Homework 8 Solutions

8-4:
a) Because we know the standard deviation of the population (sigma = 12),
Xbar +/- 1.96*(12/sqrt(n))
is a 95% confidence interval for the mean.  The "1.96" comes from the table for the z-statistics.

36 +/- 1.96*(12/sqrt(50))
36 +/- 1.96*(12/sqrt(50)) = 36 +/- 3.33
We are 95% confident that the mean age of all customers is between 32.77 and 39.33 years.

b) Want margin of error = 2
So 2 = 1.96*(12/sqrt(n))
sqrt(n) = 1.96*12/2

n = (1.96*6)²
=138.3
So let n = 139 to be sure.

c) Silly question, really, but here's the algebra:

e = (1.96 * sigma/sqrt(n))
Solving for n:
n = (1.96 * sigma/e)²

d) (1.96*12)^2 = 553.2  so n = 554

e) Let e_2 be the "new e", so e_2 = (1/16)e
e_2 = (1/16)*(1.96 * sigma/sqrt(n))

n =  ( (1/16)*(1.96*sigma/e) )^2 = .015 (sigma/e)²


8-3:  
a) Assume population standard deviation is 6000.  

3700 +/- 1.96 * 6000/sqrt(225)
3700 +/- 784

So the 95% CI is (2916, 4484).  In other words, we are 95% confident that the true mean enrollment of all 2700 institutions is in the interval 2916 to 4484.

b) In fact, we don't really know the population standard deviation.  All we really have is an estimate:  the standard deviation of the sample (which is 6000).  So we need to replace the 1.96 with the appropriate level from the t distribution with n-1 degrees of freedom.  N = 225 here, and from the table below, we see we need to choose t = 1.97.  Notice that this is *very* close to 1.96, and this is because n is so large we get about the same confidence interval whether or not we used an estimate of the SD or the true value.

3700 +/- 1.97*6000/sqrt(225)
3700 +/- 788
(2912, 4488)  The CI is slightly wider.