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The sample number of adults is \(n = 2015\). There are 1108 adults who said that they learn about medical symptoms more often from the internet than from their doctor. And the level of confidence is 95%.

As the level of confidence is 95%, the level of significance is 0.05.

From the Z-table,the critical value at 0.05 level of significance is\({z_{\frac{\alpha }{2}}} = 1.960\).

The sample proportion of adults that said that they learn about medical symptoms more often from the internet than from their doctor is

\(\begin{aligned}{c}\hat p = \frac{{1108}}{{2015}}\\ = 0.549\\ \approx 0.55.\end{aligned}\)

Therefore, the sample proportion is 0.55.

The margin of error is computed below:

\(\begin{aligned}{c}E = {z_{\frac{\alpha }{2}}}\sqrt {\frac{{\hat p\left( {1 - \hat p} \right)}}{n}} \\ = 1.960 \times \sqrt {\frac{{0.55\left( {1 - 0.55} \right)}}{{2015}}} \\ = 0.022\end{aligned}\)

Thus, the margin of error is 0.022.

The 95% confidence interval (C.I) for the estimate of population proportion of the adultswho said that they learn about medical symptoms more often from the internet than from their doctor is computed below:

\(\begin{aligned}{c}\left( {\hat p \pm E} \right) = 0.55 \pm 0.022\\ = \left( {0.528,0.572} \right)\end{aligned}\)

Therefore, the 95% confidence interval for the estimate of population proportion of the adults who said that they learn about medical symptoms more often from the internet than from their doctor is \(\left( {0.528,0.572} \right)\).

The range of the confidence interval described at 95% level suggests that the proportion of adults is more than 0.50. It can be expressed with 95% level of confidence that more than 50% adults in the population learn about medical symptoms more often from the internet than from their doctor.

Yes, it can be concluded that the majority of adults learn about medical symptoms more often from the internet than from their doctor.