Asoft 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