A local police chief claims that about 51% of all drug related arrests are ever prosecuted. A sample of 900 arrests shows that 47% of the arrests were prosecuted. Is there sufficient evidence at the 0.01 level to refute the chief's claim? State the null and alternative hypotheses for the above scenario.

The null hypothesis is .

The alternate hypothesis is .

The p-value of the test is 0.0164 > 0.01, which means that there is not sufficient evidence at the 0.01 level to refute the chief's claim.

Step-by-step explanation:

A local police chief claims that about 51% of all drug related arrests are ever prosecuted

At the null hypothesis, we test if the proportion is of 51%, that is:

At the alternate hypothesis, we test if the proportion is different from 51%, that is:

The test statistic is:

In which X is the sample mean, is the value tested at the null hypothesis, is the standard deviation and n is the size of the sample.

0.51 is tested at the null hypothesis:

This means that

A sample of 900 arrests shows that 47% of the arrests were prosecuted.

This means that

P-value of the test:

Probability that the sample proportion differs from 0.51 by at least 0.04, which is P(|z|>2.4), which is 2 multiplied by the p-value of Z = -2.4.

Looking at the z-table, the Z = -2.4 has a p-value of 0.0082.

2*0.0082 = 0.0164.

The p-value of the test is 0.0164 > 0.01, which means that there is not sufficient evidence at the 0.01 level to refute the chief's claim.

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

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step-by-step explanation: