Question

In a 1996 survey of 1000 american citizens, 300 respondents claimed to be fluent in a second language. Find 94% confidence interval for the true proportion of citizens who are not fluent in a second language

Answer 1

npq come to mind

and recall that an interval is just a form of: mean + z(sd) = x if memory serves

Answer 2

z is determined by a table, find the z score that relates closest to .9400

the mean is determined as the proportion ratio, favored divided by number surveyed

and the sd is sqrt(npq) is what i think i recall

Answer 3

correction, the z score is related to .... (1-.94)/2 in this case in order to determine the end of a tail.