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 (\( \frac{s}{\sqrt{n}}\)).

\(df=n-1=22-1=21\)

t Distribution showing the multipliers for a 95% confidence interval given 21 degrees of freedom

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, \(t^{*}=2.831\).

t Distribution showing the multipliers for a 99% confidence interval given 21 degrees of freedom

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

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


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