Mathematics, 26.11.2019 01:31, jamesleecy06

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.

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answered: alecnewman2002

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 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

$\hat p=\frac{58}{300}=0.193$ 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: p0.15

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:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:

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

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(z2.086)=0.0185

answered: Guest

step-by-step explanation:

answered: Guest

the answer would be 0.54

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Confidence interval and hypothesis test on the proportion 1. a composites manufacturer is having ser...