To solve this problem, we need to go step-by-step through the process of finding the regression line for predicting overall grades based on the hours spent unsupervised each day.

Step 1: Calculate the Mean of X and Y

The given data is:

Hours Unsupervised (X)	Overall Grades (Y)
1	93
2	88
3	80
4	72
5	68

The formula for the regression line is:

$\hat{y} = b_0 + b_1x$

Where:

• $b_0$ is the y-intercept,
• $b_1$ is the slope of the regression line,
• $x$ is the independent variable (Hours Unsupervised),
• $\hat{y}$ is the predicted value of the dependent variable (Overall Grades).

Step 1.1: Calculate the Means of X and Y

The mean of $X$ (hours unsupervised) is:

$\bar{X} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3$

The mean of $Y$ (overall grades) is:

$\bar{Y} = \frac{93 + 88 + 80 + 72 + 68}{5} = \frac{401}{5} = 80.2$

Step 1.2: Calculate the Slope (b1)

The formula for $b_1$, the slope, is:

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

First, let's compute the required sums:

$\sum (x_i - \bar{X})(y_i - \bar{Y}) = (1 - 3)(93 - 80.2) + (2 - 3)(88 - 80.2) + (3 - 3)(80 - 80.2) + (4 - 3)(72 - 80.2) + (5 - 3)(68 - 80.2)$

$= (-2)(12.8) + (-1)(7.8) + (0)(-0.2) + (1)(-8.2) + (2)(-12.2)$

$= -25.6 - 7.8 + 0 - 8.2 - 24.4 = -66$

Now, calculate:

$\sum (x_i - \bar{X})^2 = (1 - 3)^2 + (2 - 3)^2 + (3 - 3)^2 + (4 - 3)^2 + (5 - 3)^2$

$= (-2)^2 + (-1)^2 + (0)^2 + (1)^2 + (2)^2 = 4 + 1 + 0 + 1 + 4 = 10$

Thus, the slope $b_1$ is:

$b_1 = \frac{-66}{10} = -6.6$

Step 1.3: Calculate the Y-Intercept (b0)

Now that we have the slope $b_1 = -6.6$, we can calculate the y-intercept $b_0$ using the formula:

$b_0 = \bar{Y} - b_1\bar{X}$

Substituting the values:

$b_0 = 80.2 - (-6.6)(3) = 80.2 + 19.8 = 100$

Step 2: The Regression Equation

Now, we have the equation of the regression line:

$\hat{y} = 100 - 6.6x$

This is the linear model that predicts the overall grade average based on the number of hours spent unsupervised each day.

Step 3: Interpretation of the Slope

In this regression model, the slope $b_1 = -6.6$ tells us that for every additional hour spent unsupervised, the overall grade average is predicted to decrease by 6.6 points.

So, if the value of the independent variable (hours unsupervised) increases by 1 unit, the dependent variable (overall grade) is expected to decrease by 6.6 points.

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Would you like me to go into more detail about any part of the process? Feel free to ask!

Here are some follow-up questions to help you dive deeper:

1. How can you calculate the correlation coefficient for this dataset, and why is it important?
2. What would happen if we added more data points to this dataset? How might that affect the regression model?
3. What is the significance of the y-intercept in this model? Does it make sense in the context of this problem?
4. If a student spends 6 hours unsupervised, what would their predicted grade be according to the model?
5. How could you assess whether the regression model is a good fit for the data?

Tip: Always double-check the assumptions of linear regression (e.g., linearity, independence, homoscedasticity, normality of errors) before using the model for predictions!