Dr. Grey wanted to see if doctors prescribed different treatments for OCD for male patients compared to females. Please use the following frequency tables for the data collected by Dr. Grey.
Contingency table of Observed and Expected Frequencies of the different treatments (Drug vs. Cognitive Therapy) prescribed for patients with OCD. |
||||||
Therapy Totals |
Drug |
Cognitive Therapy |
Total |
|||
Gender of Patient |
Observed |
Expected |
Observed |
Expected |
||
Man |
20 |
10 |
30 |
|||
Woman |
15 |
15 |
30 |
|||
Total |
35 |
25 |
60 |
What is your Conclusion - APA statistic?
A) χ2(1) = X.XX, p > .05
B) χ2(2) = X.XX, p < .05
C) χ2(2) = X.XX, p > .05
D) χ2(1) = X.XX, p < .05
Observed Frequencies | |||
Drug | Therapy | Total | |
Man | 20 | 10 | 30 |
Woman | 15 | 15 | 30 |
Total | 35 | 25 | 60 |
Expected frequency of a cell = sum of row*sum of column / total sum | |||
Expected Frequencies | |||
Drug | Therapy | Total | |
Man | 17.5 | 12.5 | 30 |
Woman | 17.5 | 12.5 | 30 |
Total | 35 | 25 | 60 |
(fo-fe)^2/fe | ||
Man | 0.357 | 0.5 |
Woman | 0.357 | 0.5 |
Null and Alternative hypothesis:
Ho: factors are independent.
H1: factor are dependent.
Test statistic:
χ² = ∑ ((fo-fe)²/fe) = 1.7143
df = (r-1)(c-1) =1
Critical value:
χ²α = CHISQ.INV.RT(α, df) =3.841
p-value:
p-value = CHISQ.DIST.RT(χ², df) =0.1904
Decision:
p-value > 0.05, Do not reject the null hypothesis.
Answer A
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