Question

A professor at a university wants to estimate the average number of hours of sleep students get during exam week. The professor wants to find a sample mean that is within 1.939 hours of the true average for all college students at the university with 90% confidence. If the professor knows the standard deviation of the sleep times for all college students is 6.995, what sample size will need to be taken

Answer 1

Answer:

$n=(\frac{1.64(6.995)}{1.939})^2 =35.003 \approx 36$

So the answer for this case would be n=36 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma =6.995$ represent the population standard deviation

n represent the sample size  

Solution to the problem

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =1.939 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

If we replace we have this:

$n=(\frac{1.64(6.995)}{1.939})^2 =35.003 \approx 36$

So the answer for this case would be n=36 rounded up to the nearest integer