Question

In the 1990s, it was generally believed that genetic abnormalities affected about 5% of a large nation's children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities, which could have important implications for health insurance companies. A recent study examined 384 randomly selected children and found that 25 of them showed signs of a genetic abnormality.
Required:
What is the p value?

Answer 1

Answer:

The p-value of the test is 0.0869.

Step-by-step explanation:

In the 1990s, it was generally believed that genetic abnormalities affected about 5% of a large nation's children.

This means that the null hypothesis is:

$H_{0}: p = 0.05$

Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities.

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.

5% tested at the null hypothesis:

This means that:

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

A recent study examined 384 randomly selected children and found that 25 of them showed signs of a genetic abnormality.

This means that $n = 384, X = \frac{25}{384} = 0.0651$

Test statistic:

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

$z = \frac{0.0651 - 0.05}{\frac{\sqrt{0.05*0.95}}{\sqrt{384}}}$

$z = 1.36$

p-value:

Testing if the mean is bigger than a value, so the pvalue is the probability of finding a value in the z table higher than 1.36, which is 1 subtracted by the pvalue of z = 1.36.

z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

The p-value of the test is 0.0869.