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 = 80 x ¯ = 80 hours with a standard deviation of s = 4.8 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 98% 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

Answer: Sample size would be 296.

Step-by-step explanation:

Since we have given that

Standard deviation = 4.8 hours

Margin of error = 0.65 hours

At 98% level of confidence, z = 2.33

So, It becomes,

$0.65=z\times \dfrac{\sigma}{\sqrt{n}}\\\\0.65=2.33\times \dfrac{4.8}{\sqrt{n}}\\\\\dfrac{2.33\times 4.8}{0.65}=\sqrt{n}\\\\n=(\dfrac{11.184}{0.65})^2\\\\n=17.206^2\\\\n=296.051$

Hence, sample size would be 296.