Question

the annual profits for a company are given in the following table, where x represents the number of years since 2004, and y represents the profit in thousands of dollars. write the linear regression equation that represents this set of data, rounding all coefficients to the nearest hundredth. using this equation, find the projected profit (in thousands of dollars) for 2016, rounded to the nearest thousand dollars.

years since 2004 (x)	profits (y) (in thousands of dollars)
1	165
2	163
3	168

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Explanation:

Step1: Identify data points

We have the data points: \((0, 144)\), \((1, 165)\), \((2, 163)\), \((3, 168)\)

Step2: Calculate mean of \(x\) and \(y\)

The formula for the mean \(\bar{x}=\frac{\sum x}{n}\) and \(\bar{y}=\frac{\sum y}{n}\), where \(n = 4\)
\(\sum x=0 + 1+2 + 3=6\), so \(\bar{x}=\frac{6}{4} = 1.5\)
\(\sum y=144+165 + 163+168=640\), so \(\bar{y}=\frac{640}{4}=160\)

Step3: Calculate slope \(m\)

The formula for slope \(m=\frac{\sum (x_i-\bar{x})(y_i - \bar{y})}{\sum (x_i-\bar{x})^2}\)
- For \((0,144)\): \((0 - 1.5)(144 - 160)=(- 1.5)\times(-16) = 24\); \((0 - 1.5)^2=2.25\)
- For \((1,165)\): \((1 - 1.5)(165 - 160)=(-0.5)\times5=-2.5\); \((1 - 1.5)^2 = 0.25\)
- For \((2,163)\): \((2 - 1.5)(163 - 160)=(0.5)\times3 = 1.5\); \((2 - 1.5)^2=0.25\)
- For \((3,168)\): \((3 - 1.5)(168 - 160)=(1.5)\times8 = 12\); \((3 - 1.5)^2=2.25\)

\(\sum (x_i-\bar{x})(y_i - \bar{y})=24-2.5 + 1.5+12=35\)
\(\sum (x_i-\bar{x})^2=2.25 + 0.25+0.25 + 2.25 = 5\)
So \(m=\frac{35}{5}=7.00\) (Wait, let's recalculate the numerator: \(24-2.5 = 21.5\), \(21.5+1.5 = 23\), \(23 + 12=35\). Denominator: \(2.25+0.25 = 2.5\), \(2.5+0.25=2.75\), \(2.75 + 2.25 = 5\). So \(m = 7\)? Wait, but let's use another approach. Maybe using the linear regression formula. Alternatively, we can use the formula \(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}\)

First, calculate \(\sum xy\):
\((0\times144)+(1\times165)+(2\times163)+(3\times168)=0 + 165+326 + 504=995\)
\(\sum x^2=0^2 + 1^2+2^2+3^2=0 + 1+4 + 9 = 14\)
\(n = 4\)
So \(m=\frac{4\times995-6\times640}{4\times14 - 6^2}=\frac{3980 - 3840}{56 - 36}=\frac{140}{20}=7.00\)

Step4: Calculate y-intercept \(b\)

Using \(y=mx + b\) and \(\bar{y}=m\bar{x}+b\)
\(160=7\times1.5 + b\)
\(160 = 10.5+b\)
\(b=160 - 10.5 = 149.50\)

So the linear regression equation is \(y = 7.00x+149.50\)

Step5: Find \(x\) for 2016

2016 - 2004=12, so \(x = 12\)

Step6: Predict \(y\) for \(x = 12\)

\(y=7.00\times12+149.50=84 + 149.50=233.50\)
Rounded to the nearest thousand dollars (since \(y\) is in thousands of dollars), it's 234 (wait, 233.5 rounded to the nearest whole number is 234? Wait, 233.5 is halfway, but usually we round up. Wait, 233.5 thousand dollars is 233500 dollars, rounded to the nearest thousand dollars is 234 thousand dollars. Wait, but let's check the calculation again. Wait, maybe I made a mistake in slope. Wait, let's recalculate the \(\sum xy\) and \(\sum x^2\) again.

Wait, data points:
\(x: 0,1,2,3\); \(y:144,165,163,168\)

\(\sum x = 0 + 1+2 + 3=6\)

\(\sum y=144 + 165+163 + 168=640\)

\(\sum xy=0\times144 + 1\times165+2\times163+3\times168=0 + 165+326+504=995\)

\(\sum x^2=0 + 1+4 + 9 = 14\)

\(n = 4\)

\(m=\frac{n\sum xy-\sum x\sum y}{n\sum x^2-(\sum x)^2}=\frac{4\times995-6\times640}{4\times14 - 36}=\frac{3980 - 3840}{56 - 36}=\frac{140}{20}=7\). So slope is 7.

\(b=\bar{y}-m\bar{x}=\frac{640}{4}-7\times\frac{6}{4}=160 - 7\times1.5=160 - 10.5 = 149.5\). So equation is \(y = 7x + 149.5\)

For 2016, \(x=2016 - 2004 = 12\)

\(y=7\times12+149.5=84 + 149.5=233.5\)

Rounded to the nearest thousand dollars (since \(y\) is in thousands of dollars), 233.5 rounds to 234. Wait, but let's check with a calculator approach. Alternatively, maybe I made a mistake in the data. Wait, the data points: when \(x = 0\), \(y = 144\); \(x = 1\), \(y = 165\) (increase by 21); \(x = 2\), \(y = 163\) (decrease by 2); \(x = 3\), \(y = 168\) (increase by 5). So the slope of 7 seems high? Wait, maybe my calculation is wrong. Wait, let's use the formula for linear regression with these points.

Alternatively, using a calculator:

The formula for linear regression is \(y = a + bx\), where \(b=\frac{\sum (x_i-\bar{x})(y_i - \bar{y})}{\sum (x_i-\bar{x})^2}\)

\(\bar{x}=1.5\), \(\bar{y}=160\)

\((x_1-\bar{x})(y_1 - \bar{y})=(0 - 1.5)(144 - 160)=(-1.5)(-16)=24\)

\((x_2-\bar{x})(y_2 - \bar{y})=(1 - 1.5)(165 - 160)=(-0.5)(5)=-2.5\)

\((x_3-\bar{x})(y_3 - \bar{y})=(2 - 1.5)(163 - 160)=(0.5)(3)=1.5\)

\((x_4-\bar{x})(y_4 - \bar{y})=(3 - 1.5)(168 - 160)=(1.5)(8)=12\)

Sum of these: \(24-2.5 + 1.5+12=35\)

Sum of \((x_i-\bar{x})^2\): \((-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2=2.25 + 0.25+0.25 + 2.25=5\)

So \(b = 35/5 = 7\). So that's correct. Then \(a=\bar{y}-b\bar{x}=160 - 7\times1.5=160 - 10.5 = 149.5\). So the equation is correct.

Then for \(x = 12\), \(y=7\times12 + 149.5=84 + 149.5=233.5\). Rounded to the nearest thousand dollars (since \(y\) is in thousands), 233.5 rounds to 234.

Wait, but maybe the problem expects the linear regression equation first. Let's confirm the equation: \(y = 7.00x + 149.50\) (since we need to round to nearest hundredth, 7.00 and 149.50). Then for 2016, \(x = 12\), so \(y=7.00\times12+149.50=84 + 149.50=233.50\), which rounds to 234 (thousand dollars).

Answer:

The linear regression equation is \(y = 7.00x + 149.50\), and the projected profit for 2016 is \(\boxed{234}\) (in thousands of dollars).