A set of 11 (x, y) data points is least-squares fitted to a quadratic polynomial. If the sum of squares of error is 2.4, the variance of error is ________ (round off to 1decimal place).
This solution explains how to find the variance of error when fitting a quadratic polynomial using the least-squares method.
Degrees of freedom (df) for error are calculated as the number of data points minus the number of parameters:
$ df = n - p $
With $n=11$ and $p=3$:
$ df = 11 - 3 = 8 $
The variance of the error (also known as Mean Squared Error or MSE) is SSE divided by the degrees of freedom:
$ \text{Variance of Error} = \frac{SSE}{df} $
Substituting the given values:
$ \text{Variance of Error} = \frac{2.4}{8} = 0.3 $
The variance of the error, rounded to one decimal place, is 0.3.
Consider a distribution with the following probability density function $$f(x) = \begin{cases} 0.5, & 0 < x < 2 \\ 0.0, & Otherwise \end{cases}$$ Given that the mean of the above probability distribution is 1, the variance (rounded off to two decimal places) is _______________.
The unbiased sample variance for the set of numbers: $S = \{40,45,50,55,60\}$ is_____. (write answer with one decimal place)