A soft drink filling machine, when in perfect adjustment, fills the bottles with 12 ounces of soft drink. A random sample of 25 bottles is selected, and the contents are measured. The sample yielded a mean content of 11.88 ounces, with a standard deviation of 0.24 ounces. With a 0.05 level of significance, test to see if the machine is in perfect adjustment. Assume the distribution of the population is normal.

we reject H₀

Step-by-step explanation:

Normal Distribution

sample size n = 25 degees of fredom = 25 - 1 df = 24

sample standard deviation = s = 0,24

sample mean 11.88

We have a one tail test (left) investigation

1.-Test hypothesis

H₀ ⇒ null hypothesis μ₀ = 12

Hₐ ⇒ Alternative hypothesis μ₀ < 12

2.-Significance level 0,05 t (c) = - 1.7109

3.-Compute of t (s)

t (s) = (μ - μ₀) / s/√n ⇒ t (s) = [ (11.88 - 12) * √25 ]/0.24

t (s) = - 0.12*5/0.24

t (s) = - 2.5

4.-We compare t (s) with t (c)

In this case t (s) < t (c) - 2.5 < - 1.71

5.-t (s) is in the rejection region, we reject H₀

The machine is not adjusted