Final answer:
A hypothesis test was conducted to compare the proportion of patients with flulike symptoms from a new drug to that of competing drugs. The test statistic calculated was significantly higher than the critical Z-value at an alpha level of 0.01, leading to the rejection of the null hypothesis, indicating that more than 1.7% of the drug's users likely experience flulike symptoms.
Step-by-step explanation:
The student asked whether there is sufficient evidence to conclude that more than 1.7% of users of a new prescription drug experience flulike symptoms, compared to the known side effect rate of competing drugs. We need to perform a hypothesis test with the given sample data (16 out of 650 patients) and the known proportion of 1.7% at the alpha level of 0.01.
Step-by-step Hypothesis Test
- Formulate the null hypothesis (H0: p = 0.017) and the alternative hypothesis (H1: p > 0.017), where p is the true proportion of patients experiencing symptoms.
- Calculate the test statistic using the sample proportion and the known proportion:
Sample proportion (p-hat) = 16/650 = 0.0246
Standard error (SE) = sqrt[(0.017 * (1 - 0.017))/650] = 0.0016
Z = (p-hat - 0.017) / SE = (0.0246 - 0.017) / 0.0016 = 4.75 (approximately)
- Compare the calculated Z-value to the critical Z-value from the Z-table for the significance level of 0.01. If Z > Z-critical, we reject H0.
- Make the decision based on the comparison:
Since the calculated Z-value of 4.75 is greater than the critical Z-value for alpha = 0.01, we reject the null hypothesis. This suggests that there is sufficient evidence at the 1% level of significance to conclude that the proportion of patients experiencing flulike symptoms with the new drug is greater than 1.7%.