Use the survey results given in Exercise 1 and use a 0.05 significance level to test the claim that the majority of adults learn about medical symptoms more often from the internet than from their doctor.

The sample number of adults is\(n = 2015\).

Number of adults who said that they learn about medical symptoms more often from the internet than from their doctor is 1108 adults who. The level of confidence is 95%.

Let p be the true representative of the population proportionof adults who learn about medical symptoms more often from the internet than from their doctor.

The researcher claimsthat the majority of adults learn about medical symptoms more often from the internet than from their doctor.So, the statistical hypotheses are formed as shown:

\(\begin{aligned}{l}{H_0}:p = 0.50\\{H_1}:p > 0.50\end{aligned}\)

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

From the Z-table, the right-tailed critical value at 0.05 level of significance is 1.645.

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

\(\begin{aligned}{c}\hat p = \frac{{1108}}{{2015}}\\ = 0.549876.\end{aligned}\)

Therefore, the sample proportion is \(0.549876\).

The test statistic is computed below:

\(\begin{aligned}{c}Z = \frac{{\hat p - p}}{{\sqrt {\frac{{p\left( {1 - p} \right)}}{n}} }}\\ = \frac{{0.549876 - 0.50}}{{\sqrt {\frac{{0.50\left( {1 - 0.50} \right)}}{{2015}}} }}\\ = 4.4777\end{aligned}\)

Thus, the value of the test statistic is 4.48.

Since the calculated value (4.4889) of the test statistic is greater than the critical value (1.645), the null hypothesis is rejected.

There is enough evidence at 0.05 level of significance to support the claim that the majority of adults learn about medical symptoms more often from the internet than from their doctor.