Question

A local police chief claims that 31% of all drug-related arrests are never prosecuted. A sample of 500 arrests shows that 27% 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 evidence at the 0.02 level to support the officer's claim?

Answer 1

`z=\frac{0.27 -0.31}{\sqrt{(0.31(1-0.31))/(500)}}=-1.933`

`p_v =P(Z<-1.933)=0.0266`

If we compare the p value obtained and the significance level given
`\alpha=0.02` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 2% of significance the proportion of arrests that were not prosecuted is not significanlty less than 0.31.

Explanation:

1) Data given and notation

n=500 represent the random sample taken

X represent the number of arrests that were not prosecuted.

`\hat p=0.27` estimated proportion of arrests that were not prosecuted

`p_o=0.31` is the value that we want to test

`\alpha=0.02` represent the significance level

Confidence=98% or 0.98

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.31.:

Null hypothesis:
`p\geq 0.31`

Alternative hypothesis:
`p < 0.31`

When we conduct a proportion test we need to use the z statisitc, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.27 -0.31}{\sqrt{(0.31(1-0.31))/(500)}}=-1.933`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.02`. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

`p_v =P(Z<-1.933)=0.0266`

If we compare the p value obtained and the significance level given
`\alpha=0.02` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 2% of significance the proportion of arrests that were not prosecuted is not significanlty less than 0.31.