In a study of 803 randomly selected medical malpractice lawsuits, it was found that 495 of them were dropped or dismissed Use a 005 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.
C.
E.

B.
D.
F.

What is the test statistic?

(Round to two decimal places as needed.)

Ask by Parsons Carlson. 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 005 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. $$ \mathrm{H}_{0} \mathrm{p}<05 $$ $$ H_{1}: p=0.5 $$ C. $$ \mathrm{H}_{0}: \mathrm{p}>0.5 $$ $$ H_{1}: p=05 $$ E. $$ \begin{array}{l} H_{0} p=0.5 \ H_{1}: p<05 \end{array} $$ B. $$ \mathrm{H}_{0} \quad \mathrm{p}=05 $$ $$ H_{1}: p \neq 0.5 $$ D. $$ H_{0}: p \neq 0.5 $$ $$ H_{1}: p=05 $$ F. $$ \mathrm{H}_{0} \mathrm{p}=05 $$ $$ H_{1}: p>0.5 $$ What is the test statistic? $$ z=\square $$ (Round to two decimal places as needed.)

UpStudy AI · Accepted Answer

**Step 1. Formulate the hypotheses**

The claim is that most (i.e., more than 50%) medical malpractice lawsuits are dropped or dismissed. We define $$ p $$ as the proportion of lawsuits that are dropped or dismissed. The hypotheses are then:

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

Thus, we select the hypothesis given in option F.

**Step 2. Calculate the sample proportion**

The sample size is $$ n = 803 $$ and the number of dropped/dismissed cases is 495, so the sample proportion $$\hat{p}$$ is:

$$
\hat{p} = \frac{495}{803} \approx 0.6162
$$

**Step 3. Determine the test statistic**

The test statistic for a one-proportion $$ z $$-test is calculated using:

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

Here, $$ p_0 = 0.5 $$. Substitute the values:

$$
z = \frac{0.6162 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{803}}}
$$

**Step 4. Compute the denominator**

Calculate the standard error:

$$
\sqrt{\frac{0.5 \times 0.5}{803}} = \sqrt{\frac{0.25}{803}} \approx \sqrt{0.000311} \approx 0.01763
$$

**Step 5. Compute the $$ z $$-value**

$$
z = \frac{0.6162 - 0.5}{0.01763} \approx \frac{0.1162}{0.01763} \approx 6.59
$$

**Final Answer**

- The correct hypothesis test is: 
  $$
  H_0: p = 0.5 \quad \text{vs.} \quad H_1: p > 0.5
  $$
  (Option F)

- The test statistic is:
  $$
  z = 6.59
  $$
**Hypothesis Test:** - **Null Hypothesis ($$ H_0 $$):** $$ p = 0.5 $$ - **Alternative Hypothesis ($$ H_1 $$):** $$ p > 0.5 $$ **Test Statistic:** $$ z = 6.59 $$

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