A research worker wishes to estimate the mean of a population by using sufficiently large sample. At the confidence level of 99%,the sample mean will be within 35% of the standard deviation. How large a sample should be taken?
To estimate the required sample size, we need to use the formula for sampling size estimation:
n = (z * σ / E)^2
Where:
- n is the required sample size.
- z is the z-score based on the desired confidence level. Since we want a confidence level of 99%, the z-score is 2.576 (obtained from a standard normal distribution table).
- σ is the standard deviation of the population.
- E is the allowable error or the margin of error, which is expressed as a fraction of the standard deviation.
From the given information, we know that the sample mean should be within 35% of the standard deviation. Therefore, the margin of error (E) is 0.35.
Let's assume we do not know the standard deviation (σ) of the population. In this case, we can use a maximum variability estimate, which is when σ = 0.5 (0.5 is the maximum possible value for a standard deviation since it ranges from 0 to 1). However, if you have any knowledge or estimates about the population standard deviation, it is recommended to use that value instead.
Now, let's calculate the required sample size:
n = (2.576 * 0.5 / 0.35)^2
n ≈ (1.424 / 0.35)^2
n ≈ 4.069^2
n ≈ 16.55
Therefore, the research worker should take a sample size of at least 17 (rounded up) to estimate the mean of the population with a confidence level of 99% and a margin of error of 35% of the standard deviation.