Math Problem Statement

Listed below are the overhead widths (cm) of seals measured from photographs and weights (kg) of the seals. Find the regression equation, letting the overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2.4 cm, using the regression equation. Can the prediction be correct? If not, what is wrong? Use a significance level of 0.05.

Solution

To solve this problem, we need to:

  1. Find the regression equation by calculating the slope (mm) and the y-intercept (bb) of the linear regression line y^=b+mx\hat{y} = b + mx, where:

    • xx: Overhead width
    • yy: Weight
  2. Predict the weight of a seal with an overhead width of 2.4 cm using the regression equation.

  3. Determine if the prediction is valid, considering the significance level of 0.05 and the range of the data.

Step 1: Compute Regression Equation

The slope formula (mm) is: m=(xy)xyn(x2)(x)2nm = \frac{\sum(xy) - \frac{\sum x \sum y}{n}}{\sum(x^2) - \frac{(\sum x)^2}{n}} The y-intercept (bb) is: b=yˉmxˉb = \bar{y} - m \bar{x} Where:

  • nn: Number of data points
  • xˉ\bar{x}: Mean of xx values
  • yˉ\bar{y}: Mean of yy values

Step 2: Calculate Prediction

Substitute x=2.4x = 2.4 into the regression equation to find the predicted weight.

Step 3: Evaluate Validity of Prediction

The prediction is valid if (x = as np import pandas as pd

Data from the image

x = np.array([7.3, 7.4, 9.7, 9.3, 8.8, 8.5]) # Overhead Width (cm) y = np.array([139, 174, 268, 224, 224, 218]) # Weight (kg)

Number of data points

n = len(x)

Calculating means

mean_x = np.mean(x) mean_y = np.mean(y)

Calculating slope (m) and y-intercept (b)

m = (np.sum(x * y) - (np.sum(x) * np.sum(y)) / n) / (np.sum(x**2) - (np.sum(x)**2) / n) b = mean_y - m * mean_x

Regression equation

regression_equation = f"y = {round(b)} + {round(m, 1)}x"

Predicting weight for overhead width of 2.4 cm

predicted_weight = b + m * 2.4

Checking if prediction is valid (extrapolation check)

is_valid = 2.4 >= min(x) and 2.4 <= max(x)

regression_equation, predicted1. The regression equation is: y^=162+43.5x\hat{y} = -162 + 43.5x

  1. The predicted weight for an overhead width of 2.42.4 cm is approximately 57.7kg-57.7 \, \text{kg}.

  2. The prediction is not valid, as 2.4cm2.4 \, \text{cm} falls outside the range of the given overhead widths (7.3cm7.3 \, \text{cm} to 9.7cm9.7 \, \text{cm}). This is an example of extrapolation, which is unreliable.

Would you like further clarifications or step-by-step details?

Here are 5 questions to explore further:

  1. What does extrapolation mean, and why is it unreliable?
  2. How would increasing the range of data improve predictions?
  3. Can a seal realistically have a negative weight, and how should we interpret such results?
  4. What methods could be used to assess the significance of the regression model?
  5. Would transformations (e.g., logarithmic) of the data yield better predictions?

Tip: Always check whether the input values for prediction fall within

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Math Problem Analysis

Mathematical Concepts

Linear Regression
Statistics
Prediction

Formulas

Slope formula: m = (Σxy - (Σx)(Σy)/n) / (Σx^2 - (Σx)^2/n)
Regression line: ŷ = b + mx
Mean formula: mean = Σ(values)/n

Theorems

Linear Regression Model
Extrapolation Validity

Suitable Grade Level

Grade 10-12 (Statistics/High School Math)