the administration of a university has been using the following
distribution to classify the ages of their students:

Age Group                  Estimated % of Student Population

less than 18                  2.7
18-19                         29.9
20-24                         53.4
older than 24                  14 



A resect student survey provided the following data on age of
students:
                           Age Group         Frequency
                            less than 18           6
                            18-19                118
                            21-24                102
                           Older than 24          26
a. how to set up a table that compares the expected and observed
frequencies for each group?


b. Set up the hypotheses for the chi-square goodness of fit test.

c. Perform the goodness of fit test at the 0.05 level of significance.

Hi Whosher-ga:

What you ask is fairly straightforward statistics. 

Some web sites that have good information on these types of problems
include:

http://www.itl.nist.gov/div898/handbook/eda/section3/eda35f.htm
http://www.ruf.rice.edu/~bioslabs/tools/stats/chisquare.html
http://www.sjsu.edu/faculty/gerstman/EpiInfo/cat-one.htm
http://osf1.gmu.edu/~alaemmer/biostat/goodness/goodness.html

However, let's look at your particular data and your three-part
question.

First, let's look at the data that we are provided with.

Sample size = 252 students
alpha (significance level) = 0.05 (or 5%)

[I'm also assuming that you made a typo in your observed frequencies -
that where you said "21-24", you really meant "20-24".]

a) To set up a table to compare the expected and observed frequencies
for the 4 given groups, simply do as follows:

Age Group    Estimated %    Expected Freq.     Observed Frequency
  < 18           2.7                7                  6
 18 - 19        29.9               75                118
 20 - 24        53.4              135                102
  > 24          14.0               35                 26
Totals         100.0              252                252

b) In this case the Chi-squared goodness of fit hypothesis is:

H[0]: The data do follow the estimated distribution. (This is the
"null hypothesis".)
H[a]: The data do not follow the estimated distribution. (This is the
"alternate hypothesis".)

c) To perform the goodness of fit test, we must first calculate the
chi-squared value. The formula for that value is:

http://www.itl.nist.gov/div898/handbook/eda/section3/eqns/chisqgf.gif

where,
 
O[i] is the observed frequency for group i and 
E[i] is the expected frequency for group [i]. 

So, for this data, the chi-squared statistic would be:

(6-7)^2/7 + (118-75)^2/75 + (102-135)^2/135 + (26-35)^2/35 = 
.143      + 24.653        + 8.067           + 2.314        =

35.177

Since we have 4 groups, our degrees of freedom equal 3 (4-1). 

If we look up in the chi-squared table at

http://www.uvm.edu/~golivett/introbio/lab_reports/chi.html 

with three degrees of freedom and a 5% level of significance, we see
that the tabled value is:

7.81

Since the chi-squared statistic (well) exceeds the tabled value, you
can safely reject the null hypothesis and accept the alternate
hypothesis.

Therefore, in this case, the data do not follow the estimated
distribution. It is likely that the administration's distribution
estimates need to be re-examined.

I hope this answers your question. Please ask for clarification if
necessary before rating this answer.

Thanks.

websearcher-ga

---

Clarification of Answer by websearcher-ga on 29 Jul 2002 06:10 PDT

Hi Whosher-ga:

I wasn't aware from your original question that you were trying to set
this up on Excel. Some tips for that are:

* In the table in a) the values in the "Expected Freq." column can be
calculated using

=ROUND(C1*V2,0)

where C1 is the column with the "Estimated %" values and V2 is the sum
of the values in the "Observed Freq." column.

* To compute the Chi-squares statistic, you could first compute the
individual elements of the sum by:

=(Co - Ce)^2

where Ci is an element in the "Observed Freq." column and Cj is the
corresponding element in the "Expected Freq." column.

You would then sum the values of this column together to get the
chi-squared statistic.

Hope this helps. There is probably more you could do to automate these
calculation in Excel.

websearcher-ga

you have done an outstanding job on making me understand the problem thanks

This question sounds like a homework question and shouldn't be answered.

This is not a home work question/ but i am teaching my self how to do
excel just was not sure how to set up the table on excel