To find the prediction equation using the given data, we need to perform a simple linear regression where we predict one variable (dependent variable $y$) from another variable (independent variable $x$).

Given:

• $x$ (independent variable) = Study Time (hours)
• $y$ (dependent variable) = GPA

Step 1: Calculate the Means of $x$ and $y$

First, calculate the means of Study Time ($\bar{x}$) and GPA ($\bar{y}$):

$\bar{x} = \frac{\sum x_i}{n} = \frac{14 + 24 + 16 + 6 + 11 + 12 + 4 + 22}{8} = \frac{109}{8} = 13.625$

$\bar{y} = \frac{\sum y_i}{n} = \frac{2.8 + 3.7 + 3.5 + 3.2 + 3.2 + 3.2 + 2.8 + 3.8}{8} = \frac{26.2}{8} = 3.275$

Step 2: Calculate the Slope $b_1$

The slope $b_1$ is calculated using the formula:

$b_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$

Let's compute the components:

$\sum (x_i - \bar{x})(y_i - \bar{y})$

$= (14 - 13.625)(2.8 - 3.275) + (24 - 13.625)(3.7 - 3.275) + \dots + (22 - 13.625)(3.8 - 3.275)$

$= 0.375(-0.475) + 10.375(0.425) + 2.375(0.225) + (-7.625)(-0.075) + (-2.625)(-0.075) + (-1.625)(-0.075) + (-9.625)(-0.475) + 8.375(0.525)$

$\approx -0.1781 + 4.4069 + 0.5334 + 0.5719 + 0.1969 + 0.1219 + 4.5719 + 4.3988 = 14.6236$

$\sum (x_i - \bar{x})^2$

$= (14 - 13.625)^2 + (24 - 13.625)^2 + \dots + (22 - 13.625)^2$

$= 0.1406 + 107.1406 + 5.6406 + 58.1406 + 6.8906 + 2.6406 + 92.6406 + 70.1406 = 343.3725$

Now, calculate the slope $b_1$:

$b_1 = \frac{14.6236}{343.3725} \approx 0.0426$

Step 3: Calculate the Y-Intercept $b_0$

The y-intercept $b_0$ is calculated as:

$b_0 = \bar{y} - b_1\bar{x}$

$b_0 = 3.275 - 0.0426(13.625) \approx 3.275 - 0.5804 = 2.6946$

Step 4: Write the Prediction Equation

The prediction equation is:

$\hat{y} = 2.695 + 0.043x$

Interpretation of the Slope

The slope $b_1 = 0.043$ means that for each additional hour a student studies, their GPA is predicted to increase by 0.043 points.

Final Answer:

The prediction equation is:

$\hat{y} = 2.695 + 0.043x$

Would you like further details or have any more questions?

Here are some related questions:

1. What would be the predicted GPA for a student who studies 15 hours per week?
2. How would a decrease in study time affect the predicted GPA?
3. Can you explain the practical significance of a small slope like 0.043?
4. How can the accuracy of this prediction be assessed?
5. What is the y-intercept's practical meaning in this context?
6. What would happen to the prediction equation if another independent variable (e.g., skipped classes) were added?
7. How might outliers in the data affect the prediction equation?
8. How does the slope in this equation compare to the slope in a similar study with different data?

Tip: Always interpret the slope of a regression equation in the context of the data to understand its practical implications.