Question

The average weight of the entire batch of the boxes of cereal filled today was 20.5 ounces. A random sample of four boxes was selected with the following weights: 20.05, 20.56, 20.72, and 20.43. The sampling error for this sample is ________.

Answer 1

`s= \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And replacing we got:

`s= 0.286`

And then the estimator for the standard error is given by:

`SE= (0.286)/(√(4))= 0.143`

Explanation:

For this case we have the following dataset given:

20.05, 20.56, 20.72, and 20.43

We can assume that the distribution for the sample mean is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And the standard error for this case would be:

`SE= (\sigma)/(√(n))`

And we can estimate the deviation with the sample deviation:

`s= \sqrt{(\sum_(i=1)^n (X_i -\bar X)^2)/(n-1)}`

And replacing we got:

`s= 0.286`

And then the estimator for the standard error is given by:

`SE= (0.286)/(√(4))= 0.143`