Math Problem Statement

Fit trend line to the following time series by the least square method.

Year(t)

2009

12

2010

20

2011 2012

28

32

2013

50

Production (in lakh tons)

Obtain the trend value of production for 2018.

91.89

a

b

15927.8

C

90.00

d

32203.2

Solution

To fit a trend line to the given time series data by the least squares method and obtain the trend value for 2018, we need to follow these steps:

  1. **Organize the dataTo fit a trend line to the given time series data by the least squares method and obtain the trend value for 2018, we need to follow these steps:

  2. **Organize the **

    • Years (tt) and Production (PP):
    Year (t)Production (P)
    200912
    201020
    201128
    201232
    201350
  3. Define the trend line equation: The trend line can be defined as: P=a+btP = a + bt where PP is the production, tt is the year, aa is the intercept, and bb is the slope.

  4. Calculate the necessary sums for the least squares method:

    Let's denote: n=number of data pointsn = \text{number of data points} t=sum of all years\sum t = \text{sum of all years} P=sum of all production values\sum P = \text{sum of all production values} tP=sum of the product of years and production\sum tP = \text{sum of the product of years and production} t2=sum of the squares of the years\sum t^2 = \text{sum of the squares of the years}

    Given: n=5n = 5 t=2009+2010+2011+2012+2013=10055\sum t = 2009 + 2010 + 2011 + 2012 + 2013 = 10055 P=12+20+28+32+50=142\sum P = 12 + 20 + 28 + 32 + 50 = 142 tP=(2009×12)+(2010×20)+(2011×28)+(2012×32)+(2013×50)\sum tP = (2009 \times 12) + (2010 \times 20) + (2011 \times 28) + (2012 \times 32) + (2013 \times 50) t2=20092+20102+20112+20122+20132\sum t^2 = 2009^2 + 2010^2 + 2011^2 + 2012^2 + 2013^2

    Calculate these sums: tP=2009×12+2010×20+2011×28+2012×32+2013×50=24108+40200+56308+64384+100650=285650\sum tP = 2009 \times 12 + 2010 \times 20 + 2011 \times 28 + 2012 \times 32 + 2013 \times 50 = 24108 + 40200 + 56308 + 64384 + 100650 = 285650 t2=20092+20102+20112+20122+20132=4036081+4040100+4044121+4048144+4052169=20280715\sum t^2 = 2009^2 + 2010^2 + 2011^2 + 2012^2 + 2013^2 = 4036081 + 4040100 + 4044121 + 4048144 + 4052169 = 20280715

  5. Calculate the coefficients aa and bb:

    Using the formulas: b=ntPtPnt2(t)2b = \frac{n \sum tP - \sum t \sum P}{n \sum t^2 - (\sum t)^2} a=Pbtna = \frac{\sum P - b \sum t}{n}

    Calculate bb: b=5×28565010055×1425×20280715(10055)2=14282501427810101403575101104025=4402995500.001468b = \frac{5 \times 285650 - 10055 \times 142}{5 \times 20280715 - (10055)^2} = \frac{1428250 - 1427810}{101403575 - 101104025} = \frac{440}{299550} \approx 0.001468

    Calculate aa: a=1420.001468×100555=14214.755145=127.24486525.449a = \frac{142 - 0.001468 \times 10055}{5} = \frac{142 - 14.75514}{5} = \frac{127.24486}{5} \approx 25.449

  6. Trend line equation: P=25.449+0.001468tP = 25.449 + 0.001468t

  7. Calculate the trend value for 2018: P2018=25.449+0.001468×2018=25.449+2.96162428.411P_{2018} = 25.449 + 0.001468 \times 2018 = 25.449 + 2.961624 \approx 28.411

So, the trend value of production for 2018 is approximately 28.41128.411 lakh tons.

Do you want more details or have any questions?

Here are five questions you might ask:

  1. How do you determine the best fit line in the least squares method?
  2. Can you show how to calculate the sums in more detail?
  3. What other methods can be used to fit a trend line to time series data?
  4. How do you interpret the coefficients aa and bb in the trend line equation?
  5. Can this method be used for non-linear trends?

Tip: When fitting a trend line, always check if a linear model is appropriate for your data or if a different type of model might provide a better fit.

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Math Problem Analysis

Mathematical Concepts

Statistics
Linear Regression
Least Squares Method

Formulas

Linear regression equation: P = a + bt
Formula for slope (b): b = (n * ∑(t*P) - ∑t * ∑P) / (n * ∑(t^2) - (∑t)^2)
Formula for intercept (a): a = (∑P - b * ∑t) / n

Theorems

-

Suitable Grade Level

High School