For a straight line trend y_t = a+bx for n data by the method of least squares if, n=8, y=47.8, xy=4.3 and x^2=42, then y_t at x=-3.5 is _. [where y_t represents trend value of y] Option 1: 3.983 Option 2: 6.332 Option 3: -3.4458 Option 4: 5.62

Answer: 5.62. Solution: With deviations taken about the mean, a=( y)/n=47.8/8=5.975 and b=( xy)/( x^2)=4.3/420.1024. At x=-3.5: y_t=5.975+0.1024(-3.5)5.975-0.358=5.62.

Statistics & Applications

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For a straight line trend $y_t = a+bx$ for $n$ data by the method of least squares if, $n=8, \sum y=47.8, \sum xy=4.3$ and $\sum x^2=42$ , then $y_t$ at $x=-3.5$ is _____.

[where $y_t$ represents trend value of y]

3.983

6.332

-3.4458

5.62

✅ Correct Option: 4

With deviations taken about the mean, $a=\frac{\sum y}{n}=\frac{47.8}{8}=5.975$ and $b=\frac{\sum xy}{\sum x^2}=\frac{4.3}{42}\approx0.1024$ .

At $x=-3.5$ : $y_t=5.975+0.1024(-3.5)\approx5.975-0.358=5.62$ .

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Solution

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