Question

A researcher wishes to conduct a project in which they are measuring the concentration of dissolved organic carbon in the soil. They would like to estimate the mean concentration with a margin of error of ±2 mg/dl. Assuming the standard deviation is 9 mg/dl,

1. How many samples are needed to have the desired margin of error with 95% confidence?

2. How many samples are needed to have the desired margin of error with 99% confidence?

3. How many samples are needed if the assumed standard deviation is 18 mg/dl? (Assume margin of error is still ±2 mg/l and 95% confidence)

4. How many samples are needed if the desired margin of error is ±1 mg/l with 95% confidence? (Assume a standard deviation of 9 mg/dl)

Answer 1

Sample size = (Z/2 * / E)²

1)

For 95% confidence

Sample size = (1.96 * 9 / 2)²

= 77.79

Sample size = 78 (Rounded up to nearest integer)

2)

For 99% confidence

Sample size = (2.5758 * 9 / 2)²

= 134.35

Sample size = 135 (Rounded up to nearest integer)

3)

For standard deviation = 18 and 95% confidence level,

Sample size = (1.96 * 18 / 2)²

= 311.17

Sample size = 312 (Rounded up to nearest integer)

4) For margin of error E = 1

Sample size = (1.96 * 9 / 1)²

= 311.17

Sample size = 312 (Rounded up to nearest integer)