​Historically, the percentage of residents of a certain country who support stricter gun control laws has been 54%. A recent poll of 906 people showed 504 in favor of stricter gun control laws. Assume the poll was given to a random sample of people. Test the claim that the proportion of those favoring stricter gun control has charged. Perform a hypothesis test, using significance level of 0.05

Respuesta :

Answer:

[tex]z=\frac{0.556 -0.54}{\sqrt{\frac{0.54(1-0.54)}{906}}}=0.966[/tex]  

[tex]p_v =2*P(z>0.966)=0.334[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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 people in favor of stricter gun control laws is not significantly different from 0.54 or 54%

Step-by-step explanation:

Data given and notation

n=906 represent the random sample taken

X=504 represent the people  in favor of stricter gun control laws

[tex]\hat p=\frac{504}{906}=0.556[/tex] estimated proportion of people in favor of stricter gun control laws

[tex]p_o=0.54[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 different from 0.54 or no:  

Null hypothesis:[tex]p=0.54[/tex]  

Alternative hypothesis:[tex]p \neq 0.54[/tex]  

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

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

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

Calculate the statistic  

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

[tex]z=\frac{0.556 -0.54}{\sqrt{\frac{0.54(1-0.54)}{906}}}=0.966[/tex]  

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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z>0.966)=0.334[/tex]  

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] 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 people in favor of stricter gun control laws is not significantly different from 0.54 or 54%