Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- The mean is [tex]\mu = 8.5[/tex].
- The standard deviation is [tex]\sigma = 0.87[/tex].
- A sample of 30 is taken, hence [tex]n = 30, s = \frac{0.87}{\sqrt{30}} = 0.1588[/tex].
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{8 - 8.5}{0.1588}[/tex]
[tex]Z = -3.15[/tex]
[tex]Z = -3.15[/tex] 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 https://brainly.com/question/24663213
