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 2002, 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 2011, rounded to the nearest whole number.

years since 2002 (x)	new cases (y)
1	1172
2	1180
3	1279
4	1275

copy values for calculator
open statistics calculator

answer attempt 1 out of 2
regression equation:
final answer:

Explanation:

Step1: Calculate necessary sums

First, we list the data points: \((0, 1130)\), \((1, 1172)\), \((2, 1180)\), \((3, 1279)\), \((4, 1275)\)

Number of data points \(n = 5\)

Sum of \(x\) values: \(\sum x = 0 + 1 + 2 + 3 + 4 = 10\)

Sum of \(y\) values: \(\sum y = 1130 + 1172 + 1180 + 1279 + 1275 = 5036\)

Sum of \(x \cdot y\) values: \(0\times1130 + 1\times1172 + 2\times1180 + 3\times1279 + 4\times1275 = 0 + 1172 + 2360 + 3837 + 5100 = 12469\)

Sum of \(x^2\) values: \(0^2 + 1^2 + 2^2 + 3^2 + 4^2 = 0 + 1 + 4 + 9 + 16 = 30\)

Step2: Calculate slope (\(m\)) and y-intercept (\(b\))

The formula for the slope \(m\) of the linear regression line is:
\[ m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \]
Substituting the values:
\[ m = \frac{5\times12469 - 10\times5036}{5\times30 - 10^2} = \frac{62345 - 50360}{150 - 100} = \frac{11985}{50} = 239.7 \]
Wait, that can't be right. Wait, no, I must have miscalculated the sum of \(y\). Let's recalculate \(\sum y\):

\(1130 + 1172 = 2302\); \(2302 + 1180 = 3482\); \(3482 + 1279 = 4761\); \(4761 + 1275 = 6036\). Oh! I made a mistake earlier. So \(\sum y = 6036\)

Now recalculate \(m\):

\[ m = \frac{5\times12469 - 10\times6036}{5\times30 - 10^2} = \frac{62345 - 60360}{150 - 100} = \frac{1985}{50} = 39.7 \]

Now the formula for the y-intercept \(b\) is:
\[ b = \frac{\sum y - m\sum x}{n} \]
Substituting the values:
\[ b = \frac{6036 - 39.7\times10}{5} = \frac{6036 - 397}{5} = \frac{5639}{5} = 1127.8 \]

So the linear regression equation is \(y = 39.7x + 1127.8\) (Wait, but let's check with a calculator approach. Maybe using a calculator for linear regression is better. Let's list the data:

\(x\): 0,1,2,3,4

\(y\): 1130, 1172, 1180, 1279, 1275

Using a linear regression calculator (since manual calculation can be error-prone), the slope \(m\) and intercept \(b\) can be calculated as follows:

The mean of \(x\) (\(\bar{x}\)) is \(\frac{10}{5} = 2\)

The mean of \(y\) (\(\bar{y}\)) is \(\frac{6036}{5} = 1207.2\)

The slope \(m\) is also given by \(\frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}\)

Calculating \((x_i - \bar{x})\) and \((y_i - \bar{y})\):

For \(x=0\): \(0 - 2 = -2\); \(y - \bar{y} = 1130 - 1207.2 = -77.2\); product: \((-2)\times(-77.2)=154.4\); square: \((-2)^2 = 4\)

For \(x=1\): \(1 - 2 = -1\); \(y - \bar{y} = 1172 - 1207.2 = -35.2\); product: \((-1)\times(-35.2)=35.2\); square: \((-1)^2 = 1\)

For \(x=2\): \(2 - 2 = 0\); \(y - \bar{y} = 1180 - 1207.2 = -27.2\); product: \(0\times(-27.2)=0\); square: \(0^2 = 0\)

For \(x=3\): \(3 - 2 = 1\); \(y - \bar{y} = 1279 - 1207.2 = 71.8\); product: \(1\times71.8=71.8\); square: \(1^2 = 1\)

For \(x=4\): \(4 - 2 = 2\); \(y - \bar{y} = 1275 - 1207.2 = 67.8\); product: \(2\times67.8=135.6\); square: \(2^2 = 4\)

Sum of products: \(154.4 + 35.2 + 0 + 71.8 + 135.6 = 397\)

Sum of squares: \(4 + 1 + 0 + 1 + 4 = 10\)

So \(m = \frac{397}{10} = 39.7\), which matches the earlier corrected calculation. Then \(b = \bar{y} - m\bar{x} = 1207.2 - 39.7\times2 = 1207.2 - 79.4 = 1127.8\). So the regression equation is \(y = 39.7x + 1127.8\). Rounding to the nearest hundredth, since 39.7 is already to the tenth, we can write it as \(y = 39.70x + 1127.80\) (or maybe the calculator gives more precise values? Wait, let's use a calculator for linear regression. Let's input the data into a linear regression tool:

Data points:

(0, 1130), (1, 1172), (2, 1180), (3, 1279), (4, 1275)

Using a calculator, the linear regression equation is \(y = 39.7x + 1127.8\) (as we calculated). Now, for 2011, the number of years since 2002 is \(x = 2011 - 2002 = 9\).

Substitute \(x = 9\) into the equation:

\(y = 39.7\times9 + 1127.8 = 357.3 + 1127.8 = 1485.1\). Rounding to the nearest whole number, that's 1485.

Wait, but let's check the calculations again. Wait, when I first calculated \(\sum y\), I had a mistake (6036 is correct: 1130+1172=2302; 2302+1180=3482; 3482+1279=4761; 4761+1275=6036). Then the slope calculation:

\(n\sum xy = 5*12469 = 62345\)

\(\sum x \sum y = 10*6036 = 60360\)

\(n\sum x^2 = 5*30 = 150\)

\((\sum x)^2 = 100\)

So numerator: 62345 - 60360 = 1985

Denominator: 150 - 100 = 50

1985 / 50 = 39.7. Correct.

Intercept: (6036 - 39.7*10)/5 = (6036 - 397)/5 = 5639/5 = 1127.8. Correct.

So the regression equation is \(y = 39.7x + 1127.8\). Now, for x=9 (2011-2002=9):

\(y = 39.7*9 + 1127.8 = 357.3 + 1127.8 = 1485.1\), which rounds to 1485.

Answer:

Regression Equation: \(y = 39.70x + 1127.80\)
Final Answer: 1485