Question

A research center claims that at least 4545​% of adults in a certain country think the government is not aggressive enough in pursuing people who cheat on their taxes. In a random sample of 700700 adults from that​ country, 3939​% say that the government is not aggressive enough in pursuing people who cheat on their taxes. At alphaαequals=0.010.01​, is there enough evidence to reject the​ center's claim? Complete parts​ (a) through​ (e) below.

Answer 1

`z=\frac{0.39 -0.45}{\sqrt{(0.45(1-0.45))/(700)}}=-3.19`

`p_v =P(z<-3.19)=0.0007`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate

Explanation:

Data given and notation

n=700 represent the random sample taken

`\hat p=0.39` estimated proportion of interest

`p_o=0.7` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

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 at least 0.45.:

Null hypothesis:
`p\geq 0.45`

Alternative hypothesis:
`p < 0.45`

When we conduct a proportion test we need to use the z statistic, 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{0.39 -0.45}{\sqrt{(0.45(1-0.45))/(700)}}=-3.19`

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=0.01`. The next step would be calculate the p value for this test.

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

`p_v =P(z<-3.19)=0.0007`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interst is significantly lower than 0.45 and the claim is not appropiate