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 com- pleted 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?

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 com- pleted 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?

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asked 2021-05-01
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 com- pleted 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?

Answers (1)

2021-05-03
Here we have follwoing information:
\(\displaystyle{n}={165},\hat{{{p}}}={\frac{{{153}}}{{{165}}}}={0.927}\)
Hypotheses are:
\(\displaystyle{H}_{{0}}:{p}={0.92}\)
\(\displaystyle{H}_{{a}}:{p}{>}{0.92}\)
Standard deviation of the proportion is:
\(\displaystyle\sigma=\sqrt{{{\frac{{{p}{\left({1}-{p}\right)}}}{{{n}}}}}}=\sqrt{{{\frac{{{0.92}{\left({1}-{0.92}\right)}}}{{{165}}}}}}={0.0211}\)
Test statistics will be:
\(\displaystyle{z}={\frac{{\hat{{{p}}}-{p}}}{{\sigma}}}={\frac{{{0.927}-{0.92}}}{{{0.0211}}}}={0.33}\)
Alternative hypothesis shows that the test is right tailed so p-value of the test is
P(z > 0.33) = 0.3707
Since p-value of the test is greater than 0.05 so we fail to reject the null hypothesis at 0.05 level of significance. So based on this sample, there is no support for the claim that the airline’s on-time performance has improved.
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