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 ¯ = 74 hours with a standard deviation of s = 4.8 hours. You would like to estimate the mean lifespan for this species of bacteria to within a margin of error of 0.65 hours at a 90% level of confidence. What sample size should you gather to achieve a 0.65 hour margin of error? Round your answer up to the nearest whole number.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 74

sample standard deviation = s = 4.8

sample size = n = 35

Degrees of freedom = df = n - 1 = 35 - 1 = 34

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,34 = 1.691

Margin of error = E = 0.65

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

n = [1.691 * 4.8 / 0.65 ]2

n = 155.93

Sample size = n = 156