Please help! Use the following situation to answer the question below

Question

Amanda was given a 5 week old kitten for her b day. The table shows how the kitten’s weight changed over the next 5 weeks. The graph shows time with the numbers 5,6,7,8,9,10 and underneath 480,540,605,682,754,820.

1.use your calculator to write a best fit linear function that represents the given data set. Round to two decimal places.

2.what is the y intercept?

3.in what week will the kitten weigh approximately 1300 grams?

Answer 1

1.

X = 5,6,7,8,9,10

Y = 480,540,605,682,754,820

Sum of X = 45 -> Mean X = 7.5

Sum of Y = 3881 -> Mean Y = 646.83

Sum of squares = Sum of (X - Mx)^2 = 17.5

Sum of products = Sum of (X - Mx)(Y - My) = 1209.5

(1) Line Equation: Y = mX + c

m = Sum of squares / Sum of products = 17.5 / 1209.5 = 69.11

From (1) c = Y - mX = Mean Y - 69.11( Mean X) = 646.83 - (69.11*7.5) = 128.48

Y = 69.11X + 128.48

2. y intercept = 128.48

3.

Y = 1300

1300 = 69.11X + 128.48 -> X = (1300-128.48) / 69.11 = 19.95 (Week 20)

Answer 2

The slope and intercept are, respectively,

So the equation of the best fit trend line is y = 69.11429 x + 128.4762

The r-coefficient is which means this is an EXCELLENT fit

Here's the table:

x y x^2 xy y^2 line estimate error

5 480 25 2400 230400 474.047619 5.952380952

6 540 36 3240 291600 543.1619048 -3.161904762

7 605 49 4235 366025 612.2761905 -7.276190476

8 682 64 5456 465124 681.3904762 0.60952381

9 754 81 6786 568516 750.5047619 3.495238095

10 820 100 8200 672400 819.6190476 0.380952381

-----------------------------------------------------------------------------------------

45 3881 355 30317 2594065 3881 0

N 6

D 105 R-coefficient

0.99933611

M 69.11428571

B 128.4761905

Answer 3

Hello, Elizabeth,

I'll assume the numbers 5,6,7, etc are weeks and the larger numbers are grams. [Please keep the the kitten away from me if the weights are in pounds.]

The first thing to look for is whether this is a linear relationship. Does the kitten's weight gain steadily at the same rate over time. This can't hold true forever, but for the time period under question, it is a worthwhile question.

If it is a straight line, the slope should be the same for any interval. I calculated the slope for each sequential data point, but looking closely at four or five should allow a good guess.

The slope between sequential weeks is:

Week Slope

I get an average of 68.6 for all weeks up to 13, but I'll round up to 70 for this condensed version. The big point is that the line is experimentally a straight line with slope of 70.

A straight line has the form y = mx + b, where in this case y is the weight and x is the week.

So the equation is now y = 70x + b.

The intercept (b) is a bit trickier. What is the kitten's weight at birth (week 0)?

Take the basic equation of y = 70x + b and plug in week 5 data.

480 = 70*(5) + b

That allows us to compute b, which I find to be 130.

The equation is now y = 70x + 130

Try using some of the other data points in the new equation. You'll find there is a fairly good match between actual and predicted weights.

Assuming the equation is valid, use it to determine which week the kitten will weigh 1300 grams.

1300 = 70x + 130

x = 16.7 weeks. I rounded up to 17 weeks, knowing that the slope I use was slightly larger than a true calculated average.

I hope this helps,

Bob