Online NewsIn a Harris poll of 2036 adults, 40% said that they prefer to get their news online.Construct a 95% confidence interval estimate of thepercentageof all adults who say that they prefer toget 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%, whichimplies 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:

\(\begin{array}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \\ = 1.96 \times \sqrt {\frac{{0.40 \times \left( {1 - 0.40} \right)}}{{2036}}} \\ = 0.0213\end{array}\)

Therefore, the margin of error is 0.213.

The 95% confidence interval is computed as follows:

\(\begin{array}{c}\left( {\hat p - E,\hat p + E} \right) = \left( {0.40 - 0.0213,0.40 + 0.0213} \right)\\ = \left( {0.3787,0.4213} \right)\\ \approx \left( {0.379,0.421} \right)\end{array}\)

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 actualpercentage for the population of adults is within the confidence limits of 37.9% and 42.1%,it can be saidthat fewer than 50% of adults prefer to get their news online.