Solve 7(f-5) = 28<br> Help meeeee

A local police chief claims that 31% of all drug-related arrests are never prosecuted. A sample of 500 arrests shows that 27% of

Elis [28]

Answer:

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

$p_v =P(Z$

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.

Step-by-step 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:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:

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

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:

$p_v =P(Z$

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.

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