Questions: The table below gives the number of hours spent unsupervised each day as well as the overall grade averages for seven randomly selected middle school students. Using this data, consider the equation of the regression line, ŷ = b0 + b1 x, for predicting the overall grade average for a middle school student based on the number of hours spent unsupervised each day. Keep in mind, the correlation coefficient may or may not be statistically significant for the data given. Remember, in practice, it would not be appropriate to use the regression line to make a prediction if the correlation coefficient is not statistically significant. Hours Unsupervised 0 0.5 1 2 4.5 5 6 Overall Grades 100 96 88 73 71 68 67 Table Step 1 of 6: Find the estimated slope. Round your answer to three decimal places.

Solution Steps

To find the slope of the regression line, we first calculate the means of the given data:

The mean of \( x \) (hours unsupervised) is: \[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i = 2.714 \]

The mean of \( y \) (overall grades) is: \[ \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i = 80.429 \]

The correlation coefficient \( r \) is given as: \[ r = -0.917 \]

The slope \( \beta \) of the regression line is calculated using the formula: \[ \beta = \frac{\sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y}}{\sum_{i=1}^{n} x_i^2 - n \bar{x}^2} \]

Substituting the given values:

• Numerator: \(\sum_{i=1}^{n} x_i y_i - n \bar{x} \bar{y} = 1343.5 - 7 \times 2.714 \times 80.429 = -184.643\)
• Denominator: \(\sum_{i=1}^{n} x_i^2 - n \bar{x}^2 = 86.5 - 7 \times 2.714^2 = 34.929\)

Thus, the slope is: \[ \beta = \frac{-184.643}{34.929} = -5.286 \]

The intercept \( \alpha \) is calculated using the formula: \[ \alpha = \bar{y} - \beta \bar{x} \]

Substituting the values: \[ \alpha = 80.429 - (-5.286 \times 2.714) = 94.777 \]

Final Answer