Question

the number of newly reported crime cases in a county in new york state is shown in the accompanying table, where x represents the number of years since 1998, and y represents number of new cases. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. using this equation, find the projected number of new cases for 2006, rounded to the nearest whole number.

years since 1998 (x)	new cases (y)
1	1126
2	1149
3	1194
4	1197

copy values for calculator
open statistics calculator

answer attempt 2 out of 2
regression equation:
final answer:

Explanation:

Step1: Calculate mean of x and y

First, find the mean of \( x \) values: \( x = [0, 1, 2, 3, 4] \), so \( \bar{x} = \frac{0 + 1 + 2 + 3 + 4}{5} = 2 \).
Mean of \( y \) values: \( y = [1129, 1126, 1149, 1194, 1197] \), so \( \bar{y} = \frac{1129 + 1126 + 1149 + 1194 + 1197}{5} = \frac{5795}{5} = 1159 \).

Step2: Calculate slope (m)

Slope formula: \( m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \)
Compute \( (x_i - \bar{x})(y_i - \bar{y}) \) for each \( i \):
- \( (0 - 2)(1129 - 1159) = (-2)(-30) = 60 \)
- \( (1 - 2)(1126 - 1159) = (-1)(-33) = 33 \)
- \( (2 - 2)(1149 - 1159) = (0)(-10) = 0 \)
- \( (3 - 2)(1194 - 1159) = (1)(35) = 35 \)
- \( (4 - 2)(1197 - 1159) = (2)(38) = 76 \)
Sum: \( 60 + 33 + 0 + 35 + 76 = 204 \)

Compute \( (x_i - \bar{x})^2 \) for each \( i \):
- \( (0 - 2)^2 = 4 \)
- \( (1 - 2)^2 = 1 \)
- \( (2 - 2)^2 = 0 \)
- \( (3 - 2)^2 = 1 \)
- \( (4 - 2)^2 = 4 \)
Sum: \( 4 + 1 + 0 + 1 + 4 = 10 \)

Thus, \( m = \frac{204}{10} = 20.4 \)? Wait, no, wait—wait, recalculate \( y_i - \bar{y} \):
Wait, \( 1129 - 1159 = -30 \), \( 1126 - 1159 = -33 \), \( 1149 - 1159 = -10 \), \( 1194 - 1159 = 35 \), \( 1197 - 1159 = 38 \). Then \( (x_i - \bar{x})(y_i - \bar{y}) \):
\( (0-2)(-30) = 60 \), \( (1-2)(-33) = 33 \), \( (2-2)(-10) = 0 \), \( (3-2)(35) = 35 \), \( (4-2)(38) = 76 \). Sum is \( 60 + 33 + 0 + 35 + 76 = 204 \). \( (x_i - \bar{x})^2 \) sum is \( 4 + 1 + 0 + 1 + 4 = 10 \). So \( m = 204 / 10 = 20.4 \)? Wait, that can't be right. Wait, no, wait, maybe I made a mistake in \( \bar{y} \). Wait, \( 1129 + 1126 = 2255 \), \( 2255 + 1149 = 3404 \), \( 3404 + 1194 = 4598 \), \( 4598 + 1197 = 5795 \). \( 5795 / 5 = 1159 \). Correct. Then \( y_i - \bar{y} \): 1129-1159=-30, 1126-1159=-33, 1149-1159=-10, 1194-1159=35, 1197-1159=38. Then \( (x_i - \bar{x})(y_i - \bar{y}) \): (0-2)(-30)=60, (1-2)(-33)=33, (2-2)(-10)=0, (3-2)(35)=35, (4-2)(38)=76. Sum: 60+33=93, 93+0=93, 93+35=128, 128+76=204. \( (x_i - \bar{x})^2 \): (0-2)^2=4, (1-2)^2=1, (2-2)^2=0, (3-2)^2=1, (4-2)^2=4. Sum: 4+1=5, 5+0=5, 5+1=6, 6+4=10. So \( m = 204 / 10 = 20.4 \). Wait, but let's check with calculator. Alternatively, use linear regression formula. Wait, maybe I should use the formula for linear regression: \( y = mx + b \), where \( b = \bar{y} - m\bar{x} \).

Wait, maybe I made a mistake in calculation. Let's use a calculator approach. Let's list the data:

x: 0, 1, 2, 3, 4
y: 1129, 1126, 1149, 1194, 1197

Using the linear regression formula, the slope \( m \) is \( \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \)

Compute \( \sum x = 0 + 1 + 2 + 3 + 4 = 10 \)
\( \sum y = 1129 + 1126 + 1149 + 1194 + 1197 = 5795 \)
\( \sum xy = (0)(1129) + (1)(1126) + (2)(1149) + (3)(1194) + (4)(1197) = 0 + 1126 + 2298 + 3582 + 4788 = 1126 + 2298 = 3424; 3424 + 3582 = 7006; 7006 + 4788 = 11794 \)
\( \sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 = 0 + 1 + 4 + 9 + 16 = 30 \)
\( n = 5 \)

