Answer
There is not sufficient evidence to support Mendal's claim.
Work Step by Step
$H_{0}:p=25$%=0.25. $H_{a}:p\ne0.25$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{152}{152+428}=0.2621.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.2621-0.25}{\sqrt{0.25(1-0.25)/580}}=0.67.$ The P is the probability of the z-score being more than 0.67 or less than -0.67 is the sum of the probability of the z-score being less than -0.67 plus 1 minus the probability of the z-score being less than 0.67, hence:P=0.2514+1-0.7486=0.5028. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.5028 is more than $\alpha=0.05$, hence we fail to reject the null hypothesis. Hence we can say that there is not sufficient evidence to support Mendal's claim.
