Question

1. the table below shows the average sale price, p, of a house in norfolk, virginia in thousands of dollars during various year, t.

\ta) write the equation for the curve of best fit
\tb) predict the value of the house in 2015

Explanation:

Part a)

Step 1: Assign Variables

Let's let \( t \) be the number of years since 2005. So for 2005, \( t = 0 \); 2006, \( t = 1 \); 2007, \( t = 2 \); 2008, \( t = 3 \); 2009, \( t = 4 \); 2010, \( t = 5 \). The corresponding \( p \) (price in thousands) values are: \( (0, 158) \), \( (1, 145) \), \( (2, 132) \), \( (3, 130) \), \( (4, 138) \), \( (5, 142) \).

Step 2: Calculate Mean of \( t \) and \( p \)

First, find the mean of \( t \) (\( \bar{t} \)) and mean of \( p \) (\( \bar{p} \)).
\( \bar{t}=\frac{0 + 1+2 + 3+4 + 5}{6}=\frac{15}{6}=2.5 \)
\( \bar{p}=\frac{158+145 + 132+130+138+142}{6}=\frac{845}{6}\approx140.833 \)

Step 3: Calculate Slope (\( m \))

The formula for the slope of the least - squares line is \( m=\frac{\sum_{i = 1}^{n}(t_i-\bar{t})(p_i-\bar{p})}{\sum_{i = 1}^{n}(t_i-\bar{t})^2} \)
First, calculate \( (t_i-\bar{t})(p_i-\bar{p}) \) for each \( i \):

• For \( i = 1 \) (\( t = 0,p = 158 \)): \( (0 - 2.5)(158 - 140.833)=(- 2.5)(17.167)=-42.9175 \)
• For \( i = 2 \) (\( t = 1,p = 145 \)): \( (1 - 2.5)(145 - 140.833)=(-1.5)(4.167)=-6.2505 \)
• For \( i = 3 \) (\( t = 2,p = 132 \)): \( (2 - 2.5)(132 - 140.833)=(-0.5)(-8.833)=4.4165 \)
• For \( i = 4 \) (\( t = 3,p = 130 \)): \( (3 - 2.5)(130 - 140.833)=(0.5)(-10.833)=-5.4165 \)
• For \( i = 5 \) (\( t = 4,p = 138 \)): \( (4 - 2.5)(138 - 140.833)=(1.5)(-2.833)=-4.2495 \)
• For \( i = 6 \) (\( t = 5,p = 142 \)): \( (5 - 2.5)(142 - 140.833)=(2.5)(1.167)=2.9175 \)

Sum of these products: \( - 42.9175-6.2505 + 4.4165-5.4165-4.2495 + 2.9175=-51.5 \)

Now calculate \( (t_i-\bar{t})^2 \) for each \( i \):

• For \( i = 1 \): \( (0 - 2.5)^2 = 6.25 \)
• For \( i = 2 \): \( (1 - 2.5)^2 = 2.25 \)
• For \( i = 3 \): \( (2 - 2.5)^2 = 0.25 \)
• For \( i = 4 \): \( (3 - 2.5)^2 = 0.25 \)
• For \( i = 5 \): \( (4 - 2.5)^2 = 2.25 \)
• For \( i = 6 \): \( (5 - 2.5)^2 = 6.25 \)

Sum of these squares: \( 6.25+2.25 + 0.25+0.25+2.25+6.25 = 17.5 \)

Then the slope \( m=\frac{-51.5}{17.5}\approx - 2.9429 \)

Step 4: Calculate Intercept (\( b \))

Using the formula \( \bar{p}=m\bar{t}+b \), we can solve for \( b \).
\( b=\bar{p}-m\bar{t} \)
Substitute \( \bar{p}\approx140.833 \), \( m\approx - 2.9429 \), and \( \bar{t}=2.5 \)
\( b = 140.833-(-2.9429)\times2.5=140.833 + 7.35725=148.19025 \)

So the equation of the least - squares line (curve of best fit, assuming linear) is \( p=mt + b\), or \( p=-2.9429t + 148.1903 \) (where \( t \) is the number of years since 2005)

Part b)

Step 1: Determine \( t \) for 2015

Since \( t \) is the number of years since 2005, for 2015, \( t=2015 - 2005 = 10 \)

Step 2: Substitute \( t = 10 \) into the equation

We use the equation from part (a): \( p=-2.9429t + 148.1903 \)
Substitute \( t = 10 \):
\( p=-2.9429\times10+148.1903=-29.429 + 148.1903 = 118.7613 \)

So the predicted price of the house in 2015 is approximately \( 118.76 \) thousand dollars.

Final Answers

a) The equation of the curve of best fit (linear) is \( \boldsymbol{p=-2.94t + 148.19} \) (rounded to two decimal places)
b) The predicted value of the house in 2015 is approximately \( \boldsymbol{118.76} \) thousand dollars.

Answer:

Step 1: Determine \( t \) for 2015

Since \( t \) is the number of years since 2005, for 2015, \( t=2015 - 2005 = 10 \)

Step 2: Substitute \( t = 10 \) into the equation

We use the equation from part (a): \( p=-2.9429t + 148.1903 \)
Substitute \( t = 10 \):
\( p=-2.9429\times10+148.1903=-29.429 + 148.1903 = 118.7613 \)

So the predicted price of the house in 2015 is approximately \( 118.76 \) thousand dollars.

Final Answers

a) The equation of the curve of best fit (linear) is \( \boldsymbol{p=-2.94t + 148.19} \) (rounded to two decimal places)
b) The predicted value of the house in 2015 is approximately \( \boldsymbol{118.76} \) thousand dollars.