Answer: We reject H₀ we find that the machine is underflling the bottles
P ( 421 ± 4,3376 )
Step-by-step explanation:
We assume a normal distribution
Population mean 430 grs
Unknown standard deviation
We have a one tail condition "underfilling"
And our test is:
Null hypothesis H₀ X = μ₀
Alternative hypothesis Hₐ X < μ₀
We must use t student distribution and find the interval
X ± t*(s/√n)
In that expession X is the sample mean 421 grs, "s" is sample standard deviation, n is sample size, then
421 ± t * ( 15 / √21 ) (1)
We go to t table and look for t value for α = 0,1 and df = 21 - 1 df = 20
we get t (remember it is a one tail test) t = 1,325, plugging this value in equation (1) we get the interval
421 ± 1,325 * ( 15/√21 ) ⇒ 421 ± 1,325 * ( 3,2737)
421 ± 4,3376
421 + 4,3376 = 425,34
421 - 4,3376 = 416,66
As we can see the mean value of the population 430 grs is not inside the interval [ 416,66 ; 425,34 ] then we can assure the machine is underfilling the bags, and not meeting the setting spec
