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

Answer:

1.05 ± 0.05 lbs

Step-by-step 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:

$x_m \pm 1.654(\frac{\sigma}{\sqrt n})$

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

replacing these values we get:

$1.05 \pm (1.645 \times \frac{0.23}{\sqrt 65})\\1.05 \pm 0.046$

*rounded to the first decimal*

1.05 ± 0.05