Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]Step 1: Assign \( t \): Let \( t = -3, -2, -1, 0, 1, 2, 3 \) for the years 2004 to 2010.
Step 2: Tabulate the data:
\[ \begin{array}{|c|c|c|c|c|} \hline \textbf{Year} & \textbf{Profit (y)} & t & t^2 & t \cdot y \\ \hline 2004 & 114 & -3 & 9 & -342 \\ 2005 & 130 & -2 & 4 & -260 \\ 2006 & 126 & -1 & 1 & -126 \\ 2007 & 144 & 0 & 0 & 0 \\ 2008 & 138 & 1 & 1 & 138 \\ 2009 & 156 & 2 & 4 & 312 \\ 2010 & 164 & 3 & 9 & 492 \\ \hline \textbf{Total} & 972 & 0 & 28 & 214 \\ \hline \end{array} \]Step 3: Use the least squares formula:
\[ y = a + bt, \]where:
\[ a = \frac{\sum y}{N}, \quad b = \frac{\sum (t \cdot y)}{\sum t^2}. \]Substituting the values:
\[ a = \frac{972}{7} = 138.86, \quad b = \frac{214}{28} = 7.64. \]Step 4: Write the equation:
\[ y = 138.86 + 7.64t. \]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:
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \textbf{Year} & 1996 & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \textbf{Sales (in lakh ₹)} & 6.5 & 5.3 & 4.3 & 6.1 & 5.6 & 7.8 \\ \hline \end{array} \]