Respuesta :
Answer:
Parameter of interest= proportion of parts that present serious problems with porosity (defective)
z= 5.08
Based on the p value obtained 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of parts defective (problems with porosity) NOT exceeds 0.15 or 15% .
Step-by-step explanation:
1) Data given and notation n
n=300 represent the random sample taken
X=58 represent the parts that present serious problems with porosity in the sample
[tex]\hat p=\frac{58}{300}=0.193[/tex] estimated proportion of parts that present serious problems with porosity in the sample
[tex]p_o=0.15[/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)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that defective rate (proportion defective) exceeds 15%. :
Null hypothesis:[tex]p\leq 0.15[/tex]
Alternative hypothesis:[tex]p>0.15[/tex]
We assume that the proportion follows a normal distribution.
When we conduct a proportion test we need to use the z statistic, 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,higher or less) 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{0.193 -0.15}{\sqrt{\frac{0.15(1-0.15)}{300}}}=2.086[/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=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a one side test the p value would be:
[tex]p_v =P(z>2.086)=0.0185[/tex]
Based on the p value obtained 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of parts defective (problems with porosity) NOT exceeds 0.15 or 15% .
