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In a study of 803 randomly selected medical malpractice lawsuits, it was found that 495 of them were dropped or dismissed. Use a 0.05 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Which of the following is the hypothesis test to be conducted? A. \( \mathrm{H}_{0} \mathrm{p}<0.5 \) B. \( H_{0} \quad p=05 \) \( H_{1}, p=0.5 \) C. \( \mathrm{H}_{0} \mathrm{p}>05 \) D. \( H_{0}: p \neq 0.5 \) \( H_{1} p=05 \) E. \[ \begin{array}{l} H_{0} p=0.5 \\ H_{1} p<05 \end{array} \] F. \( \mathrm{H}_{0} \mathrm{p}=05 \) \( H_{1} p>0.5 \) What is the test statistic? \[ z=6.59 \] (Round to two decimal places as needed.) What is the \( P \)-value? \( P \)-value \( = \) \( \square \) (Round to three decimal places as needed.)

Ask by Wyatt Morgan. in the United States
Mar 21,2025

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**Hypothesis Test** We are testing the claim that most (more than 50%) of the lawsuits are dropped or dismissed. The hypotheses are: - **Null Hypothesis (\( H_0 \))**: \( p = 0.5 \) - **Alternative Hypothesis (\( H_1 \))**: \( p > 0.5 \) This corresponds to option **F**. **Test Statistic** The test statistic is calculated as: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \approx 6.59 \] **\( P \)-value** The \( P \)-value is: \[ P \text{-value} = 0.000 \]

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The Deep Dive

To test the claim that most medical malpractice lawsuits are dropped or dismissed, we need to set up our hypotheses clearly. The null hypothesis \( H_0: p = 0.5 \) states that 50% of the lawsuits are dropped or dismissed; the alternative hypothesis \( H_1: p < 0.5 \) suggests that less than 50% are dropped. Hence, the correct choice is E. Now, for the test statistic calculation, since we have \( n = 803 \) and \( x = 495 \), the sample proportion \( \hat{p} = \frac{495}{803} \approx 0.617 \). The standard error of the proportion is calculated as \( SE = \sqrt{ \frac{p(1-p)}{n} } \) where \( p = 0.5 \). This gives \( SE \approx 0.028 \). The test statistic is then calculated as \( z = \frac{\hat{p} - p_0}{SE} \) which results in \( z \approx 4.14 \) (rounded to two decimal places). To find the \( P \)-value, we look up or calculate the \( z \)-score in a standard normal distribution. For \( z = 4.14 \), this gives an extremely small \( P \)-value, practically near 0.000 (rounded to three decimal places).

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