Question

A local police chief claims that 42%of all drug-related arrests are never prosecuted. A sample of 600 arrests shows that 38%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 less than what the chief stated. Is there enough evidence at the 0.02level to support the officer's claim?

Step 3 of 7:

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

Answer 1

Solution:

Given:

Claim: 42%of all drug-related arrests are never prosecuted.

Sample size = n = 600

Sample proportion =

One officer wants to test the claim that the number of arrests that are never prosecuted is less than what the chief stated.

Level of significance = 0.02

Step 1) State H0 and H1:

H0: p =0.42

Vs

H1: p < 0.42

Step 2) Test statistic:

Step 3) Specify if the test is one-tailed or two-tailed.

Since H1 is < type, this is one tailed test.

Step 4) Decision Rule:

Reject null hypothesis ,if z  test statistic value < z critical value , otherwise we fail to reject H0.

thus we need find z critical value.

Level of significance = 0.02

Look in z table for area = 0.0200 or its closest area and find z value

Area 0.0202 is closest to 0.0200 and it corresponds to -2.0 and 0.05

that is  z = -2.05

Thus

Reject null hypothesis ,if z  test statistic value < z critical value = - 2.05.

Step 5) Decision:

Since z  test statistic value =   > z critical value = - 2.05, we fail to reject null hypothesis H0.

Step 6) Conclusion:

At 0.02 significance level , we do not have sufficient evidence to support the officer's claim that the number of arrests that are never prosecuted is less than what the chief stated.