Answer:
The p-value of the test is 0.9918 > 0.05, which means that we fail to reject
, as we do not have enough evidence to say that defect rate of machine is smaller than 3%.
Step-by-step explanation:
The null and alternate hypothesis are:
H0:p≥0.03
Ha:p<0.03
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)
28 out of 600 items produced by the machine were found defective.
This means that ![n = 600, X = \frac{28}{600} = 0.0467](https://tex.z-dn.net/?f=n%20%3D%20600%2C%20X%20%3D%20%5Cfrac%7B28%7D%7B600%7D%20%3D%200.0467)
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{0.0467 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{600}}}](https://tex.z-dn.net/?f=z%20%3D%20%5Cfrac%7B0.0467%20-%200.03%7D%7B%5Cfrac%7B%5Csqrt%7B0.03%2A0.97%7D%7D%7B%5Csqrt%7B600%7D%7D%7D)
![z = 2.4](https://tex.z-dn.net/?f=z%20%3D%202.4)
P-value of the test:
The p-value of the test is the probability of finding a sample proportion below 0.0467, which is the p-value of z = 2.4.
Looking at the z-table, z = 2.4 has a p-value of 0.9918.
The p-value of the test is 0.9918 > 0.05, which means that we fail to reject
, as we do not have enough evidence to say that defect rate of machine is smaller than 3%.