Question

A manufacturer of nails claims that only 4% of its nails are defective. A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective. Is it fair to reject the manufacturer's claim based on this observation?

Answer 1

Answer:

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

Step-by-step explanation:

A manufacturer of nails claims that only 4% of its nails are defective.

At the null hypothesis, we test if the proportion is of 4%, that is:

$H_0: p = 0.04$

At the alternative hypothesis, we test if the proportion is more than 4%, that is:

$H_a: p > 0.04$

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.

4% is tested at the null hypothesis

This means that $\mu = 0.04, \sigma = \sqrt{0.04*0.96}$

A random sample of 20 nails is selected, and it is found that two of them, 10%, are defective.

This means that $n = 20, X = 0.1$

Value of the test statistic:

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

$z = \frac{0.1 - 0.04}{\frac{\sqrt{0.04*0.96}}{\sqrt{20}}}$

$z = 1.37$

P-value of the test and decision:

Considering an standard significance level of 0.05.

The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.

Looking at the z-table, z = 1.37 has a p-value of 0.9147

1 - 0.9147 = 0.0853

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

Answer 2

Answer:

Considering an standard significance level of 0.05.

The p-value of the test is the probability of finding a sample proportion above 0.1, which is 1 subtracted by the p-value of z = 1.37.

Looking at the z-table, z = 1.37 has a p-value of 0.9147

1 - 0.9147 = 0.0853

The p-value of the test is 0.0853 > 0.05, which means that there is not enough evidence to reject the manufacturer's claim based on this observation.

Step-by-step explanation: