4.5 - scatter plots & lines of fit. ex.1 the table below shows the median selling price of new,...

4.5 - Scatter Plots & Lines of Fit

Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

Year 1990 1992 1994 1996 1998 2000

Price ($1000)

122.9 121.5 130 140 152.5 169

Ex.1 The table below shows the median selling price of new, privately-owned, one-family houses for some recent years.

a. Make a scatter plot of the data.

Year 1990 1992 1994 1996 1998 2000

Price ($1000)

122.9 121.5 130 140 152.5 169

Years Since 1990

Price

Years Since 1990

Price

($1000)

Years Since 1990

Median House Prices

Price

($1000)

Years Since 1990

Median House Prices

Price

($1000)

0

Years Since 1990

Median House Prices

Price

($1000)

0 2 4 6 8 10

Years Since 1990

Median House Prices

Price

($1000)

0 2 4 6 8 10

Years Since 1990

Median House Prices

Price

($1000)

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

b. Make a line of fit.

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

b. Make a line of fit.

Median House Prices

Price

($1000)

140

1200 2 4 6 8

10

Years Since 1990

b. Make a line of fit.

c. Find a prediction equation for line of fit.

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1

x2 - x1

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130

x2 - x1 8 – 4

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5

x2 - x1 8 – 4 4

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4

*So use x1 = 4

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4

*So use x1 = 4, y1 = 130

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4

*So use x1 = 4, y1 = 130, and m ≈ 5.63

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4

*So use x1 = 4, y1 = 130, and m ≈ 5.63

y – y1 = m(x – x1)

c. Find a prediction equation for line of fit.

*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)

m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4

*So use x1 = 4, y1 = 130, and m ≈ 5.63

y – y1 = m(x – x1)

c. Find a prediction equation for line of fit.*Use the best two ordered pairs from b. to find the slope for the line!

(4, 130) and (8, 152.5)m = y2 – y1 = 152.2 – 130 = 22.5 ≈ 5.63

x2 - x1 8 – 4 4*So use x1 = 4, y1 = 130, and m ≈ 5.63

y – y1 = m(x – x1)y – 130 = 5.63(x – 4)y – 130 = 5.63(x) – 5.63(4) y – 130 = 5.63x – 22.52 y = 5.63x + 107.48

d. Predict the price in 2010.

d. Predict the price in 2010.

2010 means when x=20 (yrs after 1990)

d. Predict the price in 2010.

2010 means when x=20 (yrs after 1990)

*Plug 20 in for x!

d. Predict the price in 2010.

2010 means when x=20 (yrs after 1990)

*Plug 20 in for x!

y = 5.63x + 107.48

d. Predict the price in 2010.

2010 means when x=20 (yrs after 1990)

*Plug 20 in for x!

y = 5.63x + 107.48

y = 5.63(20) + 107.48

d. Predict the price in 2010.

2010 means when x=20 (yrs after 1990)

*Plug 20 in for x!

y = 5.63x + 107.48

y = 5.63(20) + 107.48

y = 220.08

d. Predict the price in 2010.

2010 means when x=20 (yrs after 1990)

*Plug 20 in for x!

y = 5.63x + 107.48

y = 5.63(20) + 107.48

y = 220.08

So, in 2010 the price will be $220,080.