You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 35 bacteria reveals a sample mean of ¯x=72 hours with a standard deviation of s=6.6 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.6 hours at a 98% level of confidence.
What sample size should you gather to achieve a 0.6 hour margin of error? Round your answer up to the nearest whole number.
n = bacteria
A) Given data
Mean value(X')=72
Standard deviation (s)=6.6
Margin of error (E)=0.6
We need to determine the sample size (n) to achieve E=0.6 at 98% confidence level
Now Formula used is
Margin of error (E)=Z ×s/(√n)
Z value for 98% confidence level is
Z value=2.33
Now sample size (n)=((z×s)/E))^2
=((2.33×6.6)/0.6)^2
=(25.63)^2
=656.89=657
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