Chi-Squared Test for IndependenceDate: 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/ |
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