Question

A sample of 148 college students at a large university reports getting an average of 6.85 hours of sleep last night with a standard deviation of 2.12 hours. Suppose you want to conduct a similar study at your university. Assuming that the standard deviation of the above sample is a reasonable estimate of the standard deviation of sleep time at your university, how many students do you need to survey to estimate the mean sleep time of students at your university with 95% confidence and a margin of error of 0.5 hours

Answer 1

Answer:

70 students need to be surveyed.

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 $1 - \alpha$.

That is z with a pvalue of $1 - 0.025 = 0.975$, so Z = 1.96.

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.

Assuming that the standard deviation of the above sample is a reasonable estimate of the standard deviation of sleep time at your university.

This means that $\sigma = 2.12$

How many students do you need to survey to estimate the mean sleep time of students at your university with 95% confidence and a margin of error of 0.5 hours?

n students need to be surveyed, and n is found when M = 0.5. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.5 = 1.96\frac{2.12}{\sqrt{n}}$

$0.5\sqrt{n} = 1.96*2.12$

Multiplying by 2 on both sides:

$\sqrt{n} = 1.96*4.24$

$(\sqrt{n})^2 = (1.96*4.24)^2$

$n = 69.1$

Rounding up

70 students need to be surveyed.