IntegratedAssignmentOverview.aspx?homeworkld=687973034 Yvonne Manes 02/27/25 7:06 PM Question Completed: 19 of 29 |My score: \( 19 / 29 \) pts ( \( 65.52 \% \) ) Save Part 1 of 5 Researchers selected 858 patients at random among those who take a certain widely-used prescription drug daily. In a clinical trial, 23 out of the 858 patients complained of flulike symptoms. Suppose that it is known that \( 2.3 \% \) of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than \( 2.3 \% \) of this drug's users experience flulike symptoms as a side effect at the \( \alpha=0.1 \) level of significance? Because \( n p_{0}\left(1-p_{0}\right)= \) \( \square \) 10, the sample size is less than \( 5 \% \) of the population size, and the patients in the sample were selected at random, all of the requirements for testing the hypothesis are not satisfied. (Round to one decimal place as needed.)
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To tackle this hypothesis testing situation, we first need to compute the sample proportion of patients who experienced flulike symptoms, which is \( \hat{p} = \frac{23}{858} \approx 0.0268 \) or \( 2.68\% \). This is greater than the competing drug's \( 2.3\% \), but we need to consider the significance level. A z-test for proportions can be useful here, allowing us to calculate the z-value and associated p-value to determine if we can reject the null hypothesis \( H_0: p \leq 0.023 \). Make sure to check that your sample size is large enough for statistical validity! The condition \( np_0(1-p_0) \) should be greater than 10 for normal approximation. Here it seems we're okay since \( 858 \times 0.023 \times (1-0.023) \approx 19.5 \). So, with all conditions met, you might just be discovering something interesting about this new drug!
