Question

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 1

Answer:

$p_v =P(z$  

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

$\hat p=\frac{19}{380}=0.05$ estimated proportion of no-shows in the sample

$p_o=0.06$ is the value that we want to test

$\alpha$ represent the significance level  

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 the no-show rate for passengers booked on its flights is less than 6%:  

Null hypothesis:$p\geq 0.06$  

Alternative hypothesis:$p < 0.06$  

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:  

$z=\frac{0.05-0.06}{\sqrt{\frac{0.06(1-0.06)}{380}}}=-0.8208$  

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:  

$p_v =P(z$  

So the p value obtained was a very high value and using the significance level for example $\alpha=0.05$ we have $p_v>\alpha$ 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%.