Asked by Sarah
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
Answered by
MathGuru
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.
Answered by
Sarah
So the answer would be 13.4459 ?
Answered by
MathGuru
Looks OK! Round up to the next highest whole number, which would be 14.
Answered by
Edddddy
Thanks MathGuru, the formula helped with a relevant math problem.
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