Problem 1:

Assume that DePaul has 5,000 registered students. As part of a survey, 625 
of these students are chosen at random. The average age for these surveyed 
students is 24.5 years with a standard deviation of 5.0 years.

a.	Estimate the average age for all 5,000 students.

	Since the average age for the sample is 24.5 then this is a
	good estimate of the population. Thus the average age for all
	5,000 students is about 24.5 years.


b.	What is the error in your estimate of average age.

	Since s(y)=5.0 years and we are interested in s(ybar) and
	the sample is large, then s(ybar)=s(y)/sqrt(n)=5/sqrt(625)
	=5/25=0.2 


c.	Give a 95% confidence interval for the average age for all 5,000 
	students. 

	Since the sample is large we find z from the standard normal 
	table so that the area between -z and +z is 95%. The 95% CI
	is then:

		24.5 plus/minus 1.95(0.2)=[24.11, 24.89] 



Problem 2:

For a small city in southern Illinois census data indicates that there
are 100,000 married couples but the average number of years married and
the standard deviation is not known. A statistician draws a simple 
random sample of married couples from the city. Since she does 
not know the average number of years married for the population and does not 
know the population standard deviation for years married she computes 
them from her sample and uses these sample statistics as estimates of the 
population parameters. 

Suppose that a sample of size 400 was obtained for which the average
number of years of marriage is 8.5 and the standard deviation is 2.0
years.

a.	Estimate the average years of marriage for couples in this 
	city.

	As for a. in Problem 1 above the sample average is a good
	estimate of the population average. The average years of
	marriage for couples in the city is therefore about
	8.5 years.


b.	What is the error of your estimate.

	The sample size is large and therefore the error of the
	estimate is s(y)/sqrt(n)=2/sqrt(400)=2/20=0.1


c.	Give a 90% confidence interval for the average age of marriage 
	for couples in this city.

	Since the sample is large we again use the z value from our
	standard normal table but in this case we are interested in
	a 90% confidence interval and therefore the required value
	is 1.65. The 90% CI is then:

		8.5 plus/minus 1.65(0.1)=[8.335, 8.665]


d.	Suppose that the sample size had been 17 rather than 400, but 
	that all other values remain unchanged. Give the 90% confidence 
	interval in this case.

	Since the sample is small (i.e. 17) then if we assume that y is
        normally distributed then ybar is Student t with n-1 df. In this
	case we must also use the formula for small sample sizes when
	computing the standard deviation. Thus s(ybar)=s(y)/sqrt(n-1)
	=2/sqrt(16)=2/4=0.5. Hence the t value for 90% is found in the
	5% column (due to symmetry of the t distribution) and the 16 
	degrees of freedom row. Thus the required t value is 1.75. The
	90% CI is then:

		8.5 plus/minus 1.75(0.5)=[7.625, 9.375]

	Note: The interval is much wider in this case. Thus for small
	sample sizes you obtain a much wider interval for the same level
	of confidence.