.764
$$

Equation:
$$
\boxed{y = 0.3727x + 34.764}
$$

#### Part a) Answer:
$$
\boxed{y = 0.3727x + 34.764}
$$

#### Part b) Estimate age when diameter = 10 in:
$$
y = 0.3727(10) + 34.764 = 3.727 + 34.764 = \boxed{38.491} \text{ years}
$$

Answer: Approximately 38.5 years

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✔ Final Answers Summary:

Problem 1:
a) $ y = 0.0424x + 170.19 $
b) Height ≈ 171.8 cm

Problem 2:
a) $ y = 14.04x - 528.53 $
b) Sales ≈ 735 cans

Problem 3:
a) $ y = 0.3727x + 34.764 $
b) Age ≈ 38.5 years

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📌 Notes:

- These models are linear approximations; real-world relationships may not be perfectly linear.
- You can verify these results using a graphing calculator or software like Desmos, Excel, or a regression tool.
- The equations should be rounded appropriately depending on context (e.g., to nearest tenth).

Let me know if you’d like a visual explanation or help graphing them!