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
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.
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.
To find the sample size required for the given scenario, we need to use the formula for determining the sample size for estimating the population mean.
The formula is:
n = (Z * σ / E)²
Where:
n = sample size
Z = Z-score (related to the desired confidence level)
σ = standard deviation of the population
E = maximum allowable error (the difference between sample mean and population mean)
In this case, the given information tells us that the variance (σ²) is 14. To find the standard deviation (σ), we take the square root of the variance:
σ = √14
Since the confidence level is 95%, we need to find the Z-score that corresponds to that level. The Z-score can be found using a Z-table or a statistical calculator. For a confidence level of 95%, the Z-score is approximately 1.96.
The maximum allowable error (E) is given as 2 units.
Now, we can substitute the values into the formula:
n = (1.96 * √14 / 2)²
Calculating this equation will give us the sample size (n) needed to estimate the population mean with the specified confidence and allowable error.