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

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:

1. **Organize the **
   - Years (\(t\)) and Production (\(P\)):

   | Year (t) | Production (P) |
   |----------|----------------|
   | 2009     | 12             |
   | 2010     | 20             |
   | 2011     | 28             |
   | 2012     | 32             |
   | 2013     | 50             |

2. **Define the trend line equation:**
   The trend line can be defined as:
   \[
   P = a + bt
   \]
   where \(P\) is the production, \(t\) is the year, \(a\) is the intercept, and \(b\) is the slope.

3. **Calculate the necessary sums for the least squares method:**

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

   Given:
   \[
   n = 5
   \]
   \[
   \sum t = 2009 + 2010 + 2011 + 2012 + 2013 = 10055
   \]
   \[
   \sum P = 12 + 20 + 28 + 32 + 50 = 142
   \]
   \[
   \sum tP = (2009 \times 12) + (2010 \times 20) + (2011 \times 28) + (2012 \times 32) + (2013 \times 50)
   \]
   \[
   \sum t^2 = 2009^2 + 2010^2 + 2011^2 + 2012^2 + 2013^2
   \]

   Calculate these sums:
   \[
   \sum tP = 2009 \times 12 + 2010 \times 20 + 2011 \times 28 + 2012 \times 32 + 2013 \times 50 = 24108 + 40200 + 56308 + 64384 + 100650 = 285650
   \]
   \[
   \sum t^2 = 2009^2 + 2010^2 + 2011^2 + 2012^2 + 2013^2 = 4036081 + 4040100 + 4044121 + 4048144 + 4052169 = 20280715
   \]

4. **Calculate the coefficients \(a\) and \(b\):**

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

   Calculate \(b\):
   \[
   b = \frac{5 \times 285650 - 10055 \times 142}{5 \times 20280715 - (10055)^2} = \frac{1428250 - 1427810}{101403575 - 101104025} = \frac{440}{299550} \approx 0.001468
   \]

   Calculate \(a\):
   \[
   a = \frac{142 - 0.001468 \times 10055}{5} = \frac{142 - 14.75514}{5} = \frac{127.24486}{5} \approx 25.449
   \]

5. **Trend line equation:**
   \[
   P = 25.449 + 0.001468t
   \]

6. **Calculate the trend value for 2018:**
   \[
   P_{2018} = 25.449 + 0.001468 \times 2018 = 25.449 + 2.961624 \approx 28.411
   \]

So, the trend value of production for 2018 is approximately \(28.411\) 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 \(a\) and \(b\) 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.

Learn how to fit a trend line to time series data using the least squares method and calculate the production value for 2018. Detailed steps include organizing data, calculating necessary sums, determining coefficients, and interpreting results. Suitable for high school students studying statistics and linear regression.

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