Respuesta :
Answer:
The null hypothesis is [tex]H_0: p = 0.51[/tex].
The alternate hypothesis is [tex]H_1: p \neq 0.51[/tex].
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:
[tex]H_0: p = 0.51[/tex]
At the alternate hypothesis, we test if the proportion is different from 51%, that is:
[tex]H_1: p \neq 0.51[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.51 is tested at the null hypothesis:
This means that [tex]\mu = 0.51, \sigma = \sqrt{0.51*0.49}[/tex]
A sample of 900 arrests shows that 47% of the arrests were prosecuted.
This means that [tex]n = 900, X = 0.47[/tex]
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.47 - 0.51}{\frac{\sqrt{0.51*0.49}}{\sqrt{900}}}[/tex]
[tex]z = -2.4[/tex]
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.
