Question

Testing Claims About Proportions. In Exercises 9–32, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.

Medical Malpractice In a study of 1228 randomly selected medical malpractice lawsuits, it was found that 856 of them were dropped or dismissed (based on data from the Physicians Insurers Association of America). Use a 0.01 significance level to test the claim that most medical malpractice lawsuits are dropped or dismissed. Should this be comforting to physicians?

Short answer

Nullhypothesis: The proportion of medical malpractice lawsuits subjects dropped or dismissed is equal to 50%.

Alternativehypothesis: The proportion of medical malpractice lawsuits subjects dropped or dismissed is more than 50%.

Test Statistic: 13.807

Critical Value: 2.3263

P-Value: 0.000

The null hypothesis is rejected.

There is enough evidence to support the claim that most medical malpractice lawsuits subjectswere dropped or dismissed.

Since most malpractice lawsuits are either dropped or dismissed, it will be quite comforting for doctors and physicians as they would avoid any pain due to legal proceedings.

Step-by-step solution

Given information

Out of 1228 randomly selected medical malpractice lawsuits, 856 of them were dropped or dismissed.

Hypotheses

The null hypothesis is written as follows:

The proportion of medical malpractice lawsuits subjects who were dropped or dismissedequals50%.

\({H_0}:p = 0.5\)

The alternative hypothesis is written as follows:

The proportion of medical malpractice lawsuits subjects dropped or dismissed is more than 50%.

\({H_1}:p > 0.5\)

The test is right-tailed.

Sample size, sample proportion, and population proportion

The sample sizeequalsn=1228.

The sample proportion of medical malpractice lawsuits subjects dropped or dismissed isas follows:

\[\begin{array}{c}\hat p = \frac{{856}}{{1228}}\\ = 0.697\end{array}\]

The population proportion of medical malpractice lawsuits subjects dropped or dismissed is equal to 0.5.

Test statistic

The value of the test statistic is computed below:

\(\begin{array}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ = \frac{{0.697 - 0.5}}{{\sqrt {\frac{{0.5\left( {1 - 0.5} \right)}}{{1228}}} }}\\ = 13.807\end{array}\)

Thus, z=13.807.

Critical value and p-value

Referring to the standard normal distribution table, the critical value of z at\(\alpha = 0.01\)for a right-tailed test equals2.3263.

Referring to the standard normal distribution table, the p-value for the test statistic value of 13.807 equals0.000.

Since the p-value is less than 0.05, the null hypothesis is rejected.

Conclusion of the test

There is enough evidence to support the claim that the proportion of medical malpractice lawsuits subjects dropped or dismissed is greater than 0.5.

Since most malpractice lawsuits are either dropped or dismissed, it should be comforting for the physicians as they can be relieved without any trial.