In a survey of 1487 adults, 43% of them have Facebook pages. It is claimed that less than 50% of adults have Facebook pages.

The following formula of the confidence interval is used:

\(p < \hat p - E\)

Here, E is the margin of error and has the following formula:

\(E = {z_\alpha } \times \sqrt {\frac{{\hat p\hat q}}{n}} \)where

\(\hat p\)is the sample proportion of adults who have Facebook pages

\(\hat q\)is the sample proportion of adults who do not have Facebook pages

n is the sample size

\({z_\alpha }\) is the one-tailed critical value of z.

The sample size (n) is equal to 1487.

The sample proportion ofadults who have Facebook pages is equal to:

\(\begin{array}{c}\hat p = 43\% \\ = \frac{{43}}{{100}}\\ = 0.43\end{array}\)

The sample proportion of adults who do not have Facebook pages is equal to:

\(\begin{array}{c}\hat q = 1 - \hat p\\ = 1 - 0.43\\ = 0.57\end{array}\)

The value of \(\alpha \) is equal to 0.05.

Thus, \({z_\alpha }\) is equal to 1.645.

The margin of error can be computed as shown below:

\(\begin{array}{c}E = {z_\alpha } \times \sqrt {\frac{{\hat p\hat q}}{n}} \\ = 1.645\; \times \sqrt {\frac{{0.43 \times 0.57}}{{1487}}} \\ = 0.0211\end{array}\) \(\)

The one-sided confidence interval is computed below:

\(\begin{array}{c}p < \hat p - E\\ < 0.43 - 0.211\\ < 0.409\end{array}\)

Here, it can be observed that the population proportion of adults having Facebook pages is less than 0.409 or 40.9%.

Therefore, the given claim that the proportion of adults having Facebook pages is less than 50% is supported.