Question
Scores on a certain test are normally distributed with a variance of 14. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 95% confidence that the sample mean will not differ from the population mean by more than 2 units
Answers
Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = square root of 14 (standard deviation is equal to the square root of the variance), E = 2, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = square root of 14 (standard deviation is equal to the square root of the variance), E = 2, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.
So the answer would be 13.4459 ?
Looks OK! Round up to the next highest whole number, which would be 14.
Thanks MathGuru, the formula helped with a relevant math problem.
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