Respuesta :
Answer:
[tex]z=\frac{0.0433 -0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}}=-0.532[/tex]
[tex]p_v =2*P(z<-0.532)=0.595[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.08[/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 defectives is not significantly different from 0.05 .
Step-by-step explanation:
1) Data given and notation
n=300 represent the random sample taken
X=13 represent the number of defectives
[tex]\hat p=\frac{13}{300}=0.0433[/tex] estimated proportion of defectives
[tex]p_o=0.05[/tex] is the value that we want to test
[tex]\alpha=0.08[/tex] represent the significance level
Confidence=92% or 0.92
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true porportion is 0.05.:
Null hypothesis: [tex]p=0.05[/tex]
Alternative hypothesis:[tex]p \neq 0.05[/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].
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.0433 -0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}}=-0.532[/tex]
4) 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=0.08[/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<-0.532)=0.595[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.08[/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 defectives is not significantly different from 0.05 .
