Posted by **John** on Friday, July 16, 2010 at 9:22pm.

Suppose a certain species bird has an average weight of xbar 3.40 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with sigma 0.29 grams. Find the sample size necessary for a 98% confidence level with a maximal error of estimate E=0.09 for the mean weights of the hummingbirds. (Points: 5)

- Statistics -
**John**, Friday, July 16, 2010 at 9:28pm
I did 2.33x.29/.09 then sqared and got 56.3 this good?

(ZcO/E)^2 =n is this right?

- Statistics -
**MathGuru**, Monday, July 19, 2010 at 9:18pm
Formula:

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

...where n = sample size, z-value will be 2.33 using a z-table to represent the 98% confidence interval, sd = 0.29, E = 0.09, ^2 means squared, and * means to multiply.

Your answer looks correct! Good job.

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