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

Answer:

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

And replacing we got:

$s= 0.286$

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

$SE= \frac{0.286}{\sqrt{4}}= 0.143$

Step-by-step 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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the standard error for this case would be:

$SE= \frac{\sigma}{\sqrt{n}}$

And we can estimate the deviation with the sample deviation:

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

And replacing we got:

$s= 0.286$

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

$SE= \frac{0.286}{\sqrt{4}}= 0.143$