Question

problem 2
the table to the right gives the number of hours spend studying for an algebra 1 exam (x) and the final exam grade (y).
regression equation:
predict the exam grade of a student who studied for 6 hours.
predict the exam grade of a student who studied for 2.5 hours.

Explanation:

Step1: Calculate the means of x and y

Let \(n = 7\).
\(\bar{x}=\frac{2 + 5+1+0+4+2+3}{7}=\frac{17}{7}\approx2.43\)
\(\bar{y}=\frac{77 + 92+70+63+90+75+84}{7}=\frac{551}{7}=78.71\)

Step2: Calculate the numerator and denominator for the slope \(m\)

\(\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})=(2 - 2.43)(77-78.71)+(5 - 2.43)(92 - 78.71)+(1 - 2.43)(70 - 78.71)+(0 - 2.43)(63 - 78.71)+(4 - 2.43)(90 - 78.71)+(2 - 2.43)(75 - 78.71)+(3 - 2.43)(84 - 78.71)\)
\(=(- 0.43)(-1.71)+(2.57)(13.29)+(-1.43)(-8.71)+(-2.43)(-15.71)+(1.57)(11.29)+(-0.43)(-3.71)+(0.57)(5.29)\)
\(=0.7353+34.1553 + 12.4553+38.1753+17.7253+1.5953+3.0153=107.8671\)
\(\sum_{i=1}^{n}(x_i-\bar{x})^2=(2 - 2.43)^2+(5 - 2.43)^2+(1 - 2.43)^2+(0 - 2.43)^2+(4 - 2.43)^2+(2 - 2.43)^2+(3 - 2.43)^2\)
\(=(-0.43)^2+(2.57)^2+(-1.43)^2+(-2.43)^2+(1.57)^2+(-0.43)^2+(0.57)^2\)
\(=0.1849+6.6049+2.0449+5.9049+2.4649+0.1849+0.3249 = 17.7143\)
\(m=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i-\bar{x})^2}=\frac{107.8671}{17.7143}\approx6.09\)

Step3: Calculate the y - intercept \(b\)

\(b=\bar{y}-m\bar{x}=78.71-6.09\times2.43=78.71 - 14.8087=63.9013\)
The regression equation is \(y = 6.09x+63.90\)

Step4: Predict for \(x = 6\)

\(y=6.09\times6+63.90=36.54+63.90 = 100.44\approx100\)

Step5: Predict for \(x = 2.5\)

\(y=6.09\times2.5+63.90=15.225+63.90=79.125\approx79\)

Answer:

Regression Equation: \(y = 6.09x+63.90\)
Predicted grade for 6 - hour study: 100
Predicted grade for 2.5 - hour study: 79