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Sagot :
The test statistic value is -1.0328.
The null hypothesis state that the no-show rate for passengers booked on a flight is 6% and the alternative hypothesis state that the no-show rate is less than 6%. The sample proportion, in this case, can be calculated as,
ˆp = [tex]\frac{18}{360} = 0.0474[/tex]
The z-test statistic value can be calculated as,
[tex]z= \frac{^p-P}{\sqrt{\frac{P*(1-P)}{n} } }[/tex]
[tex]or, \frac{0.0474-0.06}{\sqrt{\frac{0.06*0.94}{380} } }[/tex]
[tex]or, \frac{-0.0126}{0.0122}\\ or, -1.0328[/tex]
The test statistic value is -1.0328.
Now the p-value can be calculated using the excel function =[tex]NORM.S.DIST(-1.0328,TRUE)[/tex] as 0.1508. The p-value is greater than 0.10 so we can conclude to not reject the null hypothesis at a 10% level of significance. Therefore, we do not have enough evidence to state that the no-show rate is less than 6%.
To learn more about hypothesis testing visit: brainly.com/question/17099835
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