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

Answer:

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

$p_v =P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ 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

Step-by-step 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:  

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

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

Calculate the statistic  

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

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

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:  

$p_v =P(z$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ 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