A quality control manager is concerned about variability of the net weight of his company’s individual yogurt cups. To check the consistency, he takes a random sample of sixteen 6-ounce yogurt cups and finds the mean of the sampled weights to be 5.85 ounces and the sample standard deviation to be 0.2 ounce. Requried:
a. Test the hypotheses H0: µ ≥ 6 Ha: µ < 6 at the 5% level of significance. Assume the population of yogurt-cup net weights is approximately normally distributed.
b. Based on the results of the test, will the manager be satisfied that the company is not under-filling its cups?
c. State the decision rule, the test statistic, and the manager’s decision.

H₀ should be rejected, yogurt cups are under-filling, the manager has to check the process

Step-by-step explanation:

Normal distribution and n < 30. We must use t-student test.

a) If it is required to test   H₀    μ >= 6     Hₐ   μ < 6  is a one tail test (left-tail)

At  α = 5 %     α = 0,05    and   n = 16  then degree of freedom is n -1

df = 15  find   t(c) = - 1,753

μ = 5,85     and  s  = 0,2      ( sample mean and standard deviation respectevily)

T compute   t(s)

t(s)  =  ( μ  -  μ₀ ) / s/√n

t(s)  =  ( 5,85 - 6 ) /0,2/√16

t(s)  =  - 0,15*4/0,2

t(s)  =  - 3

t(s) < t(c)       -3 < - 1,753

Then t(s) is in the rejection region we reject H₀. There is enough evidence to claim the yogurt cups are under-filling.

Manager has to check on the process

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64

x is the total time in minutes for the free response

you are given number of questions as 15 - m and time per item as 8 minutes

to find x multiply the two givens together:

x = 8 * (15-m) = 8(15-m)