Question

A researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 40 bacteria reveals a sample mean of ¯x=74x¯=74 hours with a standard deviation of s=6s=6 hours. He would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.6 hours at a 90% level of confidence.

What sample size should you gather to achieve a 0.6 hour margin of error?

He would need to sample bacteria.

Answer 1

Solution :

Given that,

standard deviation =s =   =6

Margin of error = E = 0.6

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z 0.05 = 1.645

sample size = n = [Z/2 * / E] 2

n = ( 1.645 * 6 / 0.6 )2

n =270.6025

Sample size = n =271 rounded