Question

The table shows the median home prices in Florida. What is the equation of a trend line that models a relationship between time and home prices? Use the equation to predict the median home price in 2040.

*YEAR- 1940 *MEDIAN PRICE($)-$23,800
*YEAR- 1950 *MEDIAN PRICE($)-$40,900
*YEAR- 1960 *MEDIAN PRICE($)-$58,000
*YEAR- 1970 *MEDIAN PRICE($)-$57,400
*YEAR- 1980 *MEDIAN PRICE($)-$82,400
*YEAR- 1990 *MEDIAN PRICE($)-$100,700
*YEAR- 2000 *MEDIAN PRICE($)-$113,100

y=44.65x +23,800; $172,600
y=44.65x +23,800; $142,900
y=1488.33x +23,800;$142,900
y=1488.33x +23,800;$172,600

Answer 1

Hint: use the "fit linear" function in wolfram alpha

Answer 2

What is the fit linear function? I have never heard of it.

Answer 3

@andreascott are you allowed to use technology to solve this problem or must you use just pen and paper? I was thinking of plotting everything in excel.

Answer 4

I don't have excel. I don't know how to work it anyways.

Answer 5

Ok then. Have you tried plotting all the points down on a graph yet?

Answer 6

It's multiple choice, just plug-and-chug.

Answer 7

Yes .

Answer 8

@CliffSedge ; I can't guess. I need to get this rite . I'm failing so I need to get my grade up :/

Answer 9

If it weren't multiple choice, then the suggested wolfram, or excel is a good way to go, or if you have a graphing calculator - many of them have stat-plot functions. My TI-84 does linear regression quite well.

Answer 10

I did not suggest you guess!

Answer 11

I suggested you test each equation given with the data to see which one works best.
That process is made easier if you use graph paper and actually draw the lines through the data points.

Answer 12

Oh , well I don't have one of those calculators anyways. .

Answer 13

Looks like it's graph paper to the rescue then.

Answer 14

Though, testing each equation with just arithmetic will eliminate a couple bogus options right away.

Answer 15

e.g. eq.1 y=44.65x +23,800; $172,600
44.65(2040) + 23,800 = 114,886 not 172,600 so that equation fails immediately without needing to compare it to any of the other data.

Answer 16

You can find the gradient of the graph using the two end points:
*YEAR- 1940 *MEDIAN PRICE($)-$23,800
*YEAR- 2000 *MEDIAN PRICE($)-$113,100
with the year being on the x-axis and median prices on the y-axis and since the "b" of y=mx+b is all the same, you don't need to calculate it.
The gradient should be 1488.33333333...
Do you know what to do after this?

Answer 17

Oh on another note, CliffSedge's way is much faster.

Answer 18

But when I do his way, none of the choices are right?

Answer 19

Basically, if you let

x = no of years after 1940
y = median house prices (in thousands)

You can use the fit linear function in wolframalpha using the following points:

(0, 23.8), (10,40.9 ) , (20,58 ) , (30,57.4 ) , (40,82.4 ), (50,100.7 ), (60, 113.1)
You end up with

1.47107 x+23.9107

then plug in x = 100

Answer 20

Remember x is in thousands, so you'll have to add three zeroes after it

Answer 21

I did that and I got 171.0177

Answer 22

so in reality, you have 171,017 which is approximately equal to 172,000

Answer 23

I think it's using year 1940 as a 0, so 2040 is actually x=80..

Answer 24

1940 + 100 = 2040

Answer 25

2040 is exactly 100 years after 1940

Answer 26

So it would be the 1st one. Right ?

Answer 27

No, it would not be the first one

Answer 28

You have to think outside the box to figure it out

Answer 29

*right, sorry x=100 for 2040 (mental math fail)

Answer 30

Happens to the best of us @CliffSedge. I had my share of mental math failures.

Answer 31

@CliffSedge, why do you use the year 1940 as a 0? How does it work?

Answer 32

Well I did the last one. I plugged in 100 for x. And I got 172,633 .

Answer 33

So, i'm going with the last one.

Answer 34

I could have sworn I was the one who used x = 0 as the initial year

Answer 35

Opps, sorry @Hero

Answer 36

thank you guys for helping me. I appreciate it! (:

Answer 37

(Irrelevant commentary: The actual regression line is 1471.07x + 23,910.71 but the one listed is a fine approximation.)

(More irrelevant commentary: Despite the strong linear correlation of the data, I still don't feel comfortable extrapolating out that far).

Relevant commentary: Yes, @Hero you did first, I was correcting myself when I mentioned it.
Yes, @andreascott the fourth equation seems to be the best fit.

Answer 38

duplicitycheese, by default, these kind of problems always use the first year as x = 0. It makes producing the function a lot simpler.

Answer 39

Imagine what kind of function you would produce if you used (1940, 23,800) etc as points

Answer 40

Ah! I got it. Thanks @Hero and good job for everyone working on the problem