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 2014, and y represents number of new cases. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest tenth. using this equation, estimate the calendar year in which the number of new cases would reach 579.

years since 2014 (x) | 0 | 1 | 2 | 3 | 4
new cases (y) | 904 | 900 | 833 | 837 | 754

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regression equation:
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Explanation:

Step1: Calculate necessary sums

First, we list the data points:
- \( x \): 0, 1, 2, 3, 4
- \( y \): 904, 900, 833, 837, 754

Calculate \( \sum x \), \( \sum y \), \( \sum xy \), \( \sum x^2 \):

\( \sum x = 0 + 1 + 2 + 3 + 4 = 10 \)

\( \sum y = 904 + 900 + 833 + 837 + 754 = 4228 \)

\( \sum xy = (0\times904) + (1\times900) + (2\times833) + (3\times837) + (4\times754) \)
\( = 0 + 900 + 1666 + 2511 + 3016 = 8093 \)

\( \sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 = 0 + 1 + 4 + 9 + 16 = 30 \)

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

The formula for the slope \( m \) of the linear regression line \( y = mx + b \) is:
\( m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} \)
where \( n = 5 \) (number of data points).

Substitute the values:
\( m = \frac{5\times8093 - 10\times4228}{5\times30 - 10^2} \)
\( = \frac{40465 - 42280}{150 - 100} \)
\( = \frac{-1815}{50} = -36.3 \)

The formula for the intercept \( b \) is:
\( b = \frac{\sum y - m\sum x}{n} \)

Substitute the values:
\( b = \frac{4228 - (-36.3)\times10}{5} \)
\( = \frac{4228 + 363}{5} \)
\( = \frac{4591}{5} = 918.2 \)

So the linear regression equation is \( y = -36.3x + 918.2 \)

Step3: Find \( x \) when \( y = 579 \)

Set \( y = 579 \) in the regression equation:
\( 579 = -36.3x + 918.2 \)

Solve for \( x \):
\( -36.3x = 579 - 918.2 \)
\( -36.3x = -339.2 \)
\( x = \frac{-339.2}{-36.3} \approx 9.344 \)

Step4: Find the calendar year

Since \( x \) is the number of years since 2014, the year is \( 2014 + x \approx 2014 + 9.344 \approx 2023.344 \), so we round to the nearest year, which is 2023. But wait, let's check the calculation again for \( x \):

Wait, maybe I made a mistake in the regression equation. Let's recalculate the slope and intercept more accurately.

Wait, let's use a calculator for the linear regression. Let's list the data:

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

\( y \): 904, 900, 833, 837, 754

Using the formula for linear regression:

\( \bar{x} = \frac{\sum x}{n} = \frac{10}{5} = 2 \)

\( \bar{y} = \frac{\sum y}{n} = \frac{4228}{5} = 845.6 \)

\( SS_{xx} = \sum x^2 - \frac{(\sum x)^2}{n} = 30 - \frac{100}{5} = 30 - 20 = 10 \)

\( SS_{xy} = \sum xy - \frac{\sum x \sum y}{n} = 8093 - \frac{10\times4228}{5} = 8093 - 8456 = -363 \)

Then \( m = \frac{SS_{xy}}{SS_{xx}} = \frac{-363}{10} = -36.3 \) (correct)

\( b = \bar{y} - m\bar{x} = 845.6 - (-36.3)\times2 = 845.6 + 72.6 = 918.2 \) (correct)

So the equation is \( y = -36.3x + 918.2 \)

Now, set \( y = 579 \):

\( 579 = -36.3x + 918.2 \)

\( -36.3x = 579 - 918.2 = -339.2 \)

\( x = \frac{339.2}{36.3} \approx 9.34 \)

So the year is \( 2014 + 9.34 \approx 2023.34 \), so the calendar year is 2023 (since we round to the nearest year) or maybe 2024? Wait, no, let's check the calculation again. Wait, maybe I made a mistake in the problem statement. Wait, the problem says "the number of new cases would reach 579". Wait, maybe the regression equation is correct, but let's check the calculation of \( x \):

\( x = \frac{918.2 - 579}{36.3} = \frac{339.2}{36.3} \approx 9.34 \)

So 2014 + 9.34 ≈ 2023.34, so the calendar year is 2023 (since 0.34 of a year is about 4 months, so it would reach 579 in 2023, around April). But let's confirm the regression equation again.

Wait, let's check the data points:

When x=0 (2014), y=904

x=1 (2015), y=900 (decrease by 4)

x=2 (2016), y=833 (decrease by 67)

x=3 (2017), y=837 (increase by 4)

x=4 (2018), y=754 (decrease by 83)

So the trend is decreasing, so the regression line is decreasing, which matches the slope of -36.3.

Now, solving for y=579:

\( x = \frac{918.2 - 579}{36.3} = \frac{339.2}{36.3} \approx 9.34 \)

So 2014 + 9.34 ≈ 2023.34, so the calendar year is 2023 (since we round to the nearest year) or 2024? Wait, no, 9.34 years since 2014 is 2014 + 9 = 2023, plus 0.34 years (about 4 months), so it would reach 579 in 2023, around April. So the calendar year is 2023. But let's check the calculation again.

Wait, maybe I made a mistake in the regression equation. Let's use a calculator for linear regression. Let's input the data:

x: 0,1,2,3,4

y: 904,900,833,837,754

Using a linear regression calculator:

The slope (m) is -36.3, intercept (b) is 918.2, so the equation is y = -36.3x + 918.2.

Now, set y=579:

579 = -36.3x + 918.2

-36.3x = 579 - 918.2 = -339.2

x = 339.2 / 36.3 ≈ 9.34

So 2014 + 9.34 ≈ 2023.34, so the calendar year is 2023 (since we round to the nearest year) or 2024? Wait, no, 9.34 years after 2014 is 2014 + 9 = 2023, and 0.34 of a year is about 4 months, so it would reach 579 in 2023, around April. So the final answer is 2023.

Answer:

2023