Question

this table shows the average amount of daily screen time, x, in hours,
for 5 students and the grade, y, each student currently has in their
math class.

based on the best - fit linear model for the data, which students grade is
closest to its predicted value?
a. tony
b. sam
c. rico
d. keisha

Explanation:

Step1: Calculate mean of x (screen time)

First, find the mean of \( x \) values: \( x = [5.00, 6.50, 3.75, 4.00, 1.50] \)

$$ \bar{x}=\frac{5.00 + 6.50 + 3.75 + 4.00 + 1.50}{5}=\frac{20.75}{5}=4.15 $$

Step2: Calculate mean of y (grades)

Next, find the mean of \( y \) values: \( y = [70, 68, 90, 84, 95] \)

$$ \bar{y}=\frac{70 + 68 + 90 + 84 + 95}{5}=\frac{407}{5}=81.4 $$

Step3: Calculate slope (m)

Using the formula \( m=\frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)
First, calculate \( (x_i - \bar{x}) \) and \( (y_i - \bar{y}) \) for each student:

• Rebecca: \( x - \bar{x}=5.00 - 4.15 = 0.85 \), \( y - \bar{y}=70 - 81.4=-11.4 \), product \( 0.85\times(-11.4)= -9.69 \), square \( 0.85^2 = 0.7225 \)
• Sam: \( x - \bar{x}=6.50 - 4.15 = 2.35 \), \( y - \bar{y}=68 - 81.4=-13.4 \), product \( 2.35\times(-13.4)= -31.49 \), square \( 2.35^2 = 5.5225 \)
• Tony: \( x - \bar{x}=3.75 - 4.15=-0.4 \), \( y - \bar{y}=90 - 81.4 = 8.6 \), product \( -0.4\times8.6=-3.44 \), square \( (-0.4)^2 = 0.16 \)
• Rico: \( x - \bar{x}=4.00 - 4.15=-0.15 \), \( y - \bar{y}=84 - 81.4 = 2.6 \), product \( -0.15\times2.6=-0.39 \), square \( (-0.15)^2 = 0.0225 \)
• Keisha: \( x - \bar{x}=1.50 - 4.15=-2.65 \), \( y - \bar{y}=95 - 81.4 = 13.6 \), product \( -2.65\times13.6=-36.04 \), square \( (-2.65)^2 = 7.0225 \)

Now, sum the products: \( -9.69 - 31.49 - 3.44 - 0.39 - 36.04=-81.05 \)
Sum the squares: \( 0.7225 + 5.5225 + 0.16 + 0.0225 + 7.0225 = 13.45 \)
So, \( m=\frac{-81.05}{13.45}\approx -6.026 \)

Step4: Calculate y-intercept (b)

Using \( \bar{y}=m\bar{x}+b \)

$$ 81.4=-6.026\times4.15 + b $$

$$ 81.4=-24.99 + b $$

$$ b = 81.4 + 24.99 = 106.39 $$

So the linear model is \( y = -6.026x + 106.39 \)

Step5: Predict grades and find residuals

• Rebecca: \( y = -6.026\times5 + 106.39 = -30.13 + 106.39 = 76.26 \), residual \( |70 - 76.26| = 6.26 \)
• Sam: \( y = -6.026\times6.5 + 106.39 = -39.17 + 106.39 = 67.22 \), residual \( |68 - 67.22| = 0.78 \)
• Tony: \( y = -6.026\times3.75 + 106.39 = -22.60 + 106.39 = 83.79 \), residual \( |90 - 83.79| = 6.21 \)
• Rico: \( y = -6.026\times4 + 106.39 = -24.10 + 106.39 = 82.29 \), residual \( |84 - 82.29| = 1.71 \)
• Keisha: \( y = -6.026\times1.5 + 106.39 = -9.04 + 106.39 = 97.35 \), residual \( |95 - 97.35| = 2.35 \)

The smallest residual is Sam's with 0.78.

Answer:

B. Sam