Question

Integrated circuits consist of electric channels that are etched onto silicon wafers. A certain proportion of circuits are defective because of "undercutting," which occurs when too much material is etched away so that the channels, which consist of the unetched portions of the wafers, are too narrow. A redesigned process, involving lower pressure in the etching chamber, is being investigated. The goal is to reduce the rate of undercutting to less than 5%. Out of the first 1000 circuits manufactured by the new process, only 33 show evidence of undercutting. Can you conclude that the goal has been met? Find the P-value and state a conclusion.

Answer 1

Answer:

$z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467$  

$p_v =P(Z>-2.467)=0.0068$  

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ 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 circuits that show evidence of undercutting is significantly less than 0.05.  

Step-by-step explanation:

1) Data given and notation  

n=1000 represent the random sample taken

X=33 represent the number of circuits that show evidence of undercutting

$\hat p=\frac{33}{1000}=0.033$ estimated proportion of circuits that show evidence of undercutting

$p_o=0.05$ is the value that we want to test

$\alpha$ represent the significance level

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 they reduce the rate of undercutting to less than 5%.:  

Null hypothesis:$p\geq 0.05$  

Alternative hypothesis:$p < 0.05$  

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 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:  

$z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467$  

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

Since is a left tailed test the p value would be:  

$p_v =P(Z>-2.467)=0.0068$  

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ 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 circuits that show evidence of undercutting is significantly less than 0.05.