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In recent years a town experienced an arrest rate of 25% for robberies The new sheriff compiles records showing that among 30 recent robberies the arrest rate is 30%; he claims that this arrest greater than the 25% rate in the past. Using a 0.05 significance level to test the claim, find the P-value

In recent years a town experienced an arrest rate of 25 for robberies The new sheriff compiles records showing that among 30 recent robberies the arrest rate is class=

Respuesta :

Given:

significance level = 0.05

rate = 30% = 0.30

n = 30

First, we identify the test statistic:

[tex]z=\frac{\hat{p}-p}{\sqrt{\frac{pq}{n}}}[/tex]

Substitute the values:

[tex]z=\frac{0.30-0.25}{\sqrt{\frac{0.25(1-0.25)}{30}}}[/tex]

Simplify:

[tex]z=\frac{0.05}{\sqrt{\frac{0.25(0.75)}{30}}}=0.63[/tex]

Now p-value is given by:

[tex]p_{value}=P(z>0.63)=1-P(z<0.63)[/tex]

From z table:

[tex]=1-0.7357=0.2643[/tex]

Answer: 0.2643

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