Question

A local police chief claims that about 51% of all drug related arrests are ever prosecuted. A sample of 900 arrests shows that 47% of the arrests were prosecuted. Is there sufficient evidence at the 0.01 level to refute the chief's claim? State the null and alternative hypotheses for the above scenario.

Answer 1

Answer:

The null hypothesis is $H_0: p = 0.51$.

The alternate hypothesis is $H_1: p \neq 0.51$.

The p-value of the test is 0.0164 > 0.01, which means that there is not sufficient evidence at the 0.01 level to refute the chief's claim.

Step-by-step explanation:

A local police chief claims that about 51% of all drug related arrests are ever prosecuted

At the null hypothesis, we test if the proportion is of 51%, that is:

$H_0: p = 0.51$

At the alternate hypothesis, we test if the proportion is different from 51%, that is:

$H_1: p \neq 0.51$

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.

0.51 is tested at the null hypothesis:

This means that $\mu = 0.51, \sigma = \sqrt{0.51*0.49}$

A sample of 900 arrests shows that 47% of the arrests were prosecuted.

This means that $n = 900, X = 0.47$

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

$z = \frac{0.47 - 0.51}{\frac{\sqrt{0.51*0.49}}{\sqrt{900}}}$

$z = -2.4$

P-value of the test:

Probability that the sample proportion differs from 0.51 by at least 0.04, which is P(|z|>2.4), which is 2 multiplied by the p-value of Z = -2.4.

Looking at the z-table, the Z = -2.4 has a p-value of 0.0082.

2*0.0082 = 0.0164.

The p-value of the test is 0.0164 > 0.01, which means that there is not sufficient evidence at the 0.01 level to refute the chief's claim.