Question

In a survey of n= 2015 adults, 1108 of them said that they learn about medical symptoms more often from the internet than from their doctor (based on a MerckManuals.com survey). Use the data to construct a 95% confidence interval estimate of the population proportion of all adults who say that they learn about medical symptoms more often from the internet than from their doctor. Does the result suggest that the majority of adults learn about medical symptoms more often from the internet than from their doctor?

Short answer

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

Yes, the result suggests that most adults learn about medical symptoms more often from the internet than from their doctor.

Step-by-step solution

Given information

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%.

Compute the level of significance and critical value

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\({z_{\frac{\alpha }{2}}} = 1.960\).

Compute the sample proportion

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.

Compute the margin of error

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.

Calculate the confidence interval

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 \(\left( {0.528,0.572} \right)\).

Provide the conclusion

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.