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.

\(\displaystyle{n}={165},\hat{{{p}}}={\frac{{{153}}}{{{165}}}}={0.927}\)

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.