122k views
2 votes
The process that produces Sonora Bars (a type of candy) is intended to produce bars with a mean weight of 56 gm. The process standard deviation is known to be 0.77 gm. A random sample of 49 candy bars yields a mean weight of 55.82 gm.

(a) State the hypotheses to test whether the mean is smaller than it is supposed to be.
(b) What is the test statistic?
(c) At α = .05, what is the critical value for this test?
(d) What is your conclusion?

1 Answer

4 votes

Answer:

|z(s)| < |z(c)|

Then z(s) is in the acceptance region we accept H₀.

We don´t have evidence to support that candy bar mean weight is below 56 gm

Explanation:

Goal mean μ = 56 gm

standard deviation σ = 0,77

From random sample:

mean x = 55,82

sample size n = 49

We assume a normal distribution ( is a manufacturing controlled process)

a) Hypothesis Test:

Null Hypothesis H₀ x = μ

Alternative Hypothesis Hₐ x < μ

Alternative hypothesis is telling us that the test is a one-tail test to the left

as n > 30 we use the normal distribution.

b) z(s) = ( x - μ ) / σ / √n

z(s) = 55.82 - 56 ) * 7 / 0,77

z(s) = 0,18*7/0,77

z(s) = - 1.6363

c) if significance level is α = 0.05 then α/2 = 0,025

Then from z table, z score for 0,025 z(c) = -1.96

d) Comparing z(c) and z(s) modules

|z(s)| < |z(c)|

Then z(s) is in the acceptance region we accept H₀.

We don´t have evidence to support that candy bar mean weight is below 56 gm

User RussS
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories