Question

A machine, when working properly, produces 5% or less defective items. Whenever the machine produces significantly greater than 5% defective items, it needs repair. A random sample of 300 items taken from the production line contained 27 defective items. Test at the 1% significance level if the percentage of defective items produced by this machine is greater than 5%.

Answer 1

Answer:

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.

Step-by-step explanation:

A machine, when working properly, produces 5% or less defective items.

This means that the null hypothesis is:

$H_0: p \leq 0.05$

Test if the percentage of defective items produced by this machine is greater than 5%.

This means that the alternate hypothesis is:

$H_a: p > 0.05$

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.

0.05 is tested at the null hypothesis:

This means that $\mu = 0.05, \sigma = \sqrt{0.05*0.95}$

A random sample of 300 items taken from the production line contained 27 defective items.

This means that $n = 300, X = \frac{27}{300} = 0.09$

Value of the test statistic:

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

$z = \frac{0.09 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{300}}}$

$z = 3.18$

Pvalue of the test:

Testing if the mean is greater than a value, which means that the pvalue of the test is 1 subtracted by the pvalue of Z = 3.18, which is the probability of a finding a sample proportion of 0.09 or higher.

Looking at the Z-table, Z = 3.18 has a pvalue of 0.9993

1 - 0.9993 = 0.0007

The pvalue of the test is 0.0007 < 0.01, which means that we reject the null hypothesis and accept the alternate hypothesis that the percentage of defective items produced by this machine is greater than 5%.