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Question 8 of 10 , Step 2 of 7
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A local police chief claims that $43 \%$ of all drug-related arrests are never prosecuted. A sample of 900 arrests shows that $40 \%$ 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 eviden at the 0.02 level to support the officer's claim?

Step 2 of 7: Find the value of the test statistic. Round your answer to two decimal places.

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The value of the test statistic, rounded to two decimal places, is \(\boxed{-1.82}\).

Steps

Step 1 :The question is asking for the value of the test statistic. In this case, we are dealing with a proportion test. The test statistic for a proportion test is calculated as \( (\hat{p} - p_0) / \sqrt{(p_0*(1-p_0))/n} \), where \( \hat{p} \) is the sample proportion, \( p_0 \) is the population proportion, and \( n \) is the sample size. Here, \( \hat{p} = 0.40 \), \( p_0 = 0.43 \), and \( n = 900 \).

Step 2 :Let's calculate the test statistic: \( \hat{p} = 0.4 \), \( p_0 = 0.43 \), \( n = 900 \).

Step 3 :The value of the test statistic, rounded to two decimal places, is \(\boxed{-1.82}\).

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