8.2.2.1.2- Example: Sleep Deprivation

8.2.2.1.2- Example: Sleep Deprivation

In a class survey, students are asked how many hours they sleep per night. In the sample of 22 students, the mean was 5.77 hours with a standard deviation of 1.572 hours. Let’s construct a 95% confidence interval for the mean number of hours slept per night in the population from which this sample was drawn.

This is what we know: $n=22$, $\overline{x}=5.77$, and $s=1.572$.

In order to compute the confidence interval for $\mu$ we will need the t multiplier and the standard error ($\frac{s}{\sqrt{n}}$).

$df=n-1=22-1=21$

For a 95% confidence interval with 21 degrees of freedom, $t^{*}=2.080$

$SE=\frac{s}{\sqrt{n}}=\frac{1.572}{\sqrt{22}}=0.335$

Thus, our confidence interval for $\mu$ is: $5.77\pm 2.080(0.335)=5.77\pm0.697=[5.073,\;6.467]$

We are 95% confident that the population mean is between 5.073 and 6.467 hours.

What if we wanted to be more conservative and construct a 99% confidence interval?

The only thing that would change is our multiplier. Now, $t^{*}=2.831$.

$5.77\pm 2.831(0.335)=5.77\pm0.948=[4.822,\;6.718]$

We are 99% confident that the population mean is between 4.822 and 6.718 hours.

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility