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Olenka [21]
3 years ago
8

A group of retired admirals, generals, and other senior military leaders, recently published a report, "Too Fat to Fight". The r

eport said that 75 percent of young Americans between the ages of 17 to 24 do not qualify for the military because of failure to graduate [from high school], criminal records or physical problems. The study cited Department of Defense and health data. A research group does not agree with this statistic. In a random sample of 180 young Americans 125 did not qualify for the military. Using a 5% significance level, is there sufficient evidence to conclude that less than 75% of Americans between the ages of 17 to 24 do not qualify for the military?
Mathematics
1 answer:
weqwewe [10]3 years ago
6 0

Answer:

z=\frac{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

p_v =P(z  

If we compare the p value obtained and the significance level given \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 americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

Step-by-step explanation:

1) Data given and notation  

n=180 represent the random sample taken  

X=125 represent the number of americans between 17 to 24 that not qualify for the military

\hat p=\frac{125}{180}=0.694 estimated proportion of americans between 17 to 24 that not qualify for the military

p_o=0.75 is the value that we want to test  

\alpha=0.05 represent the significance level  

Confidence=95% or 0.95  

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 less than 75% of Americans between the ages of 17 to 24 do not qualify for the military :  

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{0.694 -0.75}{\sqrt{\frac{0.75(1-0.75)}{180}}}=-1.735  

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 significance level provided \alpha=0.05. 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  

If we compare the p value obtained and the significance level given \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 americans between 17 to 24 that not qualify for the military is significantly less than 0.75 or 75% .  

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