Question

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 1

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
`\alpha=0.05` we have
`p_v<\alpha` 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% .

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

`\hat p=(58)/(300)=0.193` estimated proportion of parts that present serious problems with porosity in the sample

`p_o=0.15` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

`p_v` 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:
`p\leq 0.15`

Alternative hypothesis:
`p>0.15`

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:

`z=\frac{\hat p -p_o}{\sqrt{(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,higher or less) from a hypothesized value
`p_o`.

3) Calculate the statistic

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

`z=\frac{0.193 -0.15}{\sqrt{(0.15(1-0.15))/(300)}}=2.086`

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
`\alpha=0.05`. The next step would be calculate the p value for this test.

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

`p_v =P(z>2.086)=0.0185`

Based on the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` 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% .