Question

The management of a major diary wanted to determine the average ounces of milk consumed per resident in the state of Texas. Past data indicated that the standard deviation in milk consumption per capita across the U.S. population was 4 ounces. A 95% confidence level is required and the margin of error is not to exceed /- 0.5 ounces. Management wanted to double the level of precision and increase the level of confidence to 99%.

Required:
What sample size would you recommend?

Answer 1

The recommended sample size is 1,698

Explanation:

The given standard deviation in milk consumption per capita across the U.S., σ = 4 ounces

The required confidence level = 95%

The margin of error, e = ± 0.5

When the precision is doubled, we have the new margin of error = ±0.5/2 = ±0.25

The standard score at 99% confidence level, z = 2.576

The sample size formula is given as follows;

`S = (Z^2 * P * Q)/(E^2)`

Where;

P × Q = The variance = σ²

∴ P × Q = 4² = 16

The sample size becomes;

S = 2.576² × 16/(0.25²) = 1,698.758656

We round down to get the recommended sample size, S = 1,698