Question

A statistics professor randomly selects 31 students from their morning class and asks ``How much sleep did youget the night before?''. The sample average was 6.5 hours, and the sample standard deviation was 3.2 hours.a) Find the 95% confidence interval for the average sleep (in hours) that statistics students in the morning class gotthe night before and interpret your interval in terms of the problem.

Answer 1

Answer:

This is what we know: , , and .

In order to compute the confidence interval for  we will need the t multiplier and the standard error ().

Step-by-step explanation:

Answer 2

Answer:

The 95% confidence interval for the average sleep (in hours) that statistics students in the morning class got the night before is between 5.37 hours and 7.63 hours. This means that we are 95% that the true mean sleep that all statistics students in the morning class got the night before is between 5.37 hours and 7.63 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:

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

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.

$M = 1.96*\frac{3.2}{\sqrt{31}} = 1.13$

The lower end of the interval is the mean subtracted by M. So it is 6.5 - 1.13 = 5.37 hours

The upper end of the interval is the mean added to M. So it is 6.5 + 1.13 = 7.63 hours

The 95% confidence interval for the average sleep (in hours) that statistics students in the morning class got the night before is between 5.37 hours and 7.63 hours. This means that we are 95% that the true mean sleep that all statistics students in the morning class got the night before is between 5.37 hours and 7.63 hours.