Respuesta :
Answer:
[tex]p_v =P(z<-0.82)=0.2061[/tex] Â
B. 0.2061
Step-by-step explanation:
1) Data given and notation
n=380 represent the random sample taken
X=19 represent the no-shows in the sample
[tex]\hat p=\frac{19}{380}=0.05[/tex] estimated proportion of no-shows in the sample
[tex]p_o=0.06[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level Â
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest) Â
2) Concepts and formulas to use Â
We need to conduct a hypothesis in order to test the claim that the no-show rate for passengers booked on its flights is less than 6%: Â
Null hypothesis:[tex]p\geq 0.06[/tex] Â
Alternative hypothesis:[tex]p < 0.06[/tex] Â
When we conduct a proportion test we need to use the z statistic, and the is given by: Â
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1) Â
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 Â
Since we have all the info requires we can replace in formula (1) like this: Â
[tex]z=\frac{0.05-0.06}{\sqrt{\frac{0.06(1-0.06)}{380}}}=-0.8208[/tex] Â
4) Statistical decision Â
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. Â
The next step would be calculate the p value for this test. Â
Since is a one left tailed test the p value would be: Â
[tex]p_v =P(z<-0.82)=0.2061[/tex] Â
So the p value obtained was a very high value and using the significance level for example [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of no-show rate for passengers booked on the company flights is nor significantly less than 0.06 or 6%. Â
