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 = 76 x¯=76 hours with a standard deviation of s = 5.6 s=5.6 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?

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 76

sample standard deviation = s = 5.6

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
= t_{0.05,34} = 1.691

Margin of error = E = 0.65

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

n = [1.691 * 5.6 / 0.65 ]^{2}

n = 212.24

Sample size = n = 213

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