Question

The number of hours that a nine month old baby sleeps at night are normally distributed with a population standard deviation of 1.5 hours and an unknown population mean. A random sample of 22 nine month old babies is taken and results in a sample mean of 12 hours. Find the margin of error for a confidence interval for the population mean with a 90% confidence level.

Answer 1

Answer:

The margin of error is of 0.7123 hours.

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:

$\alpha = \frac{1-0.9}{2} = 0.05$

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 $1-0.05 = 0.95$, so $z = 1.645$

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.

Find the margin of error for a confidence interval for the population mean with a 90% confidence level.

We have that $\sigma = 1.5, n = 12$. So

$M = 1.645*\frac{1.5}{\sqrt{12}} = 0.7123$

The margin of error is of 0.7123 hours.