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Part 2 - The president of the company is uncomfortable with the precision of the estimates derived for the sample mean. They are not willing to tolerate a very large error. e = .04, how large would the sample size have to be if they specify a 98% confidence level?

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

Note: n = sample size, z-value will be 2.33 using a z-table to represent the 98% confidence interval, sd = standard deviation, E = 0.04, ^2 means squared, and * means to multiply.

Plug known values into the formula and finish the calculation. Round your answer to the next highest whole number.

I hope this will help.

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