The following table gives the production of steel (in millions of tons) for years 1976 to 1986 :Years19761977197819791980198119821983198419851986Production0442685941010Fit a trend line to the above data by the method of least squares.

Q.5 (B)(ii) The following table gives the production of steel (in millions of tons) for years 1976 to 1986 :

Years	1976	1977	1978	1979	1980	1981	1982	1983	1984	1985	1986
Production	0	4	4	2	6	8	5	9	4	10	10

Fit a trend line to the above data by the method of least squares.

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(y = 5.636 + 0.791u\), where \(u = t - 1981\)

Here \(n = 11\) (odd), so take \(u = t - 1981\) (the middle year), giving \(\Sigma u = 0\).

Year \(t\)	Production \(y\)	\(u = t - 1981\)	\(u^2\)	\(uy\)	Trend value
1976	0	-5	25	0	1.68
1977	4	-4	16	-16	2.47
1978	4	-3	9	-12	3.26
1979	2	-2	4	-4	4.05
1980	6	-1	1	-6	4.85
1981	8	0	0	0	5.64
1982	5	1	1	5	6.43
1983	9	2	4	18	7.22
1984	4	3	9	12	8.01
1985	10	4	16	40	8.80
1986	10	5	25	50	9.59
Total	62	0	110	87

Let the trend line be \(y = a + bu\). The normal equations reduce (as \(\Sigma u = 0\)) to \(\Sigma y = na\) and \(\Sigma uy = b\,\Sigma u^2\).

\(a = \dfrac{\Sigma y}{n} = \dfrac{62}{11} = 5.636\), \(b = \dfrac{\Sigma uy}{\Sigma u^2} = \dfrac{87}{110} = 0.791\)

Trend line: \(y = 5.636 + 0.791u\), where \(u = t - 1981\). The trend values are shown in the last column of the table.

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

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