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Find the regression equation for predicting final score from midterm score, based on the following information:

– Average Midterm Score = 70

– Standard Deviation of Midterm Score = 10

– Average Final Score = 70

– Standard Deviation of Final Score = 20

– Correlation Coefficient of Final Score = 0.60

The aim of this question is to use the linear regression model to find the dependence of one variable on the other and then apply this model for prediction.

The linear regression model relating a variable x with a variable y can be defined by the following formula:

\[ y \ = \ m x \ + \ c \]

The slope and intercept used in above model can be calculated using the following formula:

\[ \text{ Slope } = \ m \ = r \ \dfrac{ \sigma_{ y } }{ \sigma_{ x } } \]

\[ \text{ y-intercept } = \ c \ = \ \mu_{ y} \ – \ m \mu_{ x } \]

Expert Answer

Let’s call the midterm score $ x $, which is the independent variable, while the final score $ y $ is the dependent variable. In this case, the given data may be represented as follows:

\[ \text{ Average Midterm Score } = \ \mu_{ x } \ = \  70 \]

\[ \text{ Standard Deviation of Midterm Score } = \ \sigma_{ x } \ = \ 10 \]

\[ \text{ Average Final Score } = \ \mu_{ y } \ = \ 70 \]

\[ \text{ Standard Deviation of Final Score } = \ \sigma_{ y } \ = \ 20 \]

\[ \text{ Correlation Coefficient of Final Score } = \ r \ = \ 0.60 \]

For the case of linear regression, the slope of the equation can be calculated using the following formula:

\[ \text{ Slope } = \ m \ = r \ \dfrac{ \sigma_{ y } }{ \sigma_{ x } } \]

Substituting values in the above equation:

\[ m \ = 0.6 \ \dfrac{ 20 }{ 10 } \]

\[ m \ = 0.6 \times 2 \]

\[ m \ = 1.2 \]

For the case of linear regression, the y-intercept of the equation can be calculated using the following formula:

\[ \text{ y-intercept } = \ c \ = \ \mu_{ y} \ – \ m \mu_{ x } \]

Substituting values in the above equation:

\[ \text{ y-intercept } = \ c \ = \ 55 \ – \ ( 1.2 ) ( 70 ) \]

\[ \text{ y-intercept } = \ c \ = \ 55 \ – \ 84 \]

\[ \text{ y-intercept } = \ c \ = \ -29 \]

So the final equation of linear regression is:

\[ y \ = \ m x \ + \ c \]

Substituting values in the above equation:

\[ y \ = \ 1.2 x \ – \ 29 \]

Which is the required result.

Numerical Result

\[ y \ = \ 1.2 x \ – \ 29 \]

Example

Using the above regression equation, find the final score of a student that scored 50 marks in the mid term.

Given:

\[ x \ = \ 50 \]

Recall the linear regression equation:

\[ y \ = \ 1.2 x \ – \ 29 \]

Substituting the value of $ x $:

\[ y \ = \ 1.2 ( 50 ) \ – \ 29 \]

\[ y \ = \ 60 \ – \ 29 \]

\[ y \ = \ 31 \]

Which is the required result.

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