Question

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance alpha using the given sample statistics. ​Claim: pless than0.11​; alphaequals0.10​; Sample​ statistics: ModifyingAbove p with caretequals0.08​, nequals25.

Answer 1

Answer:

$np=25*0.11=2.75 < 10$

$n(1-p)=25*(1-0.11)=22.25 > 10$

Th second condition is satisfied but the first one on this case it's not satisfied. So for this case it's not good apply the normal approximation to the distribution of p.

Step-by-step explanation:

We need to check the conditions in order to use the normal approximation.

$np=25*0.11=2.75 < 10$

$n(1-p)=25*(1-0.11)=22.25 > 10$

Th second condition is satisfied but the first one on this case it's not satisfied. So for this case it's not good apply the normal approximation to the distribution of p.

If we have both conditions satisfied the general procedure is the following:

Data given and notation

n=23 represent the random sample taken

$\hat p=0.08$ estimated proportion

$p_o=0.11$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

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 less than 0.11.:  

Null hypothesis:$p \geq 0.11$  

Alternative hypothesis:$p < 0.11$  

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  

On this case since the normal approximation it's not satisfid it's not correct calculate the statistic.

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 $\alpha=0.1$. The next step would be calculate the p value for this test.  

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

$p_v =P(z$  

And if the $p_v$ we reject the null hypothesis. Otherwise w fail to the reject the null hypothesis.