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

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".

represent the sample mean for the sample  

population mean (variable of interest)

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 and , 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

The margin of error is given by this formula:

   (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:

  (b)

If we replace we have this:

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

-14 is the answer

hope this : ))

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!