Question

the accompanying table shows the enrollment of a preschool from 1980 through 2000. write a linear regression equation to model the data in the table.

year (x)	enrollment (y)
1985	20
1990	22
1995	28
2000	37

y = \boxed{}x + \boxed{}

Explanation:

Step1: Define variables and calculate means

Let \( x \) be the year (we can let \( x = 0 \) for 1980, \( x = 5 \) for 1985, \( x = 10 \) for 1990, \( x = 15 \) for 1995, \( x = 20 \) for 2000) and \( y \) be the enrollment.
\( n = 5 \) (number of data points)
\( \sum x = 0 + 5 + 10 + 15 + 20 = 50 \)
\( \sum y = 14 + 20 + 22 + 28 + 37 = 121 \)
\( \bar{x}=\frac{\sum x}{n}=\frac{50}{5} = 10 \)
\( \bar{y}=\frac{\sum y}{n}=\frac{121}{5}=24.2 \)

Step2: Calculate \( \sum xy \) and \( \sum x^2 \)

\( \sum xy=(0\times14)+(5\times20)+(10\times22)+(15\times28)+(20\times37) \)
\( = 0 + 100+220 + 420+740=1480 \)
\( \sum x^2=(0^2)+(5^2)+(10^2)+(15^2)+(20^2) \)
\( = 0 + 25+100 + 225+400 = 750 \)

Step3: Calculate slope \( m \)

The formula for slope \( m \) in linear regression is \( m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2 - (\sum x)^2} \)
Substitute the values:
\( m=\frac{5\times1480 - 50\times121}{5\times750-50^2} \)
\(=\frac{7400 - 6050}{3750 - 2500} \)
\(=\frac{1350}{1250}=1.08 \)

Step4: Calculate y - intercept \( b \)

Use the formula \( b=\bar{y}-m\bar{x} \)
\( b = 24.2-1.08\times10 \)
\( = 24.2 - 10.8 = 13.4 \)

Answer:

\( y = 1.08x + 13.4 \)