SOLUTION: In a recent survey of 1124 students, 980 of them would like to recommend www.helpyourmath.com to their friends. Construct a 90% confidence interval to estimate the proportion of al

Question 1204908: In a recent survey of 1124 students, 980 of them would like to recommend www.helpyourmath.com to their friends. Construct a 90% confidence interval to estimate the proportion of all students who would recommend www.helpyourmath.com to their friends.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the mean proportion is p = 980/1124 = .871886121.
q = 1-p = .128113879.
the standard error is sqrt(p * q / 1124) = sqrt(.871886121*.12811387/1124) = .0099688444.

the two tailed 90% confidence interval requires a z-score of plus or minus 1.645.

the z-score formula used is z = (x - m) / s
z is the z-score
x is the desired proportion.
m is the mean proportion.
s is the standard error.

to find the lower desired proportion, the formula becomes:
-1.645 = (x - .871886121) / .009968844.
solve for x to get x = -1.645 * .009968844 + .871886121 = .855487372.
to find the upper desired proportion, the formula becomes:
1.645 = (x - .871886121) / .009968844.
solve for x to get x = 1.645 * .009968844 + .i71886121 = .88828487.

round to 3 decimal places to get 90% confidence interval is from .855 to .888.

since the value is n * p, you get:

mean is 1124 * 980 / 1124 = 980
low threshold is 1124 * .855 = 961 roughly.
high threshold is 1124 * .888 = 998 roughly.

your solution should be that the 90% confidence interval of the proportion is .855487372 to.88828487.
round your numbers as required.

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