Question

A local police chief claims that 76% of all drug-related arrests are never prosecuted. A sample of 700 arrests shows that 72% 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 different from what the chief stated. Is there enough evidence at the 0.01level to support the officer's claim? Step 1 of 7 : State the null and alternative hypotheses. Step 2 of 7 : Find the value of the test statistic. Round your answer to two decimal places. Step 3 of 7 : Specify if the test is one-tailed or two-tailed. Step 4 of 7 : Determine the P-value of the test statistic. Round your answer to four decimal places. Step 5 of 7 : Identify the value of the level of significance. Step 6 of 7 : Make the decision to reject or fail to reject the null hypothesis. Step 7 of 7 State the conclusion of the hypothesis test.

Answer 1

We have to test whether the proportion is different from 0.76 or not. So, it is a two tailed test.

STEP1

STEP 2

Population proportion p = 0.76, sample proportion = 0.72

sample size n = 700

STEP 3

It is a two tailed test

STEP 4

Using standard normal distribution table for z = -2.48 for a two sided test, we get

p value = 0.0131

STEP 5

level of significance = 0.01 (given in question)

STEP 6

Fail to reject Ho, since p-value (0.0131) > significance level (0.01)

STEP 7

There is not sufficient evidence to support the claim that the percentage of arrests that are never prosecuted is different from 76 %