Fifty students are enrolled in a Statistics class. After the first
examination, a random sample of 5 papers was selected. The grades were
60, 75, 80, 70 and 90.

a) Determine the standard error of the mean?  

'
I used Excel to calculate this as 5. Please verify...

b) What assumption must be made before we can determine an interval
for the mean grade of all the students in the class? Please explain?

The size of the sample is large (n >= 30) vs. a small sample size (n <
30)  A small sample size requires a different approach. Please verify
this answer..

c) Assume the assumption of part b. is met Provide a 90% confidence
interval for the mean grade of all the students in the class.

d) If there were 200 students in the class, what would be the 90%
confidence interval for the mean grade of all the students in the
class?

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Clarification of Question by cecilathome-ga on 07 Aug 2002 20:15 PDT

I agree with the answer thread - and can work the answer for d. But I
do not seem to be able to concur with the answer for c.  I think that
the problem may be that gmac-ga used a 95% confidence level when I
need to use a 90% CF.  My results are

margin of error 2.600742
Low side 72.39936
High Side 77.60074

Am I way off base?

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Clarification of Question by cecilathome-ga on 07 Aug 2002 20:19 PDT

Also gmac-ga, please accept the $10. Your response has been invaluable
to me. Your commentary along with rbnn-ga has agumented my
understanding of these topics greatly. I appreciate it. I've posted
the next three questions that I'm working on. After that - I think
I'll be ready for my final..

I don't know what these lines mean: 

The size of the sample is large (n >= 30) vs. a small sample size (n <
30)  A small sample size requires a different approach. Please verify
this answer..

rbnn-ga comments: I don't know what these lines mean:  
 
The size of the sample is large (n >= 30) vs. a small sample size (n <
30)  A small sample size requires a different approach. Please verify
this answer.. 
---------------
For  n >= 30 one can generally assume that whatever the underlying
distribution for the individual data values, the Central Limit Theorem
will apply to the distribution of the means, so one can reasonably
assume the distribution of means has a normal distribution.  So one
could use z-scores to compute the confidence interval.  The 95% CI
would equal the mean +/- 1.96 standard error of mean.  For n < 30 
assuming the means would be normal could be problematic.  Even if one
did make that assumption, it would be better to use the critical
values for Student's t, instead of z, to compute the CI.

cecilathome-ga did correclty compute the standard error of the mean
using Excel.  It
is 5.

so for the CI in (c), the mean +/- t-crit(df=4,alpha=.05) * standard
error of mean =
   75 +/- 2.13185 * 5 = {64.3408, 85.6592}  

for (d) you need to figure out what the standard error of the mean be
for the
larger sample of 200 if the standard deviation were the same.  The
standard deviation in the current sample is Sqrt[5]*5.  The standard
error of the larger sample would be
that standard deviation divided by the Sqrt[200].  The
Sqrt[5]*5/Sqrt[200] is approximately
equal to 0.790569.  For a sample that large, one could safely use the
z-score for .95,
which is approximately 1.64485.  Then the 90% CI for the larger sample
would be
   75 +/- 1.64485 * 0.790569 =  {73.6996, 76.3004}

Now if you quickly withdraw your question, you can save $10.

I see, thanks for the info gmac-ga. 

Actually I think unfortunately I'm too busy  to answer this question,
but I'd still be very interested in seeing a complete answer!

One question that I never quite understood, is why in some
calculations the standard deviation and in some calculations the
sample standard deviation is used (the denominator is one smaller in
the latter case). It virtually never makes any practical difference
but in cases like this it really comes up.

By the way I think the intended answer for (b) is "the students'
scores are normally distributed"; however, this answer would  not
correct strictly speaking. There are I am sure other kinds of general
information we could know about the students' score distribution, even
if it isn't normal, that let us compute confidence intervals.

rbnn-ga asks:

One question that I never quite understood, is why in some
calculations the standard deviation and in some calculations the
sample standard deviation is used (the denominator is one smaller in
the latter case).

This is the question every intro stats teacher dreads answering.  here
goes...

the sample standard deviation is computed from the sum of the squared
deviations from the SAMPLE MEAN.  What we really would like to do
would be to compute the squared deviations from the POPULATION MEAN. 
Now the SAMPLE MEAN is guaranteed to make the sum of the squared
deviations as small as possible.  It would usually be larger if Mother
Nature (or whoever knows) would tell us what the POPULATION MEAN is so
we could use that in our calculations.  But since we do not know the
POPULATION MEAN, we instead use the SAMPLE MEAN, even though we know
that this will give us an estimate of the standard deviation that is
too small.  In one of those nice quirks of mathematics, it turns out
that dividing the squared deviations by n-1 provides just the right
amount of correction (makes the estimated standard deviation larger)
to compensate for the fact we used the SAMPLE MEAN instead of the
POPULATION MEAN when computing the squared deviations.

For a simulation showing that dividing by n-1 gives an unbiased
estimate of the true population standard deviation, visit
   http://www.seeingstatistics.com/seeing1999/spread/stdev4.html
(you will be visiting out of context and there will be some javascript
errors that you can ignore--or go to www.seeingstatistics.com and use
the Contents window to navigate to section 4.3.3 and the next several
pages.