A Social service agency plans to conduct a survey to determine the mean income of its clients. The director of the agency prefers that you measure the mean income very accurately, to within plus or minus $500. from a sample taken 2 years ago, you estimate that the standard deviation of income for this population is about $5000. I need to figure out the necessary sample size to reduce the confidence interval to plus/minus$500
To determine the necessary sample size to reduce the confidence interval to plus/minus $500, we can use the formula for sample size calculation in estimating population mean.
The formula is:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence
σ = standard deviation of the population
E = margin of error (half the width of the desired confidence interval)
In this case, the desired margin of error is $500, and the standard deviation of the population is estimated to be $5000.
To find the Z-score corresponding to the desired level of confidence, we need to determine the confidence level. Let's assume a 95% confidence level, which corresponds to a Z-score of approximately 1.96.
Substituting the values into the formula, we get:
n = (1.96 * 5000 / 500)^2
n = (9800 / 500)^2
n = 19.6^2
n ≈ 384.16
Rounding up to the nearest whole number, the necessary sample size would be 385 in order to reduce the confidence interval to plus/minus $500.