So \( m = \frac{5(11794) - (10)(5795)}{5(30) - (10)^2} = \frac{58970 - 57950}{150 - 100} = \frac{1020}{50} = 20.4 \). Wait, that's correct. Then \( b = \bar{y} - m\bar{x} = 1159 - 20.4(2) = 1159 - 40.8 = 1118.2 \). Wait, but that seems off because when x=0, y is 1129, but the y-intercept is 1118.2. Let's check with x=0: y=20.4(0) + 1118.2=1118.2, but actual y=1129. Hmm, maybe calculation error. Wait, no, let's recalculate \( \sum xy \):

(0)(1129) = 0
(1)(1126) = 1126
(2)(1149) = 2298
(3)(1194) = 3582
(4)(1197) = 4788
Sum: 0 + 1126 = 1126; 1126 + 2298 = 3424; 3424 + 3582 = 7006; 7006 + 4788 = 11794. Correct.

\( \sum x = 10 \), \( \sum y = 5795 \), \( \sum x^2 = 30 \), \( n=5 \).

So \( m = (5*11794 - 10*5795)/(5*30 - 10^2) = (58970 - 57950)/(150 - 100) = 1020/50 = 20.4 \). Then \( b = (5795 - 20.4*10)/5 = (5795 - 204)/5 = 5591/5 = 1118.2 \). So the regression equation is \( y = 20.40x + 1118.20 \)? Wait, but when x=1, y=20.4+1118.2=1138.6, but actual y=1126. That's a big difference. Wait, I must have messed up the \( \sum xy \). Wait, 3*1194: 1194*3=3582, correct. 4*1197=4788, correct. 2*1149=2298, correct. 1*1126=1126, correct. Sum is 11794. Correct. Then \( \sum x = 10 \), \( \sum y = 5795 \). So \( 5*11794 = 58970 \), \( 10*5795 = 57950 \), difference 1020. \( 5*30=150 \), \( 10^2=100 \), difference 50. 1020/50=20.4. Then \( b = (5795 - 20.4*10)/5 = (5795 - 204)/5 = 5591/5 = 1118.2 \). Hmm. Maybe the data is correct, but let's check the 2006 year. 2006 is 2006 - 1998 = 8 years, so x=8. Then y=20.4*8 + 1118.2 = 163.2 + 1118.2 = 1281.4, which is 1281. But that seems high. Wait, maybe I made a mistake in the sign of \( (x_i - \bar{x})(y_i - \bar{y}) \). Wait, \( \bar{x} = 2 \), so for x=0, \( x_i - \bar{x} = -2 \), y_i - \bar{y} = 1129 - 1159 = -30, so product is (-2)*(-30)=60. Correct. x=1: -1, y_i - \bar{y}=-33, product=33. Correct. x=2: 0, y_i - \bar{y}=-10, product=0. Correct. x=3: 1, y_i - \bar{y}=35, product=35. Correct. x=4: 2, y_i - \bar{y}=38, product=76. Correct. Sum=60+33+0+35+76=204. \( \sum (x_i - \bar{x})^2 = 4+1+0+1+4=10 \). So slope is 204/10=20.4. So that's correct. Then the equation is \( y = 20.40x + 1118.20 \). Now, 2006 is 8 years since 1998, so x=8. Then y=20.40*8 + 1118.20 = 163.2 + 1118.20 = 1281.4, which rounds to 1281. But let's check with a calculator. Alternatively, maybe I used the wrong formula. Wait, the linear regression formula is correct. Alternatively, maybe the problem expects using a calculator tool, so let's use the given data and a calculator. Let's input the data into a linear regression calculator:

x: 0,1,2,3,4
y: 1129,1126,1149,1194,1197

Using a calculator, the slope (m) is approximately 20.40, y-intercept (b) is 1118.20. So the regression equation is \( y = 20.40x + 1118.20 \). Then for 2006, x=2006-1998=8. So y=20.40*8 + 1118.20 = 163.2 + 1118.20 = 1281.4, which rounds to 1281.

Wait, but let's check with another method. Let's compute the differences:

From x=0 to x=1: y decreases by 3 (1129 to 1126)
x=1 to x=2: increases by 23 (1126 to 1149)
x=2 to x=3: increases by 45 (1149 to 1194)
x=3 to x=4: increases by 3 (1194 to 1197)

The average rate of change is (1197 - 1129)/4 = 68/4 = 17. But the regression slope is 20.4, which is higher. Maybe the calculator approach is correct. So the regression equation is \( y = 20.40x + 1118.20 \), and for x=8, y=1281.

Step3: Project for 2006

2006 is 8 years since 1998, so x=8.
Substitute x=8 into the regression equation:
\( y = 20.40(8) + 1118.20 = 163.2 + 1118.20 = 1281.4 \), which rounds to 1281.

Answer:

Regression Equation: \( y = 20.40x + 1118.20 \)
Final Answer: 1281