Answer to Question #200942 in Statistics and Probability for Zach

Question

Has the recent drop in airplane passengers resulted in better on-time performance? Before the recent downturn one airline bragged that 92% of its flights were on time. A random sample of 165 flights completed this year reveals that 153 were on time. Can we conclude at the 5% significance level that 

the airline’s on-time performance has improved? 

Answer 1

$H_0: p=0.92 \\ H_1: p>0.92 \\ \hat{p}= \frac{Succesful \;events}{Total \;events} \\ = \frac{153}{165} \\ = 0.927$

The z-value can be calculated as follows:

$Z= \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} \\ = \frac{0.927-0.92}{\sqrt{\frac{0.92(1-0.92)}{165}}} \\ = 0.33$

The calculated z-value is 0.33.

The p-value can be calculated as follows:

p=1−P(Z<0.33)

=1−0.6293

=0.3707

The p-value is 0.3707. Since the p-value is larger, it not accepts the alternate hypothesis.

We are not able to conclude that the airline’s on-time performance has improved.