In a clinical trial, 26 out of 886 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.6% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.6% of this drug's users experience flulike symptoms as a side effect at the a=0.1 level of significance? Because npo (1-Po) = 10, the sample size is 5% of the population size, and the sample satisfied. (Round to one decimal place as needed.) the requirements for testing the hypothesis

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### 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
Transcribed Image Text:### 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
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