A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It is believed that the machine is underfilling the bags. A 21 bag sample had a mean of 421 grams with a standard deviation of 15. Assume the population is normally distributed. A level of significance of 0.1 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation,

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