Question

An assembly plant orders a large shipment of electronic circuits each month. The supplier claims that the population proportion of defective circuits is rhoequals0.04. When a shipment​ arrives, the plant manager selects a random sample of 300 circuits that are tested and the sample proportion of defective circuits is computed. This result is used for a hypothesis test to determine if there is sufficient evidence to conclude that the population proportion of defectives circuits from this supplier is greater than 0.04. The hypotheses are Upper H 0​: rholess than or equals0.04 and Upper H Subscript a​: rhogreater than0.04.

Answer 1

Answer:

$p_v =P(z>2.652)=0.0040$  

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ we have $p_v$ 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 defectives is significantly higher than 0.04.  

Step-by-step explanation:

1) Data given and notation  

n=300 represent the random sample taken

X=21 represent the number of defectives (value assumed)

$\hat p=\frac{21}{300}=0.07$ estimated proportion of defectives

$p_o=0.04$ 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 the proportion of defectives it's higher than 0.04.:  

Null hypothesis:$p\leq 0.04$  

Alternative hypothesis:$p > 0.04$  

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:  

$z=\frac{0.07 -0.04}{\sqrt{\frac{0.04(1-0.04)}{300}}}=2.652$  

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 right tailed test the p value would be:  

$p_v =P(z>2.652)=0.0040$  

So the p value obtained was a very low value and using the significance level assumed for example $\alpha=0.05$ we have $p_v$ 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 defectives is significantly higher than 0.04.