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Statistics and Probability - Regression

Grade 12IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Bivariate data involves two variables, usually denoted as xx (independent/explanatory variable) and yy (dependent/response variable).

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A scatter diagram is used to visualize the relationship between xx and yy. The pattern can be linear, non-linear, or show no correlation.

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Pearson’s product-moment correlation coefficient (rr) measures the strength and direction of the linear relationship between two variables. Its value ranges from βˆ’1-1 to 11.

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Interpretation of rr: r=1r = 1 is perfect positive correlation, r=βˆ’1r = -1 is perfect negative correlation, and r=0r = 0 indicates no linear correlation.

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The Least Squares Regression Line is the line of best fit that minimizes the sum of the squares of the vertical residuals. Its equation is generally written as y=ax+by = ax + b.

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The regression line of yy on xx always passes through the mean point (xˉ,yˉ)(\bar{x}, \bar{y}).

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Interpolation is the process of predicting a yy-value for an xx-value within the range of the given data. This is generally considered reliable.

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Extrapolation is predicting a yy-value for an xx-value outside the range of the given data. This is often unreliable as the linear trend may not continue.

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The gradient aa represents the predicted change in yy for every one-unit increase in xx.

πŸ“Formulae

y=ax+by = ax + b

xΛ‰=βˆ‘i=1nxin\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}

yΛ‰=βˆ‘i=1nyin\bar{y} = \frac{\sum_{i=1}^{n} y_i}{n}

βˆ’1≀r≀1-1 \le r \le 1

yβˆ’yΛ‰=a(xβˆ’xΛ‰)y - \bar{y} = a(x - \bar{x})

πŸ’‘Examples

Problem 1:

A study finds the relationship between the number of hours spent studying (xx) and the score on a math test (yy). The mean study time is xˉ=6\bar{x} = 6 hours and the mean score is yˉ=72\bar{y} = 72. The gradient of the regression line of yy on xx is a=4.5a = 4.5. Find the equation of the regression line and predict the score for a student who studies for 88 hours.

Solution:

  1. Use the mean point (xΛ‰,yΛ‰)(\bar{x}, \bar{y}) in the equation y=ax+by = ax + b: 72=4.5(6)+b72 = 4.5(6) + b 72=27+b72 = 27 + b b=72βˆ’27=45b = 72 - 27 = 45
  2. The regression equation is y=4.5x+45y = 4.5x + 45.
  3. For x=8x = 8: y=4.5(8)+45y = 4.5(8) + 45 y=36+45=81y = 36 + 45 = 81

Explanation:

Since the regression line must pass through the mean point (xˉ,yˉ)(\bar{x}, \bar{y}), we substitute these values along with the gradient aa to find the y-intercept bb. Then, we substitute x=8x = 8 into the resulting equation to find the predicted score.

Problem 2:

Given a set of data where the correlation coefficient is r=βˆ’0.85r = -0.85, describe the relationship between the variables xx and yy. If the regression line is y=βˆ’2.5x+100y = -2.5x + 100, what is the predicted change in yy if xx increases by 44 units?

Solution:

  1. The relationship is a strong negative linear correlation because rr is close to βˆ’1-1.
  2. The gradient a=βˆ’2.5a = -2.5 represents the change in yy for an increase of 11 unit in xx.
  3. Change in y=aΓ—Ξ”xy = a \times \Delta x: Ξ”y=βˆ’2.5Γ—4=βˆ’10\Delta y = -2.5 \times 4 = -10

Explanation:

The value r=βˆ’0.85r = -0.85 indicates that as xx increases, yy tends to decrease significantly. The gradient of the regression line tells us the rate of change; multiplying the gradient by the change in xx gives the total predicted change in yy.