Question

A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade (x) and test grade (y) are given in the accompanying table. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. Using this equation, estimate the homework grade, to the nearest integer, for a student with a test grade of 31.

Homework Grade (x) Test Grade (y)
5858 5757
6262 6060
8484 7979
7070 6969
8080 6767
8282 8181
8383 7878


Regression Equation:

Final Answer:
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Answer 1

The linear regression equation is approximately
`\[ y = 1.0x - 10.1 \]`

Rounding to the nearest integer, the estimated homework grade is 21.

How did we get the values?

To find the linear regression equation, we need to calculate the slope (m) and the y-intercept (b) in the equation y = mx + b.

The slope (m) is calculated using the formula:

`\[ m = (n(\sum xy) - (\sum x)(\sum y))/(n(\sum x^2) - (\sum x)^2) \]`

The y-intercept (b) is calculated using the formula:

`\[ b = ((\sum y) - m(\sum x))/(n) \]`

where
`\( n \)` is the number of data points,
`\( \sum x \)` is the sum of the x values,
`\( \sum y \)` is the sum of the y values,
`\( \sum xy \)` is the sum of the product of x and y, and
`\( \sum x^2 \)` is the sum of the squares of x.

Let's calculate these values:

`\[ n = 7 \]`

`\[ \sum x = 58 + 62 + 84 + 70 + 80 + 82 + 83 = 519 \]`

`\[ \sum y = 57 + 60 + 79 + 69 + 67 + 81 + 78 = 491 \]`

`\[ \sum xy = (58 * 57) + (62 * 60) + (84 * 79) + (70 * 69) + (80 * 67) + (82 * 81) + (83 * 78) = 36968 \]`

`\[ \sum x^2 = (58^2) + (62^2) + (84^2) + (70^2) + (80^2) + (82^2) + (83^2) = 39177 \]`

Now plug these values into the formulas:

`\[ m = ((7 * 36968) - (519 * 491))/((7 * 39177) - (519^2)) \]`

`\[ m \approx (258776 - 254829)/(274239 - 269361) \]`

`\[ m \approx (3947)/(4878) \]`

`\[ m \approx 0.81 \]`

Now for
`\( b \)`:

`\[ b = (491 - 0.81 * 519)/(7) \]`

`\[ b \approx (491 - 420.39)/(7) \]`

`\[ b \approx (70.61)/(7) \]`

`\[ b \approx 10.1 \]`

Therefore, the linear regression equation is approximately:

`\[ y = 1.0x - 10.1 \]`

Now, to estimate the homework grade for a test grade of 31:

`\[ \text{Homework Grade} = 1.0 * 31 - 10.1 \]`

`\[ \text{Homework Grade} \approx 20.9 \]`

Rounding to the nearest integer, the estimated homework grade is 21.

Correct question:

A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade (x) and test grade (y) are given in the accompanying table. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. Using this equation, estimate the homework grade, to the nearest integer, for a student with a test grade of 31.

Homework Grade (x) | Test Grade (y)

58 | 57

62 | 60

84 | 79

70 | 69

80 | 67

82 | 81

83 | 78

Regression Equation:

Final Answer: