To find the linear regression line in the form $y = mx + b$, we can use the least-squares method. The steps involve calculating the slope ($m$) and the y-intercept ($b$).

Given dataTo find the linear regression line in the form $y = mx + b$, we can use the least-squares method. The steps involve calculating the slope ($m$) and the y-intercept ($b$).

Given

$X = [1, 2, 3, 4, 5, 6]$ $Y = [7.97, 7.86, 11.3, 10, 12.58, 15.41]$

1. Calculate the mean of $X$ and $Y$:

   $\bar{X} = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5$ $\bar{Y} = \frac{7.97 + 7.86 + 11.3 + 10 + 12.58 + 15.41}{6} \approx 10.52$

2. Calculate the slope $m$:

   $m = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{\sum (X_i - \bar{X})^2}$

3. Calculate the y-intercept $b$:

   $b = \bar{Y} - m \cdot \bar{X}$

Let's go through these calculations.

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Calculations:

1. Calculate $\sum (X_i - \bar{X})(Y_i - \bar{Y})$:

   $(X - \bar{X}) = [-2.5, -1.5, -0.5, 0.5, 1.5, 2.5]$ $(Y - \bar{Y}) = [-2.55, -2.66, 0.78, -0.52, 2.06, 4.89]$

   \approx 24.37$$

2. Calculate $\sum (X_i - \bar{X})^2$:

   = 17.5$$

3. Slope $m$:

   $m = \frac{24.37}{17.5} \approx 1.39$

4. Intercept $b$:

   $b = 10.52 - (1.39 \cdot 3.5) \approx 5.67$

Final Equation

The linear regression line is:

$y = 1.39x + 5.67$

Would you like further details or explanations?

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Related Questions

1. How is the least-squares method derived?
2. How would the linear regression change if one data point were removed?
3. What are residuals in the context of linear regression, and how are they calculated?
4. What other methods besides least-squares can be used for linear regression?
5. How do we interpret the slope and y-intercept in real-life data?

Tip

Always check for outliers in your data, as they can significantly impact the regression line, especially in small datasets.