#### Answer

There is sufficient evidence to reject that the coin toss is fair.

#### 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 $\hat{p}=\frac{x}{n}=\frac{235}{414}=0.5676.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.5676-0.5}{\sqrt{0.5(1-0.5)/414}}=2.75.$ The P is the probability of the z-score being more than 2.75 which is 1 minus the probability of the z-score being less than 2.75, hence:P=1-0.997=0.003. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.003 is less than $\alpha=0.05$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to reject that the coin toss is fair.