Answer:
Pvalue of 0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.
Step-by-step explanation:
The average weight of a package of rolled oats is supposed to be at most 18 ounces.
This means that the null hypothesis is:
![H_{0}: \mu \leq 18](https://tex.z-dn.net/?f=H_%7B0%7D%3A%20%5Cmu%20%5Cleq%2018)
Is the true mean larger than the specification?
Due to the question asked, the alternate hypothesis is:
![H_{a}: \mu > 18](https://tex.z-dn.net/?f=H_%7Ba%7D%3A%20%5Cmu%20%3E%2018)
The test statistic is:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
In which X is the sample mean,
is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
18 is tested at the null hypothesis:
This means that ![\mu = 18](https://tex.z-dn.net/?f=%5Cmu%20%3D%2018)
A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces.
This means that ![n = 18, X = 18.2, \sigma = 0.5](https://tex.z-dn.net/?f=n%20%3D%2018%2C%20X%20%3D%2018.2%2C%20%5Csigma%20%3D%200.5)
Value of the test statistic:
![z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
![z = \frac{18.2 - 18}{\frac{0.5}{\sqrt{18}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B18.2%20-%2018%7D%7B%5Cfrac%7B0.5%7D%7B%5Csqrt%7B18%7D%7D%7D)
![z = 1.7](https://tex.z-dn.net/?f=z%20%3D%201.7)
Pvalue of the test:
Probability of finding a sample mean above 18.2, which is 1 subtracted by the pvalue of z = 1.7.
Looking at the z-table, z = 1.7 has a pvalue of 0.9554.
1 - 0.9554 = 0.0446
0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.