Posted by **George** on Wednesday, May 5, 2010 at 7:02pm.

In a survey of 1037 adults, ages 65 and over, 643 were concerned about getting the flu.

a) Find the point estimate for the population proportion , p , of those concerned about getting the flu.

b) Construct a 90% Confidence Interval for the population proportion.

c) How big a sample size would be needed to allow for a 2% margin of error?

- Statistics -
**MathGuru**, Wednesday, May 5, 2010 at 9:33pm
Use a proportional confidence interval formula.

CI90 = p + or - (z-value)(√ pq/n)

Find z-value using a z-table representing 90%. p = 643/1037 (convert to a decimal). q = 1 - p. n = 1037 (sample size). Plug values into the formula and calculate the interval.

Formula to find sample size:

n = [(z-value)^2 * p * q]/E^2

... where n = sample size, z-value is the same as above, p and q are the same as above, ^2 means squared, * means to multiply, and E = .02 (for 2%).

Plug values into the formula and calculate n.

I hope this will help get you started.

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