## 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 $\hat{p}=\frac{x}{n}=\frac{152}{380}=0.4.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.4-0.5}{\sqrt{0.5(1-0.5)/380}}=-3.9.$ The P is the probability of the z-score being less than -3.9, hence:P=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 less than 50% of smartphone users say the smartphone as the only say they cannot live without. But only smartphone users were asked hence the result cannot be generalized.