Question

Determine which of the following show three biased estimators. (1 point) a. median, mean, range b. range, standard deviation, variance c. standard deviation, median, ranged. variance, proportion, mean

Answer 1

Answer: standard deviation, median, range

Explanation:

Answer 2

c. standard deviation, median, range

Explanation:

The standard deviation without the Bessel's correct is defined as:

`s= \sqrt{(\sum_(i=1)^n (x_i -\bar x)^2)/(n)`

And if we find the expected value for s we got:

`E(s^2) = (1)/(n) \sum_(i=1)^n E(x_i -\bar x)^2`

`E(s^2)= (1)/(n) E[\sum_(i=1)^n ((x_i -\mu)-(\bar x -\mu)^2)]`

We have this:

`E(\sum_(i=1)^n(x_i-\mu)^2) =n\sigma^2`

`E[\sum_(i=1)^n (x_i -\mu)(\bar x -\mu)]= \sigma^2`

`E[\sum_(i=1)^n (\bar x -\mu)^2]=\sigma^2`

`E(s^2)=(1)/(n) (n\sigma^2 -2\sigma^2 +\sigma^2)`

`E(s^2)=(n-1)/(n)\sigma^2`

as we can see the sample variance is a biased estimator since:

`E(s^2)\\eq \sigma^2`

And we see that the standard deviation is biased, since:

`E(s) = \sqrt{(n-1)/(n)} \sigma`

because
`E(s)\\eq \sigma`

The mean is not biased for this case option a is FALSE.

The proportion is not biased for this reason option d is FALSE

The range can be considered as biased since we don't have info to conclude that the range follows a distirbution in specific.

The sample median "is an unbiased estimator of the population median when the population is normal. However, for a general population it is not true that the sample median is an unbiased estimator of the population median".

And for this reason the best option is c.