The 99% confidence level corresponds to a the z-score at a probability of 99.5 = 2.58
The sample size (n) = (z-score * standard deviation/margin of error)^2 =
(2.58 * 6.4/0.45)^2 = 1346.4 ~ 1347
So you would need 1347 - 40 = 1317 more bacteria
Jafaru J.
asked 01/21/22You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 40 bacteria reveals a sample mean of 72 hours. The standard deviation is known to be 6.4 hours. 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 99% level of confidence.
How many more bacteria should you sample to achieve a 0.45-hour margin of error?
The 99% confidence level corresponds to a the z-score at a probability of 99.5 = 2.58
The sample size (n) = (z-score * standard deviation/margin of error)^2 =
(2.58 * 6.4/0.45)^2 = 1346.4 ~ 1347
So you would need 1347 - 40 = 1317 more bacteria
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