Question

3. You research the average cost of a movie ticket over several years to look for trends. The table below shows your data. What is the equation for a line of best fit? Approximately how much would you expect to pay for a movie ticket in 2021? Let x be the number of years after 1980.
Year: 1980, 1984, 1988, 1992, 1996, 2000, 2004, 2008, 2012. Price: $4.00, $4.50, $5.25, $5.50, $7.00, $7.75, $8.25, $8.78, $9.00. Answer choices: A. y=0.17x+4; $11 B. y=4x; $164 C. y=5.7x+4; $238 D. y=0.67x-4; $23

Answer 1

3)
Given
Year: 1980, 1984, 1988, 1992, 1996, 2000, 2004, 2008, 2012
Price: $4.00, $4.50, $5.25, $5.50, $7.00, $7.75, $8.25, $8.78, $9.00
Line of best fit
Ticket in 2021
Let x be 1980

The formula for the linear regression is

y = ax + b

where

a (slope) = [n(Σxy) - (Σx)(Σy)]/[n(Σx^2) - (Σx)^2]
b (intercept) = [Σy - a(Σx)]/n

n = 9
Σx = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9
= 45
Σy = 4.00 + 4.50 + 5.25 + 5.50 + 7.00 + 7.75 + 8.25 + 8.78 + 9.00
= 60.03
Σx^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2
= 285
Σxy = 1*4 + 2*4.5 + 3*5.25 + 4*5.5 + 5*7 + 6*7.75 + 7*8.25 + 8*8.78 + 9*9
= 341.24

a = [9(341.24) - (45)(60.03)]/[9(285) - (45)^2]
= 0.6848
b = [60.03 - 0.6848(45)]/9
= 3.246

The equation for a line of best fit is

y = 0.6848x + 3.246

In 2021 (2021 - 1979 = 42) the ticket price would be

y = 0.6848*42 + 3.246
= 32

Answer: It is close to D but not quite the same. Perhaps it is typographical?

Answer 2

Medal?

Answer 3

Thanks!

Answer 4

np:) thx