A universit is contemplating switching from the quarter system to the
semester system.  The administration conducts a survey of a random
sample of 400 students and finds that 240 of them prefer to remain on
the quarter system.  Construct a 95% confidence interval for the true
proportion of all students who would prefer to remain on the quarter
system.  Interpret the mathematical results.

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Request for Question Clarification by rbnn-ga on 19 Sep 2002 08:54 PDT

In order to answer this, two additional pieces of information are
necessary:

A. The total number of students (can be approximate if it is large).

B. The number of students who took the survey who did not answer the
survey question (probably assumed to be 0).

To see why A is necessary, suppose the university only has 400
students and suppose all of the students answered the survey question.
Then you now know the true and exact proportion of students who would
prefer to remain on the quarter system, so that the confidence
interval has infinitesimal width. On the other hand, if there are a
million students, then the confidence interval is smaller.

It might be possible to parameterize the interval over the total
number of students though; this seems hard to me.

technically, rbnn is correct, but seeing as you asked a standard
textbook question, here's the standard textbook solution. I'd suggest
that if you have access to a library you borrow Anderson Sweeney and
Williams (Statistics for Business and Economics) and read chapter 8.
There is a really cool flow chart at the end of the chapter which
really helps if you copy onto a crib sheet (assuming you are asking
this question because you have an exam, and assuming you are allowed a
cribsheet into your exam).

Publisher's website for ASW stats book :
http://www.swcollege.com/quant/asw/sbe_8e/sbe.html

We are not given any population statistics, hence we have to compute
the sample proportion and sample standard error and use these as
proxies for the population proportion and population standard error in
the usual confidence interval equation :

CI = mean +/- z(alpha/2) * s.e.

Because n is large (400 is way larger than the usual cutoff of 30), we
can use the normal distribution (ie. use z(alpha/2) instead of the
more general t(alpha/2).

Anyway, technicalities aside, here is the solution :

We have p-hat, the sample proportion = 240/400 = .6

                                (p-hat * (1 - p-hat))         (.4 *
.6)
The sample standard error = sqrt(-------------------)  = 
sqrt(-------)  = 0.024
                                (         n         )         (  400 
)

For a 95% Confidence Interval, alpha = 0.05. So alpha/2 = 0.025

If you have normal table which gives tail probabilities, you can look
up 0.025 and find that the corresponding z statistic is 1.96. (If you
have Anderson et al. textbook, you can use the t(infinity) table to
get these numbers - ask for clarification if you don't know where I'm
going with this one)

So the confidence interval is :

p-hat +/- z(alpha/2) * s.e. = .4 +/- 1.96 * 0.025 = [0.352, 0.448]

And there's your answer. Hope this helps and good luck.

calebu2-ga

Google search terms:

Anderson Sweeney Business Statistics
"confidence interval" "sample proportion"