Question

To Test the ability of auto mechanics to identify simple engine problems, an automobile with a single suchproblem was taken to 72 different car repair facilities. Only 42 out of 72 mechanics who worked on the carcorrectly identified the problem. Does this strongly indicate that the true proportion of mechanics who couldidentify this probelm is less than.75? Compute the p-value and reach a conclusion.

Answer 1

`z=\frac{0.583 -0.75}{\sqrt{(0.75(1-0.75))/(72)}}=-3.273`

`p_v =P(Z<-3.273)=0.00053`

So the p value obtained was a very low value and using any significance level 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 mechanics who worked on the carcorrectly identified the problem is significantly less than 0.75.

Explanation:

1) Data given and notation

n=72 represent the random sample taken

X=42 represent the mechanics who worked on the carcorrectly identified the problem

`\hat p=(42)/(72)=0.583` estimated proportion of mechanics who worked on the carcorrectly identified the problem

`p_o=0.75` 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 true proportion is less than 0.75:

Null hypothesis:
`p\geq 0.75`

Alternative hypothesis:
`p < 0.75`

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{0.583 -0.75}{\sqrt{(0.75(1-0.75))/(72)}}=-3.273`

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<-3.273)=0.00053`

So the p value obtained was a very low value and using any significance level 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 mechanics who worked on the carcorrectly identified the problem is significantly less than 0.75.