Question

ou are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 35 bacteria reveals a sample mean of �x=78 hours with a standard deviation of s=6.2hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.45 hours at a 90% level of confidence.What sample size should you gather to achieve a 0.45 hour margin of error?

Answer 1

To estimate the mean lifespan of a certain species of bacteria with a margin of error of 0.45 hours at a 90% level of confidence, you would need a sample size of 854 bacteria.

Step-by-step explanation:

To estimate the mean lifespan of a certain species of bacteria with a margin of error of 0.45 hours at a 90% level of confidence, you need to determine the required sample size. The formula for sample size calculation is:

n = (Z*s / E)^2

where n is the required sample size, Z is the Z-score for the desired confidence level (in this case, Z = 1.645 for a 90% confidence level), s is the standard deviation, and E is the desired margin of error. Plugging in the given values:

n = (1.645 * 6.2 / 0.45)^2

n = 853.948

Since you cannot have a fraction of a bacterium, you would need to round up the sample size to the nearest whole number. Therefore, the sample size needed to achieve a 0.45 hour margin of error at a 90% level of confidence is 854 bacteria.