SOLUTION: A sample of 200 people was assumed to identify their major source of news information; 110 stated that their major source was television news coverage. Construct a 99% confidence i
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Question 1205249: A sample of 200 people was assumed to identify their major source of news information; 110 stated that their major source was television news coverage. Construct a 99% confidence interval for the proportion of people in the population who consider television their major source of news information. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! sample size = 200
p = 110/200 = .55
q = 1 - p = 90/200 = .45
standard error = sqrt(.55*.45/200) = .035178
critical z-score at 99% two tailed confidence interval = plus or minus 2.5758
z = (x-m)/s
z is the critical z-score = plus or minus 2.5758.
x is the critical raw score = what you want to find.
m is the mean proportion = p = .55.
s is the standard error = .035178.
on the low end of the confidence interval, the formula becomes -2.5758 = (x - .55) / .035178.
solve for x to get x = -2.758 * .035178 + .55 = .45939.
on the high end of the confidence interval, the formula becomes 2.5758 = (x - .55) / .035178.
solve for x to get x = 2.758 * .035178 + .55 = .64061.
your 99% confidence interval is from .45939 to .64061.
here's what it looks like on a graph.
