Step 1: Use the trend equation \( y = 5.93 + 1.18t \).
Step 2: Compute the trend values for \( t = -2, -1, 0, 1, 2, 3 \):
\[ \begin{array}{|c|c|c|} \hline \textbf{Year} & t & \textbf{Trend Value (y)} \\ \hline 1996 & -2 & 5.93 + 1.18(-2) = 3.57 \\ 1997 & -1 & 5.93 + 1.18(-1) = 4.75 \\ 1998 & 0 & 5.93 + 1.18(0) = 5.93 \\ 1999 & 1 & 5.93 + 1.18(1) = 7.11 \\ 2000 & 2 & 5.93 + 1.18(2) = 8.29 \\ 2001 & 3 & 5.93 + 1.18(3) = 9.47 \\ \hline \end{array} \]Step 3: Compute the trend value for 2002 (\( t = 4 \)):
\[ y = 5.93 + 1.18(4) = 10.65. \]Final Answer:
Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]When observed over a long period of time, a time series data can predict trends that can forecast increase, decrease, or stagnation of a variable under consideration. The table below shows the sale of an item in a district during 1996–2001:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]