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A mathematics teacher wanted to see the correlation between test scores and homework. The homework grade ([tex]\( x \)[/tex]) and test grade ([tex]\( y \)[/tex]) are given in the accompanying table.

1. Write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth.
2. Using this equation, estimate the homework grade, to the nearest integer, for a student with a test grade of 42.

\begin{tabular}{|c|c|}
\hline
Homework Grade ([tex]\( x \)[/tex]) & Test Grade ([tex]\( y \)[/tex]) \\
\hline
61 & 49 \\
\hline
81 & 67 \\
\hline
86 & 86 \\
\hline
70 & 56 \\
\hline
74 & 66 \\
\hline
74 & 77 \\
\hline
55 & 52 \\
\hline
\end{tabular}



Answer :

GinnyAnswer
To find the linear regression equation that represents the given set of data, we will follow these steps:

1. Define Variables:
- Let [tex]\( x \)[/tex] represent the homework grade.
- Let [tex]\( y \)[/tex] represent the test grade.

2. Given Data:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 61 & 49 \\ 81 & 67 \\ 86 & 86 \\ 70 & 56 \\ 74 & 66 \\ 74 & 77 \\ 55 & 52 \\ \hline \end{array} \][/tex]

3. Calculate the Means [tex]\( \bar{x} \)[/tex] and [tex]\( \bar{y} \)[/tex]:
[tex]\[ \bar{x} = \frac{61 + 81 + 86 + 70 + 74 + 74 + 55}{7} = \frac{501}{7} \approx 71.6 \][/tex]
[tex]\[ \bar{y} = \frac{49 + 67 + 86 + 56 + 66 + 77 + 52}{7} = \frac{453}{7} \approx 64.7 \][/tex]

4. Calculate the Slope [tex]\( m \)[/tex]:
[tex]\[ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \][/tex]
To find [tex]\( \sum (x_i - \bar{x})(y_i - \bar{y}) \)[/tex] and [tex]\( \sum (x_i - \bar{x})^2 \)[/tex]:

[tex]\[ \sum (x_i - \bar{x})(y_i - \bar{y}) = (61 - 71.6)(49 - 64.7) + (81 - 71.6)(67 - 64.7) + (86 - 71.6)(86 - 64.7) + (70 - 71.6)(56 - 64.7) + (74 - 71.6)(66 - 64.7) + (74 - 71.6)(77 - 64.7) + (55 - 71.6)(52 - 64.7) \][/tex]
[tex]\[ = (-10.6)(-15.7) + 9.4(2.3) + 14.4(21.3) + (-1.6)(-8.7) + 2.4(1.3) + 2.4(12.3) + (-16.6)(-12.7) \][/tex]
[tex]\[ \approx 166.62 + 21.62 + 306.72 + 13.92 + 3.12 + 29.52 + 210.82 = 752.34 \][/tex]

[tex]\[ \sum (x_i - \bar{x})^2 = (61 - 71.6)^2 + (81 - 71.6)^2 + (86 - 71.6)^2 + (70 - 71.6)^2 + (74 - 71.6)^2 + (74 - 71.6)^2 + (55 - 71.6)^2 \][/tex]
[tex]\[ = (-10.6)^2 + 9.4^2 + 14.4^2 + (-1.6)^2 + 2.4^2 + 2.4^2 + (-16.6)^2 \][/tex]
[tex]\[ = 112.36 + 88.36 + 207.36 + 2.56 + 5.76 + 5.76 + 275.56 = 697.68 \][/tex]

[tex]\[ m = \frac{752.34}{697.68} \approx 1.08 \approx 1.1 \, (\text{Rounded to nearest tenth}) \][/tex]

5. Calculate the Intercept [tex]\( b \)[/tex]:
[tex]\[ b = \bar{y} - m \bar{x} \][/tex]
[tex]\[ b = 64.7 - 1.1 \times 71.6 = 64.7 - 78.76 = -14.06 \approx -14.1 \, (\text{Rounded to nearest tenth}) \][/tex]

6. Linear Regression Equation:
[tex]\[ y = 1.1x - 14.1 \][/tex]

7. Estimate Homework Grade for Test Grade of 42:
We need to find [tex]\( x \)[/tex] such that [tex]\( y = 42 \)[/tex]:
[tex]\[ 42 = 1.1x - 14.1 \][/tex]
[tex]\[ 1.1x = 42 + 14.1 \][/tex]
[tex]\[ 1.1x = 56.1 \][/tex]
[tex]\[ x = \frac{56.1}{1.1} \approx 51 \][/tex]

So, the linear regression equation is [tex]\( y = 1.1x - 14.1 \)[/tex] and the estimated homework grade for a test grade of 42 is approximately [tex]\( 51 \)[/tex].

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