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Question 15 of 41
Q.

The birth rates of a state in different years are given in the following table. Fit a linear trend for these data. Also find the estimate of birth rate for the year 2020:

Year2012201320142015201620172018
Birth rate22.221.821.320.920.620.219.9

OR

Find the trend using five yearly moving averages for the following data about yearly production (in tons) of a factory:

Year20092010201120122013201420152016201720182019
Production (tons)1121069390114159170130108113115
Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2020Subjective· 5mImportance★★★★★
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Odd nn: origin 2015, t=−3..3t=-3..3, ∑t2=28\sum t^2=28; a=146.9/7=20.99a=146.9/7=20.99, b=−10.8/28=−0.386b=-10.8/28=-0.386; y^=20.99−0.386t\hat y=20.99-0.386t; 2020 (t=5t=5): 19.0619.06. OR: 5-yr moving averages listed.

Main part — linear trend for birth rate (7 years).

Origin at the middle year 2015, t=year−2015t = \text{year} - 2015, so t=−3,−2,−1,0,1,2,3t = -3,-2,-1,0,1,2,3; ∑t=0\sum t = 0, ∑t2=28\sum t^2 = 28.

Yearyytttyty
201222.2-3-66.6
201321.8-2-43.6
201421.3-1-21.3
201520.900
201620.6120.6
201720.2240.4
201819.9359.7
Σ146.90-10.8

Because ∑t=0\sum t = 0:

a=∑yn=146.97=20.9857,b=∑ty∑t2=−10.828=−0.3857a = \frac{\sum y}{n} = \frac{146.9}{7} = 20.9857, \qquad b = \frac{\sum ty}{\sum t^2} = \frac{-10.8}{28} = -0.3857

Trend line: y^=20.99−0.3857 t\hat{y} = 20.99 - 0.3857\,t (origin 2015).

For 2020, t=2020−2015=5t = 2020 - 2015 = 5:

y^=20.9857−0.3857(5)=20.9857−1.9286=19.06\hat{y} = 20.9857 - 0.3857(5) = 20.9857 - 1.9286 = 19.06

OR — five-yearly moving averages of production (tons).

YearProduction5-yr moving total5-yr moving average
2009112––
2010106––
201193515103.0
201290562112.4

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