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Testing for Equality of Proportions Between 2 Samples

Dormitory residency is sometimes advertised as being beneficial to students because it helps foster a connectedness to the university and maybe help the students develop an academic lifestyle. In one study involving 408 entering students in 1995, students were classified according to (i) whether they lived in a campus dormitory during their first year, and (ii) whether they had graduated within 6 years of entering college. The data is presented in a $2\times 2$ cross-classification table below.


 
Table 9.2: Dorm Residency versus Graduation Rate
 



  Dorm Not Dorm
Grad $\leq$ 6 Yrs. 204 20
Grad > 6 Yrs. 160 24

We are interested in comparing 6-year graduation rate of dormitory residents versus non-dormitory residents. In terms of population parameters, we write the null and alternative hypotheses as

\begin{displaymath}H_0: p_1=p_2 \mbox{ versus } H_1: p_1 \neq p_2
\end{displaymath}

where p1 and p2 are the population graduation rates for dorm and non-dorm residents, respectively.

Note that IF we treat dorm residency as a variable, then a test for independence between graduation rate and residency also compares graduation rate between dorm and non-dorm students. Therefore, we may use the chi-square test of independence to test for equality of proportions between populations. In this case, it is called a test for equality of proportions rather than a test for independence. The null and alternative hypotheses are written differently, but the rest of the test procedure remains the same.

If the row variable has only two categories (Success-Failure), and the columns are population labels rather than levels of a second variable, the chi-square test of independence is called a test for equality of proportions.

First compute the graduation rate marginal distribution:

  Dorm Not Dorm Total
Grad $\leq$ 6 Yrs.     224 (This is 54.9% of 408)
Grad > 6 Yrs.     184 (This is 45.1% of 408)
Total 364 44 408

Extrapolating the marginal distribution to each individual column, we get the following expected frequencies:

(.549)(364)=199.84 (.549)(44)=24.16
(.451)(364)=164.16 (.451)(44)=19.84

The chi-square test compares these expected frequencies to the observed frequencies. The null hypothesis is rejected if observed and expected frequencies are too far apart.  

Here is the way a statistical report would formally present the test, in numbered stages.

1.
Hypotheses H0: p1=p2 versus $H_1: p_1 \neq p_2$, where p1=6-yr. grad rate of dorm, p2=6-yr. grad rate of non-dorm
2.
Test statistic:

\begin{displaymath}\chi^2= \begin{tabular}{c}
$\frac{(204-199.84)^2}{199.84}$\s...
...16}$\space + $\frac{(24-19.84)^2}{19.84}$ \end{tabular} = 1.78
\end{displaymath}

3.
P-value: the area greater than 1.78 under the chi-square curve with 1 degree of freedon is .18.

\epsfig{file=anniechi21at178.ps, height=4in, width=3in, angle=-90}

4.
Conclusion: Since P-value > .05, we accept H0: p1=p2. The sample does not provide enough evidence to declare a significant difference between graduation rates of dormitory and non-dormitory students.


next up previous contents index
Next: Using the TI-83 Up: Testing Equality of Frequencies Previous: The Chi-Square Test Statistic

2003-09-08