|


Chi-Squared Test for Independence
Date: 05/24/2000 at 03:34:00
From: Leanna
Subject: Probability
Saint Botulism's Hospital commissioned an independent study to
determine if there was a difference between the quality of care
provided by hospital-trained nurses and tertiary-trained nurses. Two
hundred nurses were rated as either providing Superior Service (S) or
Adequate Service (A). The results are as follows:
Superior (S) Adequate (A)
Hospital-trained (H) 48 12
Tertiary-trained (T) 76 64
Are the events H and S independent? Justify your answer
mathematically.
Date: 05/24/2000 at 11:58:06
From: Doctor Anthony
Subject: Re: Probability
Do a chi-squared test using a contingency table:
Observed
---------
| S A | Total
------|------------|---------
(H) | 48 12 | 60
(T) | 76 64 | 140
------|------------|---------
Total |124 76 | 200
Expected
---------
| S A | Total
------|------------|---------
(H) | 37.2 22.8| 60
(T) | 86.8 53.2| 140
------|------------|---------
Total |124 76 | 200
The entries in the second table are calculated as follows:
37.2 = 124 * 60/200, 22.8 = 76 * 60/200, and so on.
To calculate chi-squared we make up a third table. Since there is only
1 degree of freedom we use Yates's correction:
O E |O-E| Y = |O-E|-1/2 Y^2/E
-------------------------------------------
48 37.2 10.8 10.3 2.852
12 22.8 10.8 10.3 4.653
76 86.8 10.8 10.3 1.222
64 53.2 10.8 10.3 1.994
--------
X^2 = 10.72
We compare this with the tabular value X^2 (5%) (V = 1) = 3.84
So our value of 10.72 is very significant at the 5% level and we
conclude that H and S are NOT independent.
- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
|
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2008 The Math Forum
http://mathforum.org/dr.math/