(36÷? )−(36÷? )=5
2 answers:
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Answer:
The p value obtained was a low value and using the significance level given we have
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
estimated proportion of automobile driver fatalities in a certain county involved with an intoxicated driver
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
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:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 . The next step would be calculate the p value for this test.
Since is an unilateral lower test the p value would be:
So the p value obtained was a low value and using the significance level given we have
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%.