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lapo4ka [179]
3 years ago
7

To Test the ability of auto mechanics to identify simple engine problems, an automobile with a single suchproblem was taken to 7

2 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.
Mathematics
1 answer:
dedylja [7]3 years ago
7 0

Answer:

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

p_v =P(Z  

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

Step-by-step 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=\frac{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{\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{0.583 -0.75}{\sqrt{\frac{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  

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

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Let us assume total capacity of the container = x pounds.

Container contains 1/6th coffee of total x pounds,  that is x/6 pounds.

We also given that 1/6th part is equal to 2/3 of a pound.

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Division expression that represents the capacity of the container = 2/3 ÷ 1/6.


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