Question

​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

Answer 1

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

`p_v =2*P(z>0.966)=0.334`

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 people in favor of stricter gun control laws is not significantly different from 0.54 or 54%

Explanation:

Data given and notation

n=906 represent the random sample taken

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

`\hat p=(504)/(906)=0.556` estimated proportion of people in favor of stricter gun control laws

`p_o=0.54` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 different from 0.54 or no:

Null hypothesis:
`p=0.54`

Alternative hypothesis:
`p \\eq 0.54`

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{0.556 -0.54}{\sqrt{(0.54(1-0.54))/(906)}}=0.966`

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

Since is a bilateral test the p value would be:

`p_v =2*P(z>0.966)=0.334`

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 people in favor of stricter gun control laws is not significantly different from 0.54 or 54%