Online News In a Harris poll of 2036 adults, 40% said that they prefer to get their news online. Construct a 95% confidence interval estimate of the percentageof all adults who say that they prefer to get their news online. Can we safely say that fewer than 50% of adults prefer to get their news online?

The number of adults is $n=$2036.

The percentage of adults who prefer to get their news online is 40%.

The level of confidence is 95%.

From the given information, the following points can be drawn:

The proportion of adults who prefer to get their news online is$\hat{p}=0.40$.

The level of confidence is 95%, which implies that the level of significance is 0.05.

From the Z table, the two-tailed critical value obtained at 0.05 level of significance is 1.96.

The margin of error is computed as follows:

role="math" localid="1648195038482" $$\begin{gathered} \mathit{E}\mathbf{=}{\mathit{z}}_{\frac{\mathbf{\alpha }}{\mathbf{2}}}\sqrt{\frac{\hat{\mathbf{p}}\left(1-\hat{p}\right)}{\mathbf{n}}} \\ =1.96\times \sqrt{\frac{0.40\times \left(1-0.40\right)}{2036}} \\ =0.0213 \end{gathered}$$

Therefore, the margin of error is 0.213.

The 95% confidence interval is computed as follows:

role="math" localid="1648195006699" $$\begin{gathered} \left(\hat{p}-E,\hat{p}+E\right)=\left(0.40-0.0213,0.40+0.0213\right) \\ =\left(0.3787,0.4213\right) \\ \approx \left(0.379,0.421\right) \end{gathered}$$

Therefore, the confidence interval estimate of the percentage of all adults who say that they prefer to get their news online is37.9% <p <42.1%.

Since the actual percentage for the population of adults is within the confidence limits of 37.9% and 42.1%,it can be said that fewer than 50% of adults prefer to get their news online.