#### Answer

There is sufficient evidence to support that more than 75% of the people do not open emails and links from unfamiliar senders.

#### Work Step by Step

$H_{0}:p=75$%=0.75. $H_{a}:p>0.75.$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=0.92.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.92-0.75}{\sqrt{0.75(1-0.75)/737}}=10.66.$ The P is the probability of the z-score being more than 10.66 which is 1 minus the probability of the z-score being less than 10.66, 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 more than 75% of the people do not open emails and links from unfamiliar senders.