Question

A research scholar wants to know how many times per hour a certain strand of virus reproduces. The mean is found to be 4 reproductions and the population standard deviation is known to be 2.3. If a sample of 1083 was used for the study, construct the 99% confidence interval for the true mean number of reproductions per hour for the virus. Round your answers to one decimal place.

Answer 1

Answer:

The 99% confidence interval for the true mean number of reproductions per hour for the virus is between 3.8 and 4.2.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.99}{2} = 0.005$

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 $1 - 0.005 = 0.995$, so Z = 2.575.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.575\frac{2.3}{\sqrt{1083}} = 0.18$

Rounding to one decimal place, 0.2.

The lower end of the interval is the sample mean subtracted by M. So it is 4 - 0.2 = 3.8 reproductions

The upper end of the interval is the sample mean added to M. So it is 4 + 0.2 = 4.2 reproductions.

The 99% confidence interval for the true mean number of reproductions per hour for the virus is between 3.8 and 4.2.