Question

A local police chief claims that 66% 66 % of all drug-related arrests are never prosecuted. A sample of 700 700 arrests shows that 63% 63 % of the arrests were not prosecuted. Using this information, one officer wants to test the claim that the number of arrests that are never prosecuted is under what the chief stated. Is there enough evidence at the 0.02 0.02 level to support the officer's claim?

State the null and alternative hypotheses.

Find the value of the test statistic. Round your answer to two decimal places.

Specify if the test is one-tailed or two-tailed.

Determine the P-value of the test statistic. Round your answer to four decimal places.

Identify the value of the level of significance.

Make the decision to reject or fail to reject the null hypothesis.

State the conclusion of the hypothesis test.

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.66
Alternative Hypothesis, Ha: p < 0.66

This is left tailed test
Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.63 - 0.66)/sqrt(0.66*(1-0.66)/700)
z = -1.68

P-value Approach
P-value = 0.0465

0.02 is the level of significance

As P-value >= 0.02, fail to reject null hypothesis.

There is not sufficient evidence to conclude that the number of arrests that are never prosecuted is under what the chief stated.