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.

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 we have 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

estimated proportion of parts that present serious problems with porosity in the sample

is the value that we want to test

represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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:  

Alternative hypothesis:

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:  

(1)  

The One-Sample Proportion Test is used to assess whether a population proportion is significantly (different,higher or less) from a hypothesized value .

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

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

Since is a one side test the p value would be:  

Based on the p value obtained and using the significance level given we have 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% .