Question

60% of all robbery-related arrests are never prosecuted. A sample of 500 arrests shows that 56% 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 0.05 level to support the officer's claim?

Step 1 State the null and alternative hypotheses.

Step 2 Find the value of the test statistic. Round your answer to two decimal places

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

Step 4 Determine the P-value of the test statistic. Round your answer to four decimal places.

Step 5 Identify the value of the level of significance.

Step 6 Make the decision to reject or fail to reject the null hypothesis.

Step 7 State the conclusion of the hypothesis test.

Answer 1

Solution :

The null and alternative hypothesis is

H0 : p = 0.60

Ha : p 0.60

= 0.56

1 - P0 = 0.40

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.56 - 0.60 / [(0.60 * 0.40) / 500]

z = -1.83

This is two tailed test.

P-value = 0.0679

The level of significance () = 0.05

P-value >

Fail to reject the null hypothesis .

There is insufficient evidence at the 0.05 level to support the officer's claim