Find the P-value for the indicated hypothesis test. An airline claims that the no-show rate for passengers booked on its flights is less than 6%. Of 380 randomly selected reservations, 19 were no-shows. Find the P-value for a test of the airline's claim. A. 0.3508 B. 0.2061 C. 0.0746 D. 0.1230

Answer:[tex]p_v =P(z<-0.82)=0.2061[/tex]  B. 0.2061Step-by-step explanation:1) Data given and notation n=380 represent the random sample takenX=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  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 no-show rate for passengers booked on its flights is less than 6%:  Null hypothesis:[tex]p\geq 0.06[/tex]  Alternative hypothesis:[tex]p < 0.06[/tex]  When we conduct a proportion test we need to use the z statistic, 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{0.05-0.06}{\sqrt{\frac{0.06(1-0.06)}{380}}}=-0.8208[/tex]  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 one left tailed test the p value would be:  [tex]p_v =P(z<-0.82)=0.2061[/tex]  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%.