Answer
There is sufficient evidence to support that the majority of adults prefer window seats when flying.
Work Step by Step
$H_{0}:p=50$%=0.5. $H_{a}:p>0.5$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{492}{806}=0.6104.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.6104-0.5}{\sqrt{0.5(1-0.5)/806}}=6.27.$ The P is the probability of the z-score being more than 6.27 which is 1 minus the probability of the z-score being less than 6.27, hence:P=1-0.9999=0.0001. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0001 is less than $\alpha=0.01$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that the majority of adults prefer window seats when flying.
