Question

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.

Answer 1

a) We reject H₀

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

c) See step-by-step explanation

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