To find the estimated y-intercept $b_0$ of the regression line, we first need to determine the slope $b_1$ using the data and then use the formula for the y-intercept.

The regression line equation is given by: $\hat{y} = b_0 + b_1x$

Step 1: Calculate the slope $b_1$ using the formula:

$b_1 = \frac{n\sum{xy} - \sum{x}\sum{y}}{n\sum{x^2} - (\sum{x})^2}$

Where:

• $x$ = hours unsupervised
• $y$ = overall grades
• $n$ = number of data points = 5 in this case

Data:

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

Step 2: Compute the necessary sums:

1. $\sum{x} = 1 + 2 + 3 + 4 + 5 = 15$
2. $\sum{y} = 93 + 88 + 80 + 72 + 68 = 401$
3. $\sum{x^2} = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$
4. $\sum{xy} = (1)(93) + (2)(88) + (3)(80) + (4)(72) + (5)(68) = 93 + 176 + 240 + 288 + 340 = 1137$

Now, substitute these values into the formula for $b_1$:

$b_1 = \frac{5(1137) - (15)(401)}{5(55) - (15)^2}$ $b_1 = \frac{5685 - 6015}{275 - 225}$ $b_1 = \frac{-330}{50} = -6.6$

Step 3: Find the y-intercept $b_0$ using the formula:

$b_0 = \frac{\sum{y} - b_1\sum{x}}{n}$

Substitute the known values:

$b_0 = \frac{401 - (-6.6)(15)}{5}$ $b_0 = \frac{401 + 99}{5} = \frac{500}{5} = 100$

Conclusion:

The estimated y-intercept $b_0$ is 100.

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Would you like a more detailed explanation or additional steps for understanding? Here are some questions that might help expand your understanding:

1. What is the significance of the slope $b_1$ in this regression equation?
2. How do you interpret the y-intercept $b_0$ in the context of the problem?
3. How would you calculate the correlation coefficient $r$ for this data?
4. What does it mean if the correlation coefficient is not statistically significant in this case?
5. How would you use this regression line to predict a student's grade based on their unsupervised hours?

Tip: Always check the residuals (differences between observed and predicted values) to assess the quality of your regression model!