Answer
There is sufficient evidence to support thatmore than 75% of adults know what Twitter is.
Work Step by Step
$H_{0}:p=0.75$. $H_{a}:p\ne0.75$ $\hat{p}$ is the number of objects with a specified value divided by the sample size. Hence $\hat{p}=\frac{x}{n}=\frac{856}{1007}=0.85.$ The test statistic is:$z=\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}=\frac{0.85-0.75}{\sqrt{0.75(1-0.75)/1007}}=7.33.$ The P is the probability of the z-score being more than 7.33 or less than -7.33 is the sum of the probability of the z-score being less than -7.33 plus 1 minus the probability of the z-score being less than 7.33, hence:P=0.0001+1-0.9999=0.0002. If the P-value is less than $\alpha$, which is the significance level, then this means the rejection of the null hypothesis. Hence:P=0.0002 is less than $\alpha=0.01$, hence we reject the null hypothesis. Hence we can say that there is sufficient evidence to support thatmore than 75% of adults know what Twitter is.
