Confidence Interval and Hypothesis Test on the Proportion 1. A composites manufacturer is having serious problems with porosity in their parts. A Quality Engineer samples 300 parts and finds 58 defective. (a) Test the hypothesis that defective rate (proportion defective) exceeds 15%. Test at α = 0.05. What is the parameter of interest? What assumptions are made? Show mathematical evidence to support assumption.

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 takenX=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 levelConfidence=95% or 0.95z 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% .