Statistics

I can do problems like this if the mean and standard deviation was given but what do I do if it's like this

When asked whether they drink alcohol or not, 79 out of 441 college students said that they did not. Construct a 90% CI for the proportion of all college students that do not drink alcohol.

The mean is 79/441= .18 right?

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1. You'll need to use a different kind of formula for these proportional problems.
Here is one:
CI90 = p + or - (z-value)(√ pq/n)
Find z-value using a z-table representing 90%. p = 79/441 (convert to a decimal). q = 1 - p. n = 441 (sample size). Plug values into the formula and calculate the interval.

I hope this will help get you started.

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