The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 53 records of automobile driver fatalities in a certain county showed that 35 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use α = 0.05.

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Answer:

The p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] 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 of driver fatalities related to alcohol is less from 0.77 or 77%.  

Step-by-step explanation:

1) Data given and notation n  

n=53 represent the random sample taken

X=35 represent the automobile driver fatalities in a certain county involved with an intoxicated driver

[tex]\hat p=\frac{35}{53}=0.660[/tex] estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver

[tex]p_o=0.77[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 population proportion of driver fatalities related to alcohol is less than 77% or 0.77 in Kit Carson:  

Null hypothesis:[tex]p\geq 0.77[/tex]  

Alternative hypothesis:[tex]p < 0.77[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.660 -0.77}{\sqrt{\frac{0.77(1-0.77)}{53}}}=-1.903[/tex]

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This methos 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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is an unilateral lower test the p value would be:  

[tex]p_v =P(z<-1.903)=0.0285[/tex]  

So the p value obtained was a low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] 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 of driver fatalities related to alcohol is less from 0.77 or 77%.