Suppose a polling organization is conducting a survey to estimate the proportion of Americans who regularly attend religious services. The organization plans to gather data from a randomly selected sample of 450 individuals. On the basis of the "conservative" formula, what will be the margin of error for the survey?
+6250 I'm not sure what "conservative" means but
the 99% confidence margin of error is
$$1.29\sqrt n$$
and the 95% confidence margin of error is
$$0.98 \sqrt n$$
with n=450 these numbers are
$${\mathtt{1.29}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{450}}}} = {\mathtt{27.365\: \!032\: \!431\: \!919\: \!389\: \!2}}$$
$${\mathtt{0.98}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{450}}}} = {\mathtt{20.788\: \!939\: \!366\: \!884\: \!497\: \!2}}$$
+6250 I'm not sure what "conservative" means but
the 99% confidence margin of error is
$$1.29\sqrt n$$
and the 95% confidence margin of error is
$$0.98 \sqrt n$$
with n=450 these numbers are
$${\mathtt{1.29}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{450}}}} = {\mathtt{27.365\: \!032\: \!431\: \!919\: \!389\: \!2}}$$
$${\mathtt{0.98}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{450}}}} = {\mathtt{20.788\: \!939\: \!366\: \!884\: \!497\: \!2}}$$
