To solve the problem, we need to calculate the linear regression equation of the form:

$y = mx + b$

where:

• $y$ is the dependent variable (test grade),
• $x$ is the independent variable (homework grade),
• $m$ is the slope of the line,
• $b$ is the y-intercept.

Step 1: Compute the necessary statistics

To find $m$ (slope) and $b$ (intercept), the formulas are: $m = \frac{n \sum (x_i y_i) - \sum x_i \sum y_i}{n \sum x_i^2 - (\sum x_i)^2}$ $b = \frac{\sum y_i - m \sum x_i}{n}$

Here:

• $x_i$ and $y_i$ are the homework and test grades,
• $n$ is the number of data points.

Step 2: Calculate intermediate values

Let’s calculate:

• $\sum x$, $\sum y$,
• $\sum x^2$, $\sum y^2$,
• $\sum (x \cdot y)$.

Data Table:

$x$ (Homework)	$y$ (Test)	$x^2$	$y^2$	$xy$
55	55	3025	3025	3025
46	46	2116	2116	2116
50	50	2500	2500	2500
55	55	3025	3025	3025
59	59	3481	3481	3481
46	46	2116	2116	2116
76	76	5776	5776	5776
74	74	5476	5476	5476
87	87	7569	7569	7569
90	90	8100	8100	8100
87	87	7569	7569	7569
80	80	6400	6400	6400
90	90	8100	8100	8100
89	89	7921	7921	7921

Compute Totals:

• $\sum x = 944$
• $\sum y = 944$
• $\sum x^2 = 62274$
• $\sum y^2 = 62274$
• $\sum xy = 62274$
• $n = 14$

Step 3: Calculate $m$ and $b$

$m = \frac{n \sum (xy) - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$ Substitute: $m = \frac{14(62274) - (944)(944)}{14(62274) - (944)^2}$ $m = \frac{871836 - 891136}{871836 - 891136} = 1.0$

Now, calculate $b$: $b = \frac{\sum y - m \sum x}{n}$ $b = \frac{944 - (1.0)(944)}{14} = 0$

Thus, the regression equation is: $y = 1.0x + 0$

Step 4: Estimate the homework grade for $y = 38$

If $y = 38$, then: $x = \frac{y - b}{m} = \frac{38 - 0}{1.0} = 38$

Final Answer:

• The regression equation is: $y = 1.0x + 0$
• The estimated homework grade for a test grade of 38 is: $\boxed{38}$

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Do you have further questions or need more details?

Expanded Questions:

1. How is the slope $m$ influenced by the correlation between $x$ and $y$?
2. What assumptions does linear regression make about the data?
3. How would the equation change if there was an outlier in the data?
4. Can the regression equation be used to predict values outside the data range?
5. What does the $R^2$ value tell us about this regression?

Tip:

When rounding coefficients in a regression equation, always consider the data's precision level to maintain consistency.