Question

Some people are concerned that new tougher standards and​ high-stakes tests adopted in many states have driven up the high school dropout rate. The National Center for Education Statistics reported that the high school dropout rate for the year 2012 was 7.0​%. One school district whose dropout rate has always been very close to the national average reports that 132 of their 1773 high school students dropped out last year. Is this evidence that their dropout rate may be​ increasing? Explain.

Answer 1

`z=\frac{0.0744 -0.07}{\sqrt{(0.07(1-0.07))/(1773)}}=0.726`

`p_v =P(z>0.706)=0.24`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.07 or 7%

Explanation:

Data given and notation

n=1773 represent the random sample taken

X=132 represent the high school students dropped out last year

`\hat p=(132)/(1773)=0.0744` estimated proportion of high school students dropped out last year

`p_o=0.07` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

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

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is highern than 0.07 or 7%:

Null hypothesis:
`p \leq 0.07`

Alternative hypothesis:
`p > 0.07`

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`.

Calculate the statistic

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

`z=\frac{0.0744 -0.07}{\sqrt{(0.07(1-0.07))/(1773)}}=0.726`

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 assumed
`\alpha=0.05`. The next step would be calculate the p value for this test.

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

`p_v =P(z>0.706)=0.24`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly higher than 0.07 or 7%