Question

problem 6
the number of students in marching band has increased since the school first opened 10 years ago. the table of values represents the number of students in marching band each year.

year	0	1	2	3	4	5
number of students in the marching band	90	96	105	110	100	115

regression equation:

predict the amount of students who were in the marching band 7 years after the school opened.

predict the amount of students who were in the marching band 15 years after the school opened.

Explanation:

Step1: Assume linear regression

Let \(x\) be the year and \(y\) be the number of students. We first find the least - squares regression line \(y = ax + b\).
We have \(n=6\) data points \((x_1,y_1),(x_2,y_2),\cdots,(x_6,y_6)\) where \(x_i=\{0,1,2,3,4,5\}\) and \(y_i = \{90,96,105,110,100,115\}\).
The formulas for \(a\) and \(b\) in \(y = ax + b\) are:

$$a=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$$

$$b=\frac{\sum_{i = 1}^{n}y_i - a\sum_{i = 1}^{n}x_i}{n}$$

\(\sum_{i = 1}^{n}x_i=0 + 1+2+3+4+5 = 15\)
\(\sum_{i = 1}^{n}y_i=90 + 96+105+110+100+115 = 616\)
\(\sum_{i = 1}^{n}x_i^{2}=0^{2}+1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=0 + 1+4+9+16+25 = 55\)
\(\sum_{i = 1}^{n}x_iy_i=0\times90+1\times96 + 2\times105+3\times110+4\times100+5\times115=0 + 96+210+330+400+575 = 1611\)

$$a=\frac{6\times1611-15\times616}{6\times55 - 15^{2}}=\frac{9666-9240}{330 - 225}=\frac{426}{105}\approx4.057$$

$$b=\frac{616-4.057\times15}{6}=\frac{616 - 60.855}{6}=\frac{555.145}{6}\approx92.524$$

So the regression equation is \(y = 4.057x+92.524\)

Step2: Predict for \(x = 7\)

Substitute \(x = 7\) into \(y = 4.057x+92.524\)

$$y=4.057\times7+92.524=28.399+92.524 = 120.923\approx121$$

Step3: Predict for \(x = 15\)

Substitute \(x = 15\) into \(y = 4.057x+92.524\)

$$y=4.057\times15+92.524=60.855+92.524 = 153.379\approx153$$

Answer:

Regression Equation: \(y = 4.057x+92.524\)
Prediction for 7 years: 121 students
Prediction for 15 years: 153 students