Question

In a study of 825 randomly selected medical malpractice lawsuits, it was found that 500 of them were dropped or dismissed. Use a 0.01 significance level to test the claim that most medical
malpractice lawsuits are dropped or dismissed.

I found the z score but does anyone know how to find the p-value?

Answer 1

Using the z-distribution, since the p-value of the test is less than 0.01, there is enough evidence to conclude that the claim is correct.

What are the hypothesis tested?

At the null hypothesis, it is tested if there is not enough evidence that the proportion is above 0.5, hence:

$H_0: p \leq 0.5$

At the alternative hypothesis, it is tested if there is enough evidence that the proportion is above 0.5, hence:

$H_1: p > 0.5$

What is the test statistic?

The test statistic is given by:

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

In which:

• $\overline{p}$ is the sample proportion.

• p is the proportion tested at the null hypothesis.

• n is the sample size.

For this problem, the parameters are:

$n = 825, \overline{p} = \frac{500}{825} = 0.606, p = 0.5$

Hence the test statistic is:

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

$z = \frac{0.606 - 0.5}{\sqrt{0.5(0.5)}{825}}}$

z = 6.1.

What is the p-value?

We have a right-tailed test, as we are testing if the proportion is greater than a value. Using a z-distribution calculator, with z = 6.1, the p-value is of 0.

Since the p-value is less than 0.01, there is enough evidence to conclude that the claim is correct.

More can be learned about the z-distribution at brainly.com/question/16313918

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