I did 2.33x.29/.09 then sqared and got 56.3 this good?
(ZcO/E)^2 =n is this right?
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)
2 answers
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.
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.
1 answer