Question

A local police chief claims that 40% of all robbery-related arrests are never prosecuted. A sample of 800 arrests shows that 37% 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 0.01 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

Step 1 Null and alternative hypotheses

Ho : = 0.40

H1 : 0.40

Step 2) we have n = 800, p^ = 0.37

Test statistic Z

Z = ( p^ - p)/sqrt[ p*(1-p)/n]

Z = (0.37 - 0.40)/SQRT [ 0.40*0.60/800]

Z test statistic = -1.73

Step 3) two tailed test

Step 4) p-value = 2 * P( Z < -1.73)

p-value = 0.0836

Step 5) level of significance a = 0.01

Step 6) Decision rule : if p-value < 0.01 we reject the null hypothesis otherwise not

Here p-value = 0.0836 > 0.01

Decision : Fail to reject the null hypothesis

Step 7) conclusion : There is no sufficient evidence to support the claim the number of arrests that are never prosecuted is different from what the chief stated.