Question

You are a researcher studying the lifespan of a certain species of bacteria. A preliminary sample of 40 bacteria reveals a sample mean of ¯x=66x¯=66 hours with a standard deviation of s=6.8s=6.8 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.75 hours at a 95% level of confidence.

What sample size should you gather to achieve a 0.75 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 = 6.8

sample size = n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

Margin of error = E = 0.75

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

t/2,df = t0.025,39 = 2.023

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

n = [2.023 * 6.8 / 0.75]2

n = 336.42

Sample size = n = 337 bacteria