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. B.

C. D.

E.

F.

What is the test statistic?

(Round to two decimal places as needed.)
What is the -value?
-value
(Round to three decimal places as needed.)

Ask by Wyatt Morgan. in the United States

Mar 21,2025

Question

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.)

UpStudy AI · Accepted Answer

**Hypothesis Test**

We are testing the claim that most (i.e., more than 50%) of the lawsuits are dropped or dismissed. Let $$ p $$ be the true proportion of lawsuits dropped or dismissed. Then the hypotheses are:

$$
\begin{aligned}
H_0 &: p = 0.5 \
H_1 &: p > 0.5
\end{aligned}
$$

This corresponds to option **F**.

**Test Statistic**

The test statistic for a one‐proportion $$ z $$-test is given by

$$
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}.
$$

With $$\hat{p} = \frac{495}{803} \approx 0.616$$, $$p_0 = 0.5$$, and $$n=803$$, the test statistic is

$$
z \approx \frac{0.616 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{803}}} \approx 6.59.
$$

**$$ P $$-value**

Since this is a right-tailed test, the $$ P $$-value is

$$
P \text{-value} = P(Z > 6.59).
$$

A $$ z $$-score of $$ 6.59 $$ is extremely high, and the corresponding $$ P $$-value is very close to 0. When rounded to three decimal places, we have

$$
P \text{-value} = 0.000.
$$
**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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