Correlation Regression Numericals Page 5

Problem 1
Simple Linear Regression (Formula Method)

In this problem let us try to find the relation between Exam Score y and Study hours x.
Then we will interpret the regression equation, the correlation coefficient R, and coefficientof determination R squared.
This process is called a Basic Regression Analysis

Question

A teacher wants to check if there is any connection between the number of hours a student prepares for an exam with their score. He records study hours and exam score.
Fit the regression line of exam score y on study hours x. Interpret the regression equation, R, and R squared.

Independent variable

Study hours, x

Dependent variable

Exam score, y

Observed data

x (hours)	1	2	3	4	5	6	7	8
y (score)	50	54	57	61	62	67	69	73

Solution

Step 1 Calculate The Sums (Use a Calculator in STAT mode)

Σx = 36 Σy = 493 Σx² = 204 Σxy = 2352 n = 8

Step 2 compute the regression coefficient which is for a mathematician the slope (b_yx)

$$ b_{yx} = \frac{n\Sigma xy - (\Sigma x)(\Sigma y)}{n\Sigma x^{2} - (\Sigma x)^{2}} $$ $$ b_{yx} = \frac{8(2352) - 36(493)}{8(204) - 36^{2}} $$ $$ b_{yx} = \frac{18816 - 17748}{1632 - 1296} $$ $$ b_{yx} = \frac{1068}{336} = 3.179 $$

Step 3 Now Mean Values Of x and y

$$ \bar{x} = \frac{\Sigma x}{n} = \frac{36}{8} = 4.500 $$ $$ \bar{y} = \frac{\Sigma y}{n} = \frac{493}{8} = 61.625 $$

Step 4 Now Plugin to the formulae and simplify to get the regression equation

y − ȳ = b_yx(x − x̄) (centred form)

$$ y - 61.625 = 3.179(x - 4.500) $$ $$ y = 47.321 + 3.179x $$

Final regression equation (rounded to 3 decimals)

ŷ = 47.321 + 3.179x

Correlation coefficient and coefficient of determination

Karl Pearson’s formula

    $$ R =
    \frac{n\Sigma xy - (\Sigma x)(\Sigma y)}
    {\sqrt{\left[n\Sigma x^{2} - (\Sigma x)^{2}\right]\left[n\Sigma y^{2} - (\Sigma y)^{2}\right]}} $$

      First find Σy² = 30809
      (50² + 54² + 57² + 61² + 62² + 67² + 69² + 73²)
    
    $$ R =
    \frac{8(2352) - 36(493)}
    {\sqrt{\left[8(204) - 36^{2}\right]\left[8(30809) - 493^{2}\right]}} $$

    $$ R =
    \frac{18816 - 17748}
    {\sqrt{\left[1632 - 1296\right]\left[246472 - 243049\right]}} $$

    $$ R = \frac{1068}{\sqrt{(336)(3423)}} = 0.99586 $$

      R² = (0.99586)² = 0.99174
      which is 99.174 percent

Interpretation

The slope (regression coefficient) gives the average change in score for a one hour increase in study time. Here it is about 3.179 marks per hour.That means, every extra hour the student studies, the score increases by 3.179

The value of R is positive and very close to 1, so the relationship is strongly positive and close to linear.

The value of R squared gives us the amount of influence of study time in exam scores is explained by this linear model is about 99.174 percent.

Available Formula Sheets