SOLUTION: At a large company banquet for several thousand employees and their families, many of the attendees became ill the next day. The company doctor suspects that the illness may be rel

Question 1201173: At a large company banquet for several thousand employees and their families, many of the attendees became ill the next day. The company doctor suspects that the illness may be related to the fish, one of three options for the main course. Because all the dinner guests had to pre-order their meal, the doctor was able to randomly select and contact 80 people that ate the fish, of which 64 people got sick. The doctor also randomly selected (and contacted) 60 people that did not eat the fish, of which 39 people got sick. The doctor also knows that at least 1000 attendees ordered the fish.
a) Is this convincing evidence that the true proportion of all attendees who ate the fish that got sick is more than the true proportion of all attendees who did not eat the fish that got sick?
b) Based on your conclusion in the previous problem, which mistake (a Type I error or a Type II error) could you have made? Interpret the potential error in context.
Answer by asinus(45) (Show Source): You can put this solution on YOUR website!
**a) Hypothesis Test**
* **Hypotheses:**
* **Null Hypothesis (H0):** p1 - p2 = 0 (There is no difference in the proportion of people who got sick between those who ate fish and those who did not.)
* **Alternative Hypothesis (H1):** p1 - p2 > 0 (The proportion of people who ate fish and got sick is significantly higher.)
* **Sample Proportions:**
* p1_hat = 64/80 = 0.8 (Proportion of those who ate fish and got sick)
* p2_hat = 39/60 = 0.65 (Proportion of those who did not eat fish and got sick)
* **Pooled Proportion:**
* p̂ = (64 + 39) / (80 + 60) = 103 / 140 ≈ 0.7357
* **Standard Error:**
* SE = √[p̂(1-p̂)(1/n1 + 1/n2)]
* SE = √[0.7357 * (1 - 0.7357) * (1/80 + 1/60)]
* SE ≈ 0.0683
* **Test Statistic (z-score):**
* z = (p1_hat - p2_hat) / SE
* z = (0.8 - 0.65) / 0.0683
* z ≈ 2.20
* **P-value:**
* Using a standard normal distribution table, find the p-value associated with z = 2.20.
* P-value ≈ 0.0139
* **Decision:**
* Since the p-value (0.0139) is less than the significance level (α = 0.05), we **reject the null hypothesis**.
* **Conclusion:**
* There is sufficient evidence to suggest that the proportion of people who ate fish and got sick is significantly higher than the proportion of people who did not eat fish and got sick. This provides some evidence that the fish may have contributed to the illness.
**b) Type of Error**
* Since we rejected the null hypothesis, we could have made a **Type I error**.
* **Type I Error:**
* This occurs when we reject the null hypothesis when it is actually true.
* In this context, a Type I error would mean concluding that the fish caused the illness when in reality, there was no significant difference in the proportion of people who got sick between those who ate fish and those who did not.
* **Interpretation of Type I Error:**
* This could lead to unnecessary concern and potentially harmful actions, such as unnecessarily restricting the sale or consumption of fish, even if it was not the cause of the illness.
**Important Notes:**
* This analysis provides some evidence of a possible link between eating the fish and illness, but it does not definitively prove causation.
* Other factors could have contributed to the illness, such as other food items, cross-contamination, or other environmental factors.
* Further investigation and a more comprehensive analysis would be necessary to establish a definitive cause-and-effect relationship.
This analysis provides a framework for understanding the potential implications of a Type I error in this specific context.

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