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

`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<\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 defectives is significantly higher than 0.04.

Explanation:

1) Data given and notation

n=300 represent the random sample taken

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

`\hat p=(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{(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{(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<\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 defectives is significantly higher than 0.04.