Question

A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 30 bacteria reveals a sample mean of ¯ x = 76 hours with a standard deviation of s = 6.8 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.5 hours at a 99% level of confidence. What sample size should you gather to achieve a 0.5 hour margin of error? He would need to sample___ bacteria.

Answer 1

To estimate the sample size needed to achieve a margin of error of 0.5 hours at a 99% level of confidence, use the formula n = (Z * σ / E)², where n is the sample size, Z is the z-score, σ is the standard deviation of the population, and E is the margin of error.

Step-by-step explanation:

To estimate the sample size needed to achieve a margin of error of 0.5 hours at a 99% level of confidence, we can use the formula:

n = (Z * σ / E)²

Where:

• n is the sample size
• Z is the z-score corresponding to the desired level of confidence
• σ is the standard deviation of the population
• E is the margin of error

For a 99% level of confidence, the z-score is approximately 2.576. The standard deviation of the population is given as 6.8 hours. The margin of error is 0.5 hours.

Substituting these values into the formula, we get:

n = (2.576 * 6.8 / 0.5)²

Solving for n, we find:

n ≈ 661.35

Therefore, the researcher would need to sample approximately 661 bacteria.