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### 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:

Hospital-trained (H)     48                12
Tertiary-trained (T)     76                64

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/
```
Associated Topics:
College Statistics
High School Statistics

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