Question

the table shows the population of center city in various years. use the model that predicts the population of the city (y) in a given year (x). in which year is the city’s population most different from the value predicted by this model?

year	city population
1990	197,800
1992	199,532
2000	203,750
2005	206,561
2012	210,600

○ 1985
○ 1992
○ 2000
○ 2012

Answer:

To solve this, we first need to determine the linear model (or observe the trend) of the population over time. Let's assume a linear model \( y = mx + b \), where \( x \) is the year (we can set \( x = 0 \) for 1985, \( x = 5 \) for 1990, \( x = 7 \) for 1992, \( x = 15 \) for 2000, \( x = 20 \) for 2005, \( x = 27 \) for 2012) and \( y \) is the population.

Step 1: Calculate the slope between 1985 (x=0, y=194957) and 1990 (x=5, y=197800)

The slope \( m \) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \)
\( m=\frac{197800 - 194957}{5 - 0}=\frac{2843}{5} = 568.6 \)

Step 2: Predict population for other years

• For 1992 (x = 7):

Predicted \( y=194957+568.6\times7=194957 + 3980.2=198937.2 \)
Actual population is 199532. Difference: \( |199532 - 198937.2| = 594.8 \)

• For 2000 (x = 15):

Predicted \( y=194957+568.6\times15=194957+8529 = 203486 \)
Actual population is 203750. Difference: \( |203750 - 203486| = 264 \)

• For 2012 (x = 27):

Predicted \( y=194957+568.6\times27=194957+15352.2 = 210309.2 \)
Actual population is 210600. Difference: \( |210600 - 210309.2| = 290.8 \)

• For 1985, the predicted value is the actual value (since we set \( x = 0 \) for 1985), so the difference is 0.

Among the differences (594.8 for 1992, 264 for 2000, 290.8 for 2012, 0 for 1985), the largest difference is for the year 1992. But wait, let's re - check the calculation for 2012. Wait, maybe we made a wrong assumption about the linear model. Let's try another approach. Let's check the average rate of change between consecutive years:

• From 1985 to 1990: \( \frac{197800 - 194957}{5}=\frac{2843}{5}=568.6 \)
• From 1990 to 1992: \( \frac{199532 - 197800}{2}=\frac{1732}{2} = 866 \)
• From 1992 to 2000: \( \frac{203750 - 199532}{8}=\frac{4218}{8}=527.25 \)
• From 2000 to 2005: \( \frac{206561 - 203750}{5}=\frac{2811}{5}=562.2 \)
• From 2005 to 2012: \( \frac{210600 - 206561}{7}=\frac{4039}{7}\approx577 \)

The rate of change from 1990 to 1992 (866) is much higher than the other rates. Let's calculate the predicted value for 1992 using the rate from 1985 - 1990: \( 197800+568.6\times2 = 197800 + 1137.2=198937.2 \), and the actual is 199532, difference is 594.8.

For 2012, using the rate from 2005 - 2012: \( 206561+577\times0 = 206561 \) (no, better to use the overall trend). Wait, maybe the model is linear regression. But since we can see that the population in 2012: Let's check the difference between actual and what we would expect from a more consistent trend.

Wait, another way: Let's list the years as \( x \) (let \( x = 0 \) for 1985, \( x = 5 \) for 1990, \( x = 7 \) for 1992, \( x = 15 \) for 2000, \( x = 20 \) for 2005, \( x = 27 \) for 2012)

We can also use the two - point formula for linear regression. But maybe a simpler way is to check the residuals (difference between actual and predicted).

If we consider the line passing through (1985, 194957) and (2005, 206561). The slope \( m=\frac{206561 - 194957}{2005 - 1985}=\frac{11604}{20}=580.2 \)

Equation: \( y - 194957=580.2(x - 0)\) (since \( x = 0 \) for 1985), so \( y = 580.2x+194957 \)

• For 1990 (\( x = 5 \)): \( y=580.2\times5 + 194957=2901+194957 = 197858 \). Actual: 197800. Difference: \( |197800 - 197858| = 58 \)
• For 1992 (\( x = 7 \)): \( y=580.2\times7+194957 = 4061.4+194957=199018.4 \). Actual: 199532. Difference: \( |199532 - 199018.4| = 513.6 \)
• For 2000 (\( x = 15 \)): \( y=580.2\times15+194957 = 8703+194957=203660 \). Actual: 203750. Difference: \( |203750 - 203660| = 90 \)
• For 2012 (\( x = 27 \)): \( y=580.2\times27+194957=15665.4 + 194957=210622.4 \). Actual: 210600. Difference: \( |210600 - 210622.4| = 22.4 \)

Now, if we consider the line passing through (1990, 197800) and (2005, 206561). Slope \( m=\frac{206561 - 197800}{2005 - 1990}=\frac{8761}{15}\approx584.07 \)

Equation: \( y - 197800=584.07(x - 5) \)

For 1985 (\( x = 0 \)): \( y=197800-584.07\times5=197800 - 2920.35 = 194879.65 \). Actual: 194957. Difference: \( |194957 - 194879.65| = 77.35 \)

For 1992 (\( x = 7 \)): \( y=197800+584.07\times2=197800 + 1168.14 = 198968.14 \). Actual: 199532. Difference: \( |199532 - 198968.14| = 563.86 \)

For 2000 (\( x = 15 \)): \( y=197800+584.07\times10=197800+5840.7 = 203640.7 \). Actual: 203750. Difference: \( |203750 - 203640.7| = 109.3 \)

For 2012 (\( x = 27 \)): \( y=197800+584.07\times17=197800+9929.19 = 207729.19 \)? Wait, no, 2005 is \( x = 15 \) (if 1990 is \( x = 0 \)), 2012 is \( x = 22 \). I made a mistake in \( x \) assignment.

Let's re - assign \( x \) as the number of years since 1985. So:

• 1985: \( x = 0 \), \( y = 194957 \)
• 1990: \( x = 5 \), \( y = 197800 \)
• 1992: \( x = 7 \), \( y = 199532 \)
• 2000: \( x = 15 \), \( y = 203750 \)
• 2005: \( x = 20 \), \( y = 206561 \)
• 2012: \( x = 27 \), \( y = 210600 \)

Let's calculate the linear regression equation. The formula for the slope \( m \) of a linear regression line \( y=mx + b \) is \( m=\frac{n\sum xy-\sum x\sum y}{n\sum x^{2}-(\sum x)^{2}} \) and \( b=\frac{\sum y - m\sum x}{n} \), where \( n = 6 \)

First, calculate \( \sum x=0 + 5+7 + 15+20+27=74 \)

\( \sum y=194957+197800+199532+203750+206561+210600 = 1213200 \)

\( \sum xy=0\times194957+5\times197800+7\times199532+15\times203750+20\times206561+27\times210600 \)

\( = 0+989000+1396724+3056250+4131220+5686200 \)

\( = 989000+1396724 = 2385724; 2385724+3056250 = 5441974; 5441974+4131220 = 9573194; 9573194+5686200 = 15259394 \)

\( \sum x^{2}=0^{2}+5^{2}+7^{2}+15^{2}+20^{2}+27^{2}=0 + 25+49+225+400+729 = 1428 \)

Now, \( m=\frac{6\times15259394-74\times1213200}{6\times1428-(74)^{2}} \)

\( 6\times15259394 = 91556364 \); \( 74\times1213200 = 89776800 \); \( 6\times1428 = 8568 \); \( 74^{2}=5476 \)

\( m=\frac{91556364 - 89776800}{8568 - 5476}=\frac{1779564}{3092}\approx575.5 \)

\( b=\frac{1213200-575.5\times74}{6}=\frac{1213200 - 42587}{6}=\frac{1170613}{6}\approx195102.17 \)

So the regression equation is \( y = 575.5x+195102.17 \)

Now, calculate the predicted values and differences:

• 1985 (\( x = 0 \)): \( y = 575.5\times0+195102.17 = 195102.17 \). Difference: \( |194957 - 195102.17| = 145.17 \)
• 1990 (\( x = 5 \)): \( y = 575.5\times5+195102.17 = 2877.5+195102.17 = 197979.67 \). Difference: \( |197800 - 197979.67| = 179.67 \)
• 1992 (\( x = 7 \)): \( y = 575.5\times7+195102.17 = 4028.5+195102.17 = 199130.67 \). Difference: \( |199532 - 199130.67| = 401.33 \)
• 2000 (\( x = 15 \)): \( y = 575.5\times15+195102.17 = 8632.5+195102.17 = 203734.67 \). Difference: \( |203750 - 203734.67| = 15.33 \)
• 2005 (\( x = 20 \)): \( y = 575.5\times20+195102.17 = 11510+195102.17 = 206612.17 \). Difference: \( |206561 - 206612.17| = 51.17 \)
• 2012 (\( x = 27 \)): \( y = 575.5\times27+195102.17 = 15538.5+195102.17 = 210640.67 \). Difference: \( |210600 - 210640.67| = 40.67 \)

Among all the differences, the largest difference is for the year 1992? Wait, no, wait in the first calculation when we used the line from 1985 - 1990, the difference for 1992 was 594.8, for 2012 it was 22.4. But when we use linear regression, the difference for 1992 is 401.33, for 1985 is 145