Question

A professor wants to estimate the average time it takes his students to finish a computer project. Based on previous evidence, he believes that the standard deviation is approximately 3.6 hours. He would like to be 96% confident that his estimate is within 5 hours of the true population mean. Use RStudio to determine how large of a sample size is required without rounding any interim calculations.

Answer 1

Answer:

A sample size of 3 is required.

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:

$\alpha = \frac{1 - 0.96}{2} = 0.02$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$.

That is z with a pvalue of $1 - 0.02 = 0.98$, so Z = 2.054.

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.

Based on previous evidence, he believes that the standard deviation is approximately 3.6 hours.

This means that $\sigma = 3.6$

He would like to be 96% confident that his estimate is within 5 hours of the true population mean. Use RStudio to determine how large of a sample size is required without rounding any interim calculations.

The sample size needed is of n, and n is found when M = 5. So

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

$5 = 2.054\frac{3.6}{\sqrt{n}}$

$5\sqrt{n} = 2.054*3.6$

$\sqrt{n} = \frac{2.054*3.6}{5}$

$(\sqrt{n})^2 = (\frac{2.054*3.6}{5})^2$

$n = 2.19$

Rounding up:

A sample size of 3 is required.