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

Expert verified

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 subjects were 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

01

Given information

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

02

Hypotheses

The null hypothesis is written as follows:

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

H0:p=0.5

The alternative hypothesis is written as follows:

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

H0:p=0.5

The test is right-tailed.

03

Sample size, sample proportion, and population proportion

The sample size equals n=1228.

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

p^=8561228=0.697

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

04

Test statistic

The value of the test statistic is computed below:

z=p^-ppqn=0.697-0.50.51-0.51228=13.807

Thus, z=13.807.

05

Critical value and p-value

Referring to the standard normal distribution table, the critical value of z at α=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.

06

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.

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Most popular questions from this chapter

Critical Values. In Exercises 21–24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of = 0.05, should we reject H0or should we fail to reject H0?

Exercise 20

P-Values. In Exercises 17–20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

c. Using a significance level of α = 0.05, should we reject H0or should we fail to reject H0?

The test statistic of z = -2.50 is obtained when testing the claim that p<0.75

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.

Store Checkout-Scanner Accuracy In a study of store checkout-scanners, 1234 items were checked for pricing accuracy; 20 checked items were found to be overcharges, and 1214 checked items were not overcharges (based on data from “UPC Scanner Pricing Systems: Are They Accurate?” by Goodstein, Journal of Marketing, Vol. 58). Use a 0.05 significance level to test the claim that with scanners, 1% of sales are overcharges. (Before scanners were used, the overcharge rate was estimated to be about 1%.) Based on these results, do scanners appear to help consumers avoid overcharges?

Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been selected and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Cans of Coke Data Set 26 “Cola Weights and Volumes” in Appendix B includes volumes (ounces) of a sample of cans of regular Coke. The summary statistics are n = 36, x = 12.19 oz, s = 0.11 oz. Use a 0.05 significance level to test the claim that cans of Coke have a mean volume of 12.00 ounces. Does it appear that consumers are being cheated?

t Test Exercise 2 refers to a t test. What is the t test? Why is the letter t used? What is unrealistic about the z test methods in Part 2 of this section?

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