Question

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=66x¯=66 hours with a standard deviation of s=4.8s=4.8 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 98% level of confidence.

What sample size should you gather to achieve a 0.45 hour margin of error? Round your answer up to the nearest whole number.

n =   bacteria

Answer 1

Solution :

Given that,

standard deviation =s =   =4.8

Margin of error = E = 0.45

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

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

n = ( 2.326* 4.8 / 0.45)2

n =615.569

Sample size = n =616 rounded