91影视

A survey consisted of 1007 U.S. adults. 85% of those who were surveyed know what Twitter is.

The formula of the confidence interval for a population proportion is given as follows:

$$\begin{gathered} CI=\left(\hat{p}-E,\hat{p}+E\right) \\ =\left(\hat{p}-z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}},\hat{p}+z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}\right) \end{gathered}$$

Where, $\hat{p}$ be the sample proportion, E is the margin of error, n is the sample size, $z_{\frac{伪}{2}}$is the two-tailed critical value obtained from standard normal table.

Also,

$\hat{q}=1-\hat{p}$

The proportion of adults who know what Twitter is is shown below:

$$\begin{gathered} \hat{p}=85\% \\ =\frac{85}{100} \\ =0.85 \end{gathered}$$

The sample size (n) is equal to 1007.

The confidence level is given to be equal to 95%. This implies that the level of significance is equal to 0.05.

The value of $z_{\frac{伪}{2}}$becomes equal to 1.96.

The following computation is made to construct the confidence interval estimate of the proportion of all adults who know what Twitter is:

$$\begin{gathered} CI=\left(\hat{p}-E,\hat{p}+E\right) \\ =\left(\hat{p}-z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}},\hat{p}+z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}\hat{q}}{n}}\right) \\ =\left(0.85-\left(1.96\right)\sqrt{\frac{\left(0.85\right)\left(1-0.85\right)}{1007}},0.85+\left(1.96\right)\sqrt{\frac{\left(0.85\right)\left(1-0.85\right)}{1007}}\right) \\ =\left(0.828,0.872\right) \end{gathered}$$

In terms of percentage, the confidence interval becomes as follows:

$$\begin{gathered} CI=\left(0.828,0.872\right) \\ =\left(82.8\%,87.2\%\right) \end{gathered}$$

Thus, the 95% confidence interval estimate of the percentage of all adults, who know what Twitter is, is equal to (82.8%, 87.2%).