SOLUTION: A survey of 300 students at York University revealed that 112 favour an NDP candidate for M.P. Construct a 98% confidence interval for the percentage of all York University studen

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Question 1205357: A survey of 300 students at York University revealed that 112 favour an NDP candidate for M.P. Construct a 98% confidence interval for the percentage of all York University students who favour an NDP candidate.

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
p = 112/300
q = 1 - p = 1 - 112/300 = 300 /300 - 112/300 = (300 - 112)/300 = 188/300.
standard error = sqrt(112/300 * 188/300 / 300) = .027926.
98% two tail confidence interval requires a z-score of plus or minus 2.326.
z = (x-m)/s
z = plus or minus 2.326.
x = x
m = p = 112/300 = .373333
s = .027963.
on the low end of the confidence interval, -2.326 = (x-.373333)/.027963.
solve for x to get x = -2.326 * .027963 + .373333 = .30829.
on the high end of the confidence interval, 2.326 = (x - .373333)/.027963.
solve for x to get x = 3.236 * .027963 + .373333 = .43837.
your 98% confidence interval is from .30829 to .43837.

this is what it looks like using z-scores.

this is what it looks like using raw scores.

in the z-score formula:
z = the z-score
x = the sample proportion
m = the mean proportion which is equal to p
s = the standard error.

p is equal to the proportion of the sample size that favor an NDP candidate.
q = 1-p is equal to the proportion of the sample size that do not favor an NDP candidate.
the mean proportion is equal to p which is equal to m in the z-score formula.