To determine the equation of the regression line ($y = b_0 + b_1x$) for predicting grades ($y$) based on hours unsupervised ($x$), we proceed step by step:

Step 1: Organize the given data

Step 2: Recall the formulas

1. The slope of the regression line ($b_1$) is: $b_1 = \frac{n\sum{xy} - (\sum{x})(\sum{y})}{n\sum{x^2} - (\sum{x})^2}$
2. The intercept ($b_0$) is: $b_0 = \bar{y} - b_1\bar{x}$ where $\bar{x}$ and $\bar{y}$ are the means of $x$ and $y$.

Step 3: Perform calculations

1. Find the necessary sums:

To compute $\sum{x}, \sum{y}, \sum{xy}, \sum{x^2}$:

Now, calculate totals:

$\sum{x} = 0.5 + 1 + 1.5 + 2.5 + 3 + 4.5 + 5 = 18$ $\sum{y} = 94 + 87 + 85 + 79 + 74 + 72 + 69 = 560$ $\sum{x^2} = 0.25 + 1 + 2.25 + 6.25 + 9 + 20.25 + 25 = 64$ $\sum{xy} = 47 + 87 + 127.5 + 197.5 + 222 + 324 + 345 = 1350$ Number of data points $n = 7$.

2. Calculate $b_1$ (slope):

Substitute values into the slope formula: $b_1 = \frac{n\sum{xy} - (\sum{x})(\sum{y})}{n\sum{x^2} - (\sum{x})^2}$ $b_1 = \frac{7(1350) - (18)(560)}{7(64) - (18)^2}$ Simplify step by step:

• $7(1350) = 9450$
• $(18)(560) = 10080$
• $7(64) = 448$
• $(18)^2 = 324$

$b_1 = \frac{9450 - 10080}{448 - 324} = \frac{-630}{124} \approx -5.08$

3. Calculate $b_0$ (intercept):

Using: $b_0 = \bar{y} - b_1\bar{x}$ First, find the means: $\bar{x} = \frac{\sum{x}}{n} = \frac{18}{7} \approx 2.57, \quad \bar{y} = \frac{\sum{y}}{n} = \frac{560}{7} = 80$ Substitute: $b_0 = 80 - (-5.08)(2.57)$ Simplify:

• $-5.08 \times 2.57 \approx -13.06$ $b_0 = 80 + 13.06 \approx 93.06$

Step 4: Write the regression equation

The regression line is: $y = 93.06 - 5.08x$

Step 5: Interpret the slope

The slope ($b_1 = -5.08$) indicates that for every additional hour unsupervised, the overall grade decreases by approximately 5.08 points.

Let me know if you want further details or calculations!

5 Related Questions:

1. How do you determine the correlation coefficient ($r$) from the data?
2. What does a negative slope indicate in a real-world context?
3. How can you check if the regression line is statistically significant?
4. Can you predict a grade for 4 hours unsupervised using the regression line?
5. How would the regression change if one outlier data point were removed?

Tip:

Always check the correlation coefficient $r$ before using a regression line to ensure the relationship is strong and meaningful.