Question

A research scientist wants to know how many times per hour a certain strand of bacteria reproduces. He believes that the mean is 12.6. Assume the variance is known to be 3.61. How large of a sample would be required in order to estimate the mean number of reproductions per hour at the 95% confidence level with an error of at most 0.19 reproductions

Answer 1

A sample of 385 is needed.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population, which is the square root of the variance, and n is the size of the sample.

Assume the variance is known to be 3.61.

This means that
`\sigma = √(3.61)`

How large of a sample would be required in order to estimate the mean number of reproductions per hour at the 95% confidence level with an error of at most 0.19 reproductions?

A sample of n is needed.

n is found when
`M = 0.19`. So

`M = z(\sigma)/(√(n))`

`0.19 = 1.96(√(3.61))/(√(n))`

`0.19√(n) = 1.96√(3.61)`

`√(n) = (1.96√(3.61))/(0.19)`

`(√(n))^2 = ((1.96√(3.61))/(0.19))^2`

`n = 384.16`

Rounding up

A sample of 385 is needed.