The following table presents the number of hours a group of school students played video games during the weekends and the test scores attained by each of them in a test, the following Monday. \begin{tabular}{|c|c|} \hline Time (in hours) & Test score \\ \hline 0 & 96 \\ \hline 1 & 85 \\ \hline 2 & 82 \\ \hline 3 & 74 \\ \hline 3 & 95 \\ \hline 5 & 68 \\ \hline 5 & 76 \\ \hline 5 & 84 \\ \hline 6 & 58 \\ \hline 7 & 65 \\ \hline 7 & 75 \\ \hline 10 & 50 \\ \hline \end{tabular} (a.) It is believed that a linear relationship exists between the time spent on playing video games and test score attained. Find out the strength of this linear relationship. (b.) Estimate the line of best fit in the scenario. Use this line to find the expected test score for a student who plays video games for 9 hours.

Part (a): Finding the Strength of the Linear Relationship

Step 1: Calculating the Means

First, we calculate the mean of the hours played ((\bar{x})) and the mean of the test scores ((\bar{y})).

• Mean of hours played ((\bar{x})):
  [
  \bar{x} = \frac{\sum x_i}{n} = \frac{0 + 1 + 2 + 3 + 3 + 5 + 5 + 5 + 6 + 7 + 7 + 10}{12} = \frac{54}{12} = 4.5
  ]

• Mean of test scores ((\bar{y})):
  [
  \bar{y} = \frac{\sum y_i}{n} = \frac{96 + 85 + 82 + 74 + 95 + 68 + 76 + 84 + 58 + 65 + 75 + 50}{12} = \frac{908}{12} = 75.6667
  ]

Step 2: Using Assumed Means

Since (\bar{x} = 4.5) and (\bar{y} = 75.6667) are not integers, we use assumed means (A = 5) and (B = 76), respectively.

Step 3: Calculating Deviations

We calculate the deviations from the assumed means ((d x = x – A) and (d y = y – B)) and their products and squares.

• The table of deviations and their products and squares is as follows:

[
\begin{array}{|c|c|c|c|c|c|c|}
\hline x & y & d x=x-A=x-5 & d y=y-B=y-76 & d x^2 & d y^2 & d x \cdot d y \
\hline 0 & 96 & -5 & 20 & 25 & 400 & -100 \
\hline 1 & 85 & -4 & 9 & 16 & 81 & -36 \
\hline 2 & 82 & -3 & 6 & 9 & 36 & -18 \
\hline 3 & 74 & -2 & -2 & 4 & 4 & 4 \
\hline 3 & 95 & -2 & 19 & 4 & 361 & -38 \
\hline 5 & 68 & 0 & -8 & 0 & 64 & 0 \
\hline 5 & 76 & 0 & 0 & 0 & 0 & 0 \
\hline 5 & 84 & 0 & 8 & 0 & 64 & 0 \
\hline 6 & 58 & 1 & -18 & 1 & 324 & -18 \
\hline 7 & 65 & 2 & -11 & 4 & 121 & -22 \
\hline 7 & 75 & 2 & -1 & 4 & 1 & -2 \
\hline 10 & 50 & 5 & -26 & 25 & 676 & -130 \
\hline \text {— } & \text {— } & \text {— } & — & \text {— } & \text {— } & \text {— } \
\hline 54 & 908 & \sum d x=-6 & \sum d y=-4 & \sum d x^2=92 & \sum d y^2=2132 & \sum d x \cdot d y=-360 \
\hline
\end{array}
]

After calculating, we get:

[
\sum d x = -6, \quad \sum d y = -4, \quad \sum d x^2 = 92, \quad \sum d y^2 = 2132, \quad \sum d x \cdot d y = -360
]

Step 4: Calculating the Regression Coefficient

The regression coefficient ((b_{yx})) is calculated as follows:

[
b_{yx} = \frac{n \sum d x d y – (\sum d x)(\sum d y)}{n \sum d x^2 – (\sum d x)^2}
]

Substituting the values:

[
b_{yx} = \frac{12 \times -360 – (-6) \times -4}{12 \times 92 – (-6)^2} = \frac{-4320 – 24}{1104 – 36}
]

After calculating, we find:

[
b_{yx} = -4.0674
]

Part (b): Estimating the Line of Best Fit and Expected Test Score

Step 1: Formulating the Regression Line

The regression line of (y) on (x) is given by:

[
y – \bar{y} = b_{yx}(x – \bar{x})
]

Substituting the means and the regression coefficient:

[
y – 75.6667 = -4.0674(x – 4.5)
]

Expanding and rearranging:

[
y = -4.0674x + 18.3034 + 75.6667
]

Simplifying:

[
y = -4.0674x + 93.97
]

Step 2: Estimating the Test Score for 9 Hours of Gameplay

Now, we estimate the test score ((y)) for a student who plays video games for 9 hours ((x = 9)):

[
y = -4.0674 \times 9 + 93.97
]

After calculating, we find:

[
y = 57.3633
]

Summary

• The strength of the linear relationship between time spent on playing video games and test scores is represented by the regression coefficient (b_{yx} = -4.0674).
• The line of best fit is (y = -4.0674x + 93.97).
• For a student playing video games for 9 hours, the expected test score is approximately 57.36.