SOLUTION: In 2010, Dr. Bob decided to gather research on the type of disorders that present among his patients. His data collection resulted in the following breakdown of patients by disorde

Question 1179576: In 2010, Dr. Bob decided to gather research on the type of disorders that present among his patients. His data collection resulted in the following breakdown of patients by disorder: 54.9% Schizophrenia; 21.1% Major Depression; 7.9% Obsessive-Compulsive Disorder; 4.5% Anxiety Disorder; 2.9% Personality Disorder; 8.8% Other. Information was collected from a random sample 0f 300 patients in 2018 to determine whether or not the data has changed significantly. The sample data is given in the table below. At the α = 0.01 level of significance, test the claim that the disorder breakdown of patients at Dr. Bob's hospital has not changed significantly since 2010.

Type of disorder per patient in sample:
Disorder Count
Schizophrenia 158
Major Dispression 65
Obsessive-Compulsive Disorder 29
Anxiety Disorder 11
Personality Disorder 14
Other 23

Test Statistic:

Give the P-value:

Which is the correct result:
Do not Reject the Null Hypothesis
Reject the Null Hypothesis

Which would be the appropriate conclusion?
There is enough evidence to suggest that the breakdown of patients by disorder has changed significantly since 2010.
There is not enough evidence to suggest that the breakdown of patients by disorder has changed significantly since 2010.
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
It appears you're dealing with a Chi-Square Goodness of Fit test. Here's how to approach the problem:
**1. State the Hypotheses**
* Null Hypothesis (H0): The distribution of disorders in 2018 is the same as in 2010.
* Alternative Hypothesis (H1): The distribution of disorders in 2018 is different from 2010.
**2. Calculate Expected Frequencies**
Multiply the percentages from 2010 by the sample size in 2018 (300) to get the expected counts for each disorder:
| Disorder | 2010 % | Expected Count |
|---|---|---|
| Schizophrenia | 54.9% | 164.7 |
| Major Depression | 21.1% | 63.3 |
| Obsessive-Compulsive | 7.9% | 23.7 |
| Anxiety | 4.5% | 13.5 |
| Personality | 2.9% | 8.7 |
| Other | 8.8% | 26.4 |
**3. Calculate the Chi-Square Test Statistic**
The formula for the Chi-Square test statistic is:
χ² = Σ [(Observed Count - Expected Count)² / Expected Count]
| Disorder | Observed | Expected | (O - E)² / E |
|---|---|---|---|
| Schizophrenia | 158 | 164.7 | 0.27 |
| Major Depression | 65 | 63.3 | 0.05 |
| Obsessive-Compulsive | 29 | 23.7 | 1.18 |
| Anxiety | 11 | 13.5 | 0.46 |
| Personality | 14 | 8.7 | 3.22 |
| Other | 23 | 26.4 | 0.44 |
χ² = 0.27 + 0.05 + 1.18 + 0.46 + 3.22 + 0.44 = **5.62**
**4. Determine Degrees of Freedom**
Degrees of freedom (df) = Number of categories - 1 = 6 - 1 = 5
**5. Find the P-value**
Using a Chi-Square distribution table or calculator with df = 5 and χ² = 5.62, you'll find the P-value. The P-value is approximately 0.346.
**6. Compare P-value to Significance Level (α)**
* α = 0.01 (given)
* P-value (0.346) > α (0.01)
**7. Make a Decision**
Since the P-value is greater than α, we **Do Not Reject the Null Hypothesis**.
**8. Conclusion**
**There is not enough evidence to suggest that the breakdown of patients by disorder has changed significantly since 2010.**
**Important Note:** While the observed counts in the 2018 sample might look different from the 2010 percentages, the Chi-Square test tells us that this difference is not statistically significant. The variation could be due to random chance in the sample.
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