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.

a) We reject H₀

b) The manager won´t be satisfied with nominal filling its cup

c) See step-by-step explanation

Step-by-step explanation:

Normal distribution  n <  30, therefore, we should use t - student table

Sample size  n  =  16

degree of freedom  =   df =  n - 1     df = 15

Sample mean    μ  = 5,85 ou

Sample standard deviation  is    s = 0,2 ou

Hypothesis test

Null hypothesis                               H₀              μ   >=  μ₀

Alternative hypothesis                   Hₐ               μ   <   μ₀

CI  = 95 %    then   α = 5 %     α = 0,05     α/2  =  0,025

Then in t-student table we find  t(c) = 1,753

To calculate  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

To compare  t(s) and t(c)

|t(s)| > |t(c)|        3 > 1,753

Then t(s) is in the rejection region. We should reject H₀. Data indicate that at 95 % of CI μ seems to be smaller than 6 ou

b) The manager won´t be satisfied with nominal filling its cup

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6

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step-by-step explanation: