Question

You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 30 bacteria reveals a sample mean of ¯x=66x¯=66 hours with a standard deviation of s=6.6s=6.6 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.7 hours at a 99% level of confidence.

What sample size should you gather to achieve a 0.7 hour margin of error? Round your answer up to the nearest whole number.

n =   bacteria

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 66

sample standard deviation = s = 6.6

sample size = n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

Margin of error = E = 0.7

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,29 = 2.756

sample size = n = [ t/2,df * s / E] 2

n = [2.756 * 6.6 / 0.7 ]2

n = 675.22

Sample size = n = 676 bacteria.