Question

A company considers buying a machine to manufacture a certain item. When tested, 28 out of 600 items produccd by the machine were found defective. Do the data support the hypothesis that the defect rate of the machine is smaller than 3%, at the 5% significance level?
H0:p≥0.03
Ha:p<0.03
α=0.05
Critical value z=−1.645
Reject H0 if z<−1.645
Test statistic
⇒Z0=X¯¯¯¯¯−μoσ/n√=>0.0467−0.030.03∗0.97/600√=14.057
Therefore, we fail to reject H0. There isn't enough evidence to say that defect rate of machine is smaller than 3%.

Answer 1

Answer:

The p-value of the test is 0.9918 > 0.05, which means that we fail to reject $H_0$, 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:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.03 is tested at the null hypothesis:

This means that $\mu = 0.03, \sigma = \sqrt{0.03*0.97}$

28 out of 600 items produced by the machine were found defective.

This means that $n = 600, X = \frac{28}{600} = 0.0467$

Value of the test-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.0467 - 0.03}{\frac{\sqrt{0.03*0.97}}{\sqrt{600}}}$

$z = 2.4$

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 $H_0$, as we do not have enough evidence to say that defect rate of machine is smaller than 3%.