## Elementary Statistics (12th Edition)

$H_{0}:p=50$%=0.5. $H_{a}:p<0.5.$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{5938}{5938+6062}=0.4948.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.4948-0.5}{\sqrt{0.5(1-0.5)/12000}}=-1.14.$ The P is the probability of the z-score being less than -1.14, hence:P=0.1271. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.1271 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 that people can postpone death after Thanksgiving.