Question

consider the following scenario: Assume daily protein intake values in a population of athletes are known to have a standard deviation of 58.6 grams. From a random sample of size 267 athletes, the mean daily protein intake of 77.0 grams. Find a 95% confidence interval for the population mean daily protein intake. Round final answers to 2 decimal places.

Answer 1

Answer:

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

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:

$\alpha = \frac{1-0.95}{2} = 0.025$

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

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find 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.

$M = 1.96*\frac{58.6}{\sqrt{267}} = 7.03$

The lower end of the interval is the sample mean subtracted by M. So it is 77 - 7.03 = 69.97g.

The upper end of the interval is the sample mean added to M. So it is 77 + 7.03 = 84.03g.

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.