#### Answer

There is sufficient evidence to support that more than 25% of women purchase books online.

#### Work Step by Step

$H_{0}:p=25$%=0.25. $H_{a}:p>0.25$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=0.29.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.29-0.25}{\sqrt{0.25(1-0.25)/427}}=1.91.$ The P is the probability of the z-score being more than 1.91 is 1 minus the probability of the z-score being less than 1.91, hence:P=1-0.9719=0.0281. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0281 is less than $\alpha=0.05$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support that more than 25% of women purchase books online.