Answer:
The p-value of the test is 0.9918 > 0.05, which means that we fail to reject [tex]H_0[/tex], 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:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.03 is tested at the null hypothesis:
This means that [tex]\mu = 0.03, \sigma = \sqrt{0.03*0.97}[/tex]
28 out of 600 items produced by the machine were found defective.
This means that [tex]n = 600, X = \frac{28}{600} = 0.0467[/tex]
Value of the test-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.0467 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{600}}}[/tex]
[tex]z = 2.4[/tex]
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 [tex]H_0[/tex], as we do not have enough evidence to say that defect rate of machine is smaller than 3%.
