Question

Lucky Charms are sold in boxes with an advertised mean weight of 20.5 oz. The standard deviation is known to be 2.25 oz. A consumer group claims that the makers of Lucky Charms are cheating consumers by underfilling the boxes. They take a sample of 50 boxes and find an average weight of 20.4 oz. What is the set of hypotheses that should be used to test the accuracy of the advertised weight?

Answer 1

Set of hypothesis.

Explanation:

We are given the following in the question:

Population mean, μ = 20.5 oz

Sample mean,
`\bar{x}` = 20.4 oz

Sample size, n = 50

Population standard deviation, σ = 2.25 oz

First, we design the null and the alternate hypothesis

`H_(0): \mu = 20.5\text{ oz}\\H_A: \mu < 20.5\text{ oz}`

We use One-tailed z test(left tailed) to perform this hypothesis.

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(20.4 - 20.5)/((2.25)/(√(50)) ) = -0.3142`

Based on the significance level, we can make our decision and complete the hypothesis testing.

We may accept the null hypothesis or fail to accept depending on different significance level.