### Hypothesis Testing for Proportion**Problem Statement:**In a clinical trial, 26 out of 886 patients taking a prescription drug daily complained of flu-like symptoms. Suppose that it is known that 2.6% of patients taking competing drugs complain of flu-like symptoms. Is there sufficient evidence to conclude that more than 2.6% of this drug’s users experience flu-like symptoms as a side effect at the $\alpha = 0.1$ level of significance?**Details and Calculation:**1. **Statement Explanation:** This is a hypothesis testing problem for a population proportion. We want to determine if the proportion of patients experiencing flu-like symptoms from the new drug exceeds the known rate of 2.6% from competing drugs. 2. **Graph or Diagram Explanation:** There is no graph or diagram in this context, but the problem involves several important statistical concepts.3. **Calculation Steps:** **Step 1: State the null and alternative hypotheses.** \[ H_0: p = 0.026 \quad \text{(null hypothesis: proportion is 2.6%)} \] \[ H_1: p > 0.026 \quad \text{(alternative hypothesis: proportion is more than 2.6%)} \] **Step 2: Calculating Test Statistic.** Given, - $ n = 886 $ (sample size) - $ x = 26 $ (number of patients with symptoms) - $ p_0 = 0.026 $ (population proportion) **Sample Proportion ($ \hat{p} $):** \[ \hat{p} = \frac{x}{n} = \frac{26}{886} \approx 0.0293 \] **Standard Error (SE):** \[ SE = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.026 \times (1 - 0.026)}{886}} \approx 0.0054 \] **Z-Statistic:** \[ Z = \frac{\hat{p} - p_0}{SE} = \frac{0.0293 - 0.026}{0.0054} \approx 0

The value of p0 is 0.026 and the sample size n is 800.

Answered: In a clinical trial, 26 out of 886…

MATLAB: An Introduction with Applications

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Publisher:Amos Gilat

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