Question

The meat department of a supermarket sells ground beef in approximate 1 lb packages, but there is some variability. A random sample of 65 packages yielded a mean of 1.05 lbs and a standard deviation of .23 lbs. What is the 90% Confidence Interval for this problem?

Answer 1

1.05 ± 0.05 lbs

Explanation:

Hi!

We can calculate this interval with the z-score of the 90% which is (by convention) 1.645

The interval is calculated as follows:

`x_m \pm 1.654((\sigma)/(\sqrt n))`

where x_m is the mean, σ the standar deviation and n is the number of samples:

replacing these values we get:

`1.05 \pm (1.645 * (0.23)/(\sqrt 65))\\1.05 \pm 0.046`

*rounded to the first decimal*

1.05 ± 0.05