Question

A manufacturer makes chocolate squares that have a target weight of 8 g. Quality control engineers sample 30 chocolate squares to see whether H0: μ = 8 vs. Ha: μ ≠ 8 where μ is the true average weight of the chocolate squares. They calculate the sample mean of the 30 squares to be 8.3 g. Suppose the population standard deviation is 0.87 g and the true weight is 8.5 g. At the 5% level of significance, calculate the power for the test. For full credit, include all steps as described in class.

Answer 1

Using the normal distribution and the central limit theorem, it is found that the power of the test is of 0.9992 = 99.92%.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

In this problem:

• The mean is $\mu = 8.5$.

• The standard deviation is $\sigma = 0.87$.

• A sample of 30 is taken, hence $n = 30, s = \frac{0.87}{\sqrt{30}} = 0.1588$.

The power of the test is given by the probability of a sample mean above 8, which is 1 subtracted by the p-value of Z when X = 8, so:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

$Z = \frac{8 - 8.5}{0.1588}$

$Z = -3.15$

$Z = -3.15$ has a p-value of 0.0008.

1 - 0.0008 = 0.9992.

The power of the test is of 0.9992 = 99.92%.

To learn more about the normal distribution and the central limit theorem, you can check brainly.com/question/24663213