Question

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

Answer 1

Given:

significance level = 0.05

rate = 30% = 0.30

n = 30

First, we identify the test statistic:

`z=\frac{\hat{p}-p}{\sqrt{(pq)/(n)}}`

Substitute the values:

`z=\frac{0.30-0.25}{\sqrt{(0.25(1-0.25))/(30)}}`

Simplify:

`z=\frac{0.05}{\sqrt{(0.25(0.75))/(30)}}=0.63`

Now p-value is given by:

`p_(value)=P(z>0.63)=1-P(z<0.63)`

From z table:

`=1-0.7357=0.2643`

Answer: 0.2643