Answer:
The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.
Step-by-step explanation:
A manufacturer considers his production process to be out of control when defects exceed 3%.
At the null hypothesis, we test if the production process is in control, that is, the defective proportion is of 3% or less. So
![H_0: p \leq 0.03](https://tex.z-dn.net/?f=H_0%3A%20p%20%5Cleq%200.03)
At the alternate hypothesis, we test if the production process is out of control, that is, the defective proportion exceeds 3%. So
![H_1: p > 0.03](https://tex.z-dn.net/?f=H_1%3A%20p%20%3E%200.03)
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.
0.03 is tested at the null hypothesis
This means that ![\mu = 0.03, \sigma = \sqrt{0.03*0.97}](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.03%2C%20%5Csigma%20%3D%20%5Csqrt%7B0.03%2A0.97%7D)
In a random sample of 100 items, the defect rate is 4%.
This means that ![n = 100, X = 0.04](https://tex.z-dn.net/?f=n%20%3D%20100%2C%20X%20%3D%200.04)
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{0.04 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{100}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.04%20-%200.03%7D%7B%5Cfrac%7B%5Csqrt%7B0.03%2A0.97%7D%7D%7B%5Csqrt%7B100%7D%7D%7D)
![z = 0.59](https://tex.z-dn.net/?f=z%20%3D%200.59)
P-value of the test
The p-value of the test is the probability of finding a sample proportion above 0.04, which is 1 subtracted by the p-value of z = 0.59.
Looking at the z-table, z = 0.59 has a p-value of 0.7224
1 - 0.7224 = 0.2776
The p-value of the test is of 0.2776 > 0.01, which means that the we accept the null hypothesis, that is, the manager's claim that this is only a sample fluctuation and production is not really out of control.