Question

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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=78x¯=78 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 0.5 hours at a 95% level of confidence.

What sample size should you gather to achieve a 0.5 hour margin of error? Round your answer up to the nearest whole number.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 78

sample standard deviation = s = 6.8

sample size = n = 40

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

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

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

margin of error = E = 0.5

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

n = [2.023 * 6.8 / 0.5 ]2

n = 756.95

Sample size = n = 